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For this Question: Anjali has 3 balls of sizes X, Y, and Z respectively, She puts the balls in boxes of size A (X ≤ Y ≤ Z ≤ A). Determine the bare minimum of boxes Anjali requires to fit each ball inside a box. If the total size of the balls inside the box does not exceed the size of the box, then the box may hold more than one ball.The first line contains T denoting the number of test cases. Then the test cases follow. Each test case contains four integers X, Y, Z, and A on a single line denoting the sizes of the balls and boxes.
<b>Constraints</b>
1 ≤ T ≤ 100
1 ≤ X ≤ Y ≤ Z ≤ A ≤ 100For each test case, output on a single line the minimum number of boxes Anjali needs.Sample Input :
3
2 3 5 10
1 2 3 5
3 3 4 4
Sample Output :
1
2
3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
int t, a, b, c, d;
cin >> t;
assert(1 <= t <= 100);
while (t--) {
cin >> a >> b >> c >> d;
assert(1 <= a <= 100);
assert(1 <= b <= 100);
assert(1 <= c <= 100);
assert(1 <= d <= 100);
if (a + b + c <= d)
cout << "1\n";
else if ((a + b) <= d || (a + c) <= d || (b + c) <= d)
cout << "2\n";
else
cout << "3\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a grid having N*M elements. Find the index of the row (1-base indexing) which has the maximum sum of elements.
Note: if more than two rows have the same sum of elements, then return the smallest index.The first line of the input contains two integers N and M.
The next N lines each contain M space separated integers.
<b>Constraints:</b>
1 <= N, M <= 10<sup>3</sup>
1 <= A<sub>i, j</sub> <= 10<sup>9Print the index of the row (1-base indexing) which has the maximum sum of elements.Sample Input:
4 3
3 4 2
5 1 7
2 8 1
2 3 3
Sample Output:
2
<b>Explaination:</b>
Row number 2 has sum = 5 + 1 + 7 = 13 which is the maximum among all rows., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
if(st.hasMoreTokens()){
str = st.nextToken("\n");
}
else{
str = br.readLine();
}
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main (String[] args) {
FastReader sc=new FastReader();
int n=sc.nextInt();
int m=sc.nextInt();
int arr[][]=new int[n][m];
for(int i=0; i<n; i++)
{
for(int j=0; j<m; j++)
{
arr[i][j]= sc.nextInt();
}
}
long maxsum=0;
int index=0;
for(int i=0; i<n; i++)
{
long sum=0;
for(int j=0; j<m; j++)
{
sum += arr[i][j];
}
if(sum > maxsum)
{
maxsum=sum;
index=i+1;
}
}
System.out.print(index);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a grid having N*M elements. Find the index of the row (1-base indexing) which has the maximum sum of elements.
Note: if more than two rows have the same sum of elements, then return the smallest index.The first line of the input contains two integers N and M.
The next N lines each contain M space separated integers.
<b>Constraints:</b>
1 <= N, M <= 10<sup>3</sup>
1 <= A<sub>i, j</sub> <= 10<sup>9Print the index of the row (1-base indexing) which has the maximum sum of elements.Sample Input:
4 3
3 4 2
5 1 7
2 8 1
2 3 3
Sample Output:
2
<b>Explaination:</b>
Row number 2 has sum = 5 + 1 + 7 = 13 which is the maximum among all rows., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n, m;
cin >> n >> m;
vector<vector<int>> a(n, vector<int> (m));
vector<int> sum(n);
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
cin >> a[i][j];
sum[i] += a[i][j];
}
}
int mx = 0, ans = 0;
for(int i = 0; i < n; i++){
if(mx < sum[i]){
mx = sum[i];
ans = i;
}
}
cout << ans + 1;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a stack of integers and N queries. Your task is to perform these operations:-
<b>push:-</b>this operation will add an element to your current stack.
<b>pop:-</b>remove the element that is on top
<b>top:-</b>print the element which is currently on top of stack
<b>Note:-</b>if the stack is already empty then the pop operation will do nothing and 0 will be printed as a top element of the stack if it is empty.<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions:
<b>push()</b>:- that takes the stack and the integer to be added as a parameter.
<b>pop()</b>:- that takes the stack as parameter.
<b>top()</b> :- that takes the stack as parameter.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>3</sup>You don't need to print anything else other than in <b>top</b> function in which you require to print the topmost element of your stack in a new line, if the stack is empty you just need to print 0.Input:
7
push 1
push 2
top
pop
top
pop
top
Output:
2
1
0
, I have written this Solution Code:
public static void push (Stack < Integer > st, int x)
{
st.push (x);
}
// Function to pop element from stack
public static void pop (Stack < Integer > st)
{
if (st.isEmpty () == false)
{
int x = st.pop ();
}
}
// Function to return top of stack
public static void top(Stack < Integer > st)
{
int x = 0;
if (st.isEmpty () == false)
{
x = st.peek ();
}
System.out.println (x);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Andrew loves to solve problems related to prime numbers. One of Andrew's friend has asked him to solve below problem for him.
Given two positive integers <b>N</b> and <b>M</b>, the task is to check that N is the Mth power of a <b>prime number</b> or not.First line of input contains testcases <b>T</b>. For each testcase, there will be two positive integers N and M.
Constraints :
1 <= T <= 100
2 <= N <= 10^6
1 <= M <= 10For each testcase you need to print "<b>Yes</b>" if N is the Mth power of a prime number otherrwise "<b>No</b>". Input :
2
16 4
16 3
Output:
Yes
No
Explanation :
16 is m-th (4th) power of 2, where 2 is prime., I have written this Solution Code: import math
def mroot(n,m):
check = round(n**(1/m))
# print(check)
if check**m == n:
return check
else:
return -1
def prime(n,m):
check = mroot(n,m)
if check == -1:
return "No"
if check == 2:
return "Yes"
for i in range(2, int(math.sqrt(check))+1):
if n%i == 0:
return "No"
return "Yes"
T = int(input())
for _ in range(T):
n,m = list(map(int, input().split()))
print(prime(n,m)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Andrew loves to solve problems related to prime numbers. One of Andrew's friend has asked him to solve below problem for him.
Given two positive integers <b>N</b> and <b>M</b>, the task is to check that N is the Mth power of a <b>prime number</b> or not.First line of input contains testcases <b>T</b>. For each testcase, there will be two positive integers N and M.
Constraints :
1 <= T <= 100
2 <= N <= 10^6
1 <= M <= 10For each testcase you need to print "<b>Yes</b>" if N is the Mth power of a prime number otherrwise "<b>No</b>". Input :
2
16 4
16 3
Output:
Yes
No
Explanation :
16 is m-th (4th) power of 2, where 2 is prime., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader rd=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(rd.readLine());
while(t-->0){
String b[]=rd.readLine().split(" ");
int n=Integer.parseInt(b[0]);
int m=Integer.parseInt(b[1]);
if(m==1){
int a=(int)Math.sqrt(n);
int p=1;
for(int k=2;k<=a;k++){
if(n%k==0){
p=0;
break;
}
}
if(p==1)
System.out.println("Yes");
else
System.out.println("No");
}
else{
for(int i=2;i<=n/2;i++){
boolean p=true;
for(int j=2;j<=(int)Math.sqrt(i);j++){
if(i%j==0){
p=false;
break;}
}
if(p){
if((int)Math.pow(i,m)==n){
System.out.println("Yes");
break;}
else if((int)Math.pow(i,m)>n){
System.out.println("No");
break;
}
}
}
}}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Andrew loves to solve problems related to prime numbers. One of Andrew's friend has asked him to solve below problem for him.
Given two positive integers <b>N</b> and <b>M</b>, the task is to check that N is the Mth power of a <b>prime number</b> or not.First line of input contains testcases <b>T</b>. For each testcase, there will be two positive integers N and M.
Constraints :
1 <= T <= 100
2 <= N <= 10^6
1 <= M <= 10For each testcase you need to print "<b>Yes</b>" if N is the Mth power of a prime number otherrwise "<b>No</b>". Input :
2
16 4
16 3
Output:
Yes
No
Explanation :
16 is m-th (4th) power of 2, where 2 is prime., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
const int n = 1000001;
bool a[n];
int main(){
for(int i=0;i<n;i++){
a[i]=false;
}
for(int i=2;i<n;i++){
if(a[i]==false){
for(int j=i+i;j<n;j+=i){
a[j]=true;
}
}
}
int t;
cin>>t;
while(t--){
int x,y;
cin>>x>>y;
if(y>1){
for(int i=2;i<1000;i++){
if(a[i]==false){
int s=1;
for(int j=0;j<y;j++){
s*=i;
}
if(s==x){
cout<<"Yes"<<endl;
goto f;
}
}
}
cout<<"No"<<endl;
f:;
}
else{
if(a[x]==false){cout<<"Yes"<<endl;}
else{cout<<"No"<<endl;}
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Blue Ranger hates palindromic strings and he is also jealous of Red Ranger. So to test Red Ranger he gives him a task. Given N and M, Red Ranger must find a string of length N and containing at max M distinct english alphabets such that the string contains minimum substrings that are palindromes. As there can be many such strings Blue Ranger asks him to simply tell the minimum possible number of substrings in the optimal string that are palindromes.
Palindrome is a string that can be read in the same way in both left-to-right and right-to-left directions.Input contains two integers N and M.
Constraints:
1 <= N <= 100000
1 <= M <= 100Print a single integer that is the minimum possible number of substrings in the optimal string that are palindromes.Sample Input 1
3 15
Sample Output 1
3
Explanation: optimal string can be "abc" which has three substrings that are palindromes.
Sample Input 2
3 1
Sample Output 2
6
Explanation: optimal string can be "aaa" which has 6 substrings that are pallindromes., I have written this Solution Code:
import java.io.*;
import java.util.*;
public class Main
{
public static void main(String[] args)throws Exception{ new Main().run();}
long mod=1000000000+7;
void solve() throws Exception
{
long N=nl();
long M=nl();
if(N==1)
{
out.println("1");
}
else if(M==1)
{
long sum=(N*(N+1))/2;
out.println(sum);
}
else if(M==2)
{
out.println((N-1)*2);
}
else
{
out.println(N);
}
}
private byte[] buf=new byte[1024];
private int index;
private InputStream in;
private int total;
private SpaceCharFilter filter;
PrintWriter out;
int min(int... ar){int min=Integer.MAX_VALUE;for(int i:ar)min=Math.min(min, i);return min;}
long min(long... ar){long min=Long.MAX_VALUE;for(long i:ar)min=Math.min(min, i);return min;}
int max(int... ar) {int max=Integer.MIN_VALUE;for(int i:ar)max=Math.max(max, i);return max;}
long max(long... ar) {long max=Long.MIN_VALUE;for(long i:ar)max=Math.max(max, i);return max;}
void reverse(int a[]){for(int i=0;i<a.length>>1;i++){int tem=a[i];a[i]=a[a.length-1-i];a[a.length-1-i]=tem;}}
void reverse(long a[]){for(int i=0;i<a.length>>1;i++){long tem=a[i];a[i]=a[a.length-1-i];a[a.length-1-i]=tem;}}
String reverse(String s){StringBuilder sb=new StringBuilder(s);sb.reverse();return sb.toString();}
void shuffle(int a[]) {
ArrayList<Integer> al = new ArrayList<>();
for(int i=0;i<a.length;i++)
al.add(a[i]);
Collections.sort(al);
for(int i=0;i<a.length;i++)
a[i]=al.get(i);
}
long lcm(long a,long b)
{
return (a*b)/(gcd(a,b));
}
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
long expo(long p,long q)
{
long z = 1;
while (q>0) {
if (q%2 == 1) {
z = (z * p)%mod;
}
p = (p*p)%mod;
q >>= 1;
}
return z;
}
void run()throws Exception
{
in=System.in; out = new PrintWriter(System.out);
solve();
out.flush();
}
private int scan()throws IOException
{
if(total<0)
throw new InputMismatchException();
if(index>=total)
{
index=0;
total=in.read(buf);
if(total<=0)
return -1;
}
return buf[index++];
}
private int ni() throws IOException
{
int c = scan();
while (isSpaceChar(c))
c = scan();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = scan();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = scan();
} while (!isSpaceChar(c));
return res * sgn;
}
private long nl() throws IOException
{
long num = 0;
int b;
boolean minus = false;
while ((b = scan()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
;
if (b == '-') {
minus = true;
b = scan();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = scan();
}
}
private double nd() throws IOException{
return Double.parseDouble(ns());
}
private String ns() throws IOException {
int c = scan();
while (isSpaceChar(c))
c = scan();
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c))
res.appendCodePoint(c);
c = scan();
} while (!isSpaceChar(c));
return res.toString();
}
private String nss() throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
return br.readLine();
}
private char nc() throws IOException
{
int c = scan();
while (isSpaceChar(c))
c = scan();
return (char) c;
}
private boolean isWhiteSpace(int n)
{
if(n==' '||n=='\n'||n=='\r'||n=='\t'||n==-1)
return true;
return false;
}
private boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return isWhiteSpace(c);
}
private interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Blue Ranger hates palindromic strings and he is also jealous of Red Ranger. So to test Red Ranger he gives him a task. Given N and M, Red Ranger must find a string of length N and containing at max M distinct english alphabets such that the string contains minimum substrings that are palindromes. As there can be many such strings Blue Ranger asks him to simply tell the minimum possible number of substrings in the optimal string that are palindromes.
Palindrome is a string that can be read in the same way in both left-to-right and right-to-left directions.Input contains two integers N and M.
Constraints:
1 <= N <= 100000
1 <= M <= 100Print a single integer that is the minimum possible number of substrings in the optimal string that are palindromes.Sample Input 1
3 15
Sample Output 1
3
Explanation: optimal string can be "abc" which has three substrings that are palindromes.
Sample Input 2
3 1
Sample Output 2
6
Explanation: optimal string can be "aaa" which has 6 substrings that are pallindromes., I have written this Solution Code: n,m = [int(x) for x in input().split()]
if m>2 or m>=n:
print(n)
elif m==1:
res = n*(n+1)//2
print(res)
else:
res = 2*(n-1)
print(res), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Blue Ranger hates palindromic strings and he is also jealous of Red Ranger. So to test Red Ranger he gives him a task. Given N and M, Red Ranger must find a string of length N and containing at max M distinct english alphabets such that the string contains minimum substrings that are palindromes. As there can be many such strings Blue Ranger asks him to simply tell the minimum possible number of substrings in the optimal string that are palindromes.
Palindrome is a string that can be read in the same way in both left-to-right and right-to-left directions.Input contains two integers N and M.
Constraints:
1 <= N <= 100000
1 <= M <= 100Print a single integer that is the minimum possible number of substrings in the optimal string that are palindromes.Sample Input 1
3 15
Sample Output 1
3
Explanation: optimal string can be "abc" which has three substrings that are palindromes.
Sample Input 2
3 1
Sample Output 2
6
Explanation: optimal string can be "aaa" which has 6 substrings that are pallindromes., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n,m;
cin>>n>>m;
if(n==1)
cout<<1;
else
if(m==1){
cout<<(n*(n+1))/2;
}
else{
if(m==2)
cout<<(2*n-2);
else
cout<<n;
}
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For an integer N, your task is to calculate sum of first N natural numbers.<b>User Task:</b>
Since this will be a functional problem, you don't have to worry about input. You just have to complete the function <b>sum()</b> which takes the integer N as a parameter.
Constraints:
1 <= N < = 100000000Print the sum of first N natural numbers.Sample Input:-
5
Sample Output:-
15
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code:
static void sum(int N){
long x=N;
x=x*(x+1);
x=x/2;
System.out.print(x);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: They say friendship is greater than love. Why not play the famous game "FLAMES".
The rules are super simple. Given two strings (all lowercase), remove all the letters that are common to both the strings from both the strings. You cannot erase a character in first string whichever corresponding same character in other string not exist.
For example, in the case
String 1: saumya
String 2: ansh
You can remove only 1 'a' and 1 's' from both the strings.
Remaining strings are:
String 1: umya
String 2: nh
Now all you need to do is find the sum of the remaining strings length % 6.
Output:
If obtained value is 1, output "Friends"
If obtained value is 2, output "Love"
If obtained value is 3, output "Affection"
If obtained value is 4, output "Marriage"
If obtained value is 5, output "Enemy"
If obtained value is 0, output "Siblings"You will be given two strings on different lines.
Constraints
1 <= Length of both the strings <= 100000Output a single string, the result of FLAMES test.Sample Input:-
saumya
ansh
Sample Output:-
Siblings
Explanation:-
after deleting characters :-
str1 = umya
str2 = nh
sum = 4+2
sum%6=0
, I have written this Solution Code:
import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str1 = br.readLine();
String str2 = br.readLine();
countCommonCharacters(str1, str2);
}
static void countCommonCharacters(String s1, String s2) {
int[] arr1 = new int[26];
int[] arr2 = new int[26];
for (int i = 0; i < s1.length(); i++)
arr1[s1.codePointAt(i) - 97]++;
for (int i = 0; i < s2.length(); i++)
arr2[s2.codePointAt(i) - 97]++;
int lenToCut = 0;
for (int i = 0; i < 26; i++) {
int leastOccurrence = Math.min(arr1[i], arr2[i]);
lenToCut += (2 * leastOccurrence);
}
System.out.println(findRelation((s1.length() + s2.length() - lenToCut) % 6));
}
static String findRelation(int value) {
switch (value) {
case 1:
return "Friends";
case 2:
return "Love";
case 3:
return "Affection";
case 4:
return "Marriage";
case 5:
return "Enemy";
default:
return "Siblings";
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: They say friendship is greater than love. Why not play the famous game "FLAMES".
The rules are super simple. Given two strings (all lowercase), remove all the letters that are common to both the strings from both the strings. You cannot erase a character in first string whichever corresponding same character in other string not exist.
For example, in the case
String 1: saumya
String 2: ansh
You can remove only 1 'a' and 1 's' from both the strings.
Remaining strings are:
String 1: umya
String 2: nh
Now all you need to do is find the sum of the remaining strings length % 6.
Output:
If obtained value is 1, output "Friends"
If obtained value is 2, output "Love"
If obtained value is 3, output "Affection"
If obtained value is 4, output "Marriage"
If obtained value is 5, output "Enemy"
If obtained value is 0, output "Siblings"You will be given two strings on different lines.
Constraints
1 <= Length of both the strings <= 100000Output a single string, the result of FLAMES test.Sample Input:-
saumya
ansh
Sample Output:-
Siblings
Explanation:-
after deleting characters :-
str1 = umya
str2 = nh
sum = 4+2
sum%6=0
, I have written this Solution Code: name1 = input().strip().lower()
name2 = input().strip().lower()
listA = [0]*26
listB = [0]*26
for i in name1:
listA[ord(i)-ord('a')] = listA[ord(i)-ord('a')] + 1
for i in name2:
listB[ord(i)-ord('a')] = listB[ord(i)-ord('a')] + 1
count = 0
for i in range(0,26):
count = count+abs(listA[i]-listB[i])
res = ["Siblings","Friends","Love","Affection", "Marriage", "Enemy"]
print(res[count%6]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: They say friendship is greater than love. Why not play the famous game "FLAMES".
The rules are super simple. Given two strings (all lowercase), remove all the letters that are common to both the strings from both the strings. You cannot erase a character in first string whichever corresponding same character in other string not exist.
For example, in the case
String 1: saumya
String 2: ansh
You can remove only 1 'a' and 1 's' from both the strings.
Remaining strings are:
String 1: umya
String 2: nh
Now all you need to do is find the sum of the remaining strings length % 6.
Output:
If obtained value is 1, output "Friends"
If obtained value is 2, output "Love"
If obtained value is 3, output "Affection"
If obtained value is 4, output "Marriage"
If obtained value is 5, output "Enemy"
If obtained value is 0, output "Siblings"You will be given two strings on different lines.
Constraints
1 <= Length of both the strings <= 100000Output a single string, the result of FLAMES test.Sample Input:-
saumya
ansh
Sample Output:-
Siblings
Explanation:-
after deleting characters :-
str1 = umya
str2 = nh
sum = 4+2
sum%6=0
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 10000001
int main(){
string s1,s2;
cin>>s1>>s2;
string a[6];
a[1]= "Friends";
a[2]= "Love";
a[3]="Affection";
a[4]= "Marriage";
a[5]= "Enemy";
a[0]= "Siblings";
int b[26],c[26];
for(int i=0;i<26;i++){b[i]=0;c[i]=0;}
for(int i=0;i<s1.length();i++){
b[s1[i]-'a']++;
}
for(int i=0;i<s2.length();i++){
c[s2[i]-'a']++;
}
int sum=0;
for(int i=0;i<26;i++){
sum+=abs(b[i]-c[i]);
}
cout<<a[sum%6];
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: They say friendship is greater than love. Why not play the famous game "FLAMES".
The rules are super simple. Given two strings (all lowercase), remove all the letters that are common to both the strings from both the strings. You cannot erase a character in first string whichever corresponding same character in other string not exist.
For example, in the case
String 1: saumya
String 2: ansh
You can remove only 1 'a' and 1 's' from both the strings.
Remaining strings are:
String 1: umya
String 2: nh
Now all you need to do is find the sum of the remaining strings length % 6.
Output:
If obtained value is 1, output "Friends"
If obtained value is 2, output "Love"
If obtained value is 3, output "Affection"
If obtained value is 4, output "Marriage"
If obtained value is 5, output "Enemy"
If obtained value is 0, output "Siblings"You will be given two strings on different lines.
Constraints
1 <= Length of both the strings <= 100000Output a single string, the result of FLAMES test.Sample Input:-
saumya
ansh
Sample Output:-
Siblings
Explanation:-
after deleting characters :-
str1 = umya
str2 = nh
sum = 4+2
sum%6=0
, I have written this Solution Code:
import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str1 = br.readLine();
String str2 = br.readLine();
countCommonCharacters(str1, str2);
}
static void countCommonCharacters(String s1, String s2) {
int[] arr1 = new int[26];
int[] arr2 = new int[26];
for (int i = 0; i < s1.length(); i++)
arr1[s1.codePointAt(i) - 97]++;
for (int i = 0; i < s2.length(); i++)
arr2[s2.codePointAt(i) - 97]++;
int lenToCut = 0;
for (int i = 0; i < 26; i++) {
int leastOccurrence = Math.min(arr1[i], arr2[i]);
lenToCut += (2 * leastOccurrence);
}
System.out.println(findRelation((s1.length() + s2.length() - lenToCut) % 6));
}
static String findRelation(int value) {
switch (value) {
case 1:
return "Friends";
case 2:
return "Love";
case 3:
return "Affection";
case 4:
return "Marriage";
case 5:
return "Enemy";
default:
return "Siblings";
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: They say friendship is greater than love. Why not play the famous game "FLAMES".
The rules are super simple. Given two strings (all lowercase), remove all the letters that are common to both the strings from both the strings. You cannot erase a character in first string whichever corresponding same character in other string not exist.
For example, in the case
String 1: saumya
String 2: ansh
You can remove only 1 'a' and 1 's' from both the strings.
Remaining strings are:
String 1: umya
String 2: nh
Now all you need to do is find the sum of the remaining strings length % 6.
Output:
If obtained value is 1, output "Friends"
If obtained value is 2, output "Love"
If obtained value is 3, output "Affection"
If obtained value is 4, output "Marriage"
If obtained value is 5, output "Enemy"
If obtained value is 0, output "Siblings"You will be given two strings on different lines.
Constraints
1 <= Length of both the strings <= 100000Output a single string, the result of FLAMES test.Sample Input:-
saumya
ansh
Sample Output:-
Siblings
Explanation:-
after deleting characters :-
str1 = umya
str2 = nh
sum = 4+2
sum%6=0
, I have written this Solution Code: name1 = input().strip().lower()
name2 = input().strip().lower()
listA = [0]*26
listB = [0]*26
for i in name1:
listA[ord(i)-ord('a')] = listA[ord(i)-ord('a')] + 1
for i in name2:
listB[ord(i)-ord('a')] = listB[ord(i)-ord('a')] + 1
count = 0
for i in range(0,26):
count = count+abs(listA[i]-listB[i])
res = ["Siblings","Friends","Love","Affection", "Marriage", "Enemy"]
print(res[count%6]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: They say friendship is greater than love. Why not play the famous game "FLAMES".
The rules are super simple. Given two strings (all lowercase), remove all the letters that are common to both the strings from both the strings. You cannot erase a character in first string whichever corresponding same character in other string not exist.
For example, in the case
String 1: saumya
String 2: ansh
You can remove only 1 'a' and 1 's' from both the strings.
Remaining strings are:
String 1: umya
String 2: nh
Now all you need to do is find the sum of the remaining strings length % 6.
Output:
If obtained value is 1, output "Friends"
If obtained value is 2, output "Love"
If obtained value is 3, output "Affection"
If obtained value is 4, output "Marriage"
If obtained value is 5, output "Enemy"
If obtained value is 0, output "Siblings"You will be given two strings on different lines.
Constraints
1 <= Length of both the strings <= 100000Output a single string, the result of FLAMES test.Sample Input:-
saumya
ansh
Sample Output:-
Siblings
Explanation:-
after deleting characters :-
str1 = umya
str2 = nh
sum = 4+2
sum%6=0
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 10000001
int main(){
string s1,s2;
cin>>s1>>s2;
string a[6];
a[1]= "Friends";
a[2]= "Love";
a[3]="Affection";
a[4]= "Marriage";
a[5]= "Enemy";
a[0]= "Siblings";
int b[26],c[26];
for(int i=0;i<26;i++){b[i]=0;c[i]=0;}
for(int i=0;i<s1.length();i++){
b[s1[i]-'a']++;
}
for(int i=0;i<s2.length();i++){
c[s2[i]-'a']++;
}
int sum=0;
for(int i=0;i<26;i++){
sum+=abs(b[i]-c[i]);
}
cout<<a[sum%6];
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Power Rangers have to trap a evil spirit monster into a cuboid shape box with dimensions A*B*C, where A, B and C are positive integers. The volume of the box should be exactly X cubic units (that is A*B*C should be equal to X). To allow minimum evil aura to leak from the box the surface area of the box should be minimized. Given X, find the minimum surface area of the optimal box.The first and the only line of input contains a single integer X.
Constraints:
1 <= X <= 1000000Print the minimum surface area of the optimal box.Sample Input 1
125
Sample Output 1
150
Explanation: Optimal dimensions are 5*5*5.
Sample Input 2
100
Sample Output 1
130
Explanation: Optimal dimensions are 5*4*5., I have written this Solution Code:
import java.io.*;
import java.util.*;
public class Main
{
public static void main(String[] args)throws Exception{ new Main().run();}
long mod=1000000000+7;
long tsa=Long.MAX_VALUE;
void solve() throws Exception
{
long X=nl();
div(X);
out.println(tsa);
}
long cal(long a,long b, long c)
{
return 2l*(a*b + b*c + c*a);
}
void div(long n)
{
for (long i=1; i*i<=n; i++)
{
if (n%i==0)
{
if (n/i == i)
{
all_div(i, i);
}
else
{
all_div(i, n/i);
all_div(n/i, i);
}
}
}
}
void all_div(long n , long alag)
{
ArrayList<Long> al = new ArrayList<>();
for (long i=1; i*i<=n; i++)
{
if (n%i==0)
{
if (n/i == i)
tsa=min(tsa,cal(i,i,alag));
else
{
tsa=min(tsa,cal(i,n/i,alag));
}
}
}
}
private byte[] buf=new byte[1024];
private int index;
private InputStream in;
private int total;
private SpaceCharFilter filter;
PrintWriter out;
int min(int... ar){int min=Integer.MAX_VALUE;for(int i:ar)min=Math.min(min, i);return min;}
long min(long... ar){long min=Long.MAX_VALUE;for(long i:ar)min=Math.min(min, i);return min;}
int max(int... ar) {int max=Integer.MIN_VALUE;for(int i:ar)max=Math.max(max, i);return max;}
long max(long... ar) {long max=Long.MIN_VALUE;for(long i:ar)max=Math.max(max, i);return max;}
void reverse(int a[]){for(int i=0;i<a.length>>1;i++){int tem=a[i];a[i]=a[a.length-1-i];a[a.length-1-i]=tem;}}
void reverse(long a[]){for(int i=0;i<a.length>>1;i++){long tem=a[i];a[i]=a[a.length-1-i];a[a.length-1-i]=tem;}}
String reverse(String s){StringBuilder sb=new StringBuilder(s);sb.reverse();return sb.toString();}
void shuffle(int a[]) {
ArrayList<Integer> al = new ArrayList<>();
for(int i=0;i<a.length;i++)
al.add(a[i]);
Collections.sort(al);
for(int i=0;i<a.length;i++)
a[i]=al.get(i);
}
long lcm(long a,long b)
{
return (a*b)/(gcd(a,b));
}
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}
long expo(long p,long q)
{
long z = 1;
while (q>0) {
if (q%2 == 1) {
z = (z * p)%mod;
}
p = (p*p)%mod;
q >>= 1;
}
return z;
}
void run()throws Exception
{
in=System.in; out = new PrintWriter(System.out);
solve();
out.flush();
}
private int scan()throws IOException
{
if(total<0)
throw new InputMismatchException();
if(index>=total)
{
index=0;
total=in.read(buf);
if(total<=0)
return -1;
}
return buf[index++];
}
private int ni() throws IOException
{
int c = scan();
while (isSpaceChar(c))
c = scan();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = scan();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = scan();
} while (!isSpaceChar(c));
return res * sgn;
}
private long nl() throws IOException
{
long num = 0;
int b;
boolean minus = false;
while ((b = scan()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
;
if (b == '-') {
minus = true;
b = scan();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = scan();
}
}
private double nd() throws IOException{
return Double.parseDouble(ns());
}
private String ns() throws IOException {
int c = scan();
while (isSpaceChar(c))
c = scan();
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c))
res.appendCodePoint(c);
c = scan();
} while (!isSpaceChar(c));
return res.toString();
}
private String nss() throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
return br.readLine();
}
private char nc() throws IOException
{
int c = scan();
while (isSpaceChar(c))
c = scan();
return (char) c;
}
private boolean isWhiteSpace(int n)
{
if(n==' '||n=='\n'||n=='\r'||n=='\t'||n==-1)
return true;
return false;
}
private boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return isWhiteSpace(c);
}
private interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Power Rangers have to trap a evil spirit monster into a cuboid shape box with dimensions A*B*C, where A, B and C are positive integers. The volume of the box should be exactly X cubic units (that is A*B*C should be equal to X). To allow minimum evil aura to leak from the box the surface area of the box should be minimized. Given X, find the minimum surface area of the optimal box.The first and the only line of input contains a single integer X.
Constraints:
1 <= X <= 1000000Print the minimum surface area of the optimal box.Sample Input 1
125
Sample Output 1
150
Explanation: Optimal dimensions are 5*5*5.
Sample Input 2
100
Sample Output 1
130
Explanation: Optimal dimensions are 5*4*5., I have written this Solution Code: m =int(input())
def print_factors(x):
factors=[]
for i in range(1, x + 1):
if x % i == 0:
factors.append(i)
return(factors)
area=[]
factors=print_factors(m)
for a in factors:
for b in factors:
for c in factors:
if(a*b*c==m):
area.append((2*a*b)+(2*b*c)+(2*a*c))
print(min(area)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Power Rangers have to trap a evil spirit monster into a cuboid shape box with dimensions A*B*C, where A, B and C are positive integers. The volume of the box should be exactly X cubic units (that is A*B*C should be equal to X). To allow minimum evil aura to leak from the box the surface area of the box should be minimized. Given X, find the minimum surface area of the optimal box.The first and the only line of input contains a single integer X.
Constraints:
1 <= X <= 1000000Print the minimum surface area of the optimal box.Sample Input 1
125
Sample Output 1
150
Explanation: Optimal dimensions are 5*5*5.
Sample Input 2
100
Sample Output 1
130
Explanation: Optimal dimensions are 5*4*5., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int x;
cin >> x;
int ans = 6*x*x;
for (int i=1;i*i*i<=x;++i)
if (x%i==0)
for (int j=i;j*j<=x/i;++j)
if (x/i % j==0) {
int k=x/i/j;
int cur=0;
cur=i*j+i*k+k*j;
cur*=2;
ans=min(ans,cur);
}
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The coolness of the subarray of an array is the sum of elements in the subarray. The coolest subarray is the subarray having the maximum coolness, while the hottest subarray is the one having minimum coolness (obviously it can be negative). The coolest and the hottest subarrays are always <b>non-empty</b>.
Tono believes that cool and hot are happy together. The happiness of the array is the absolute difference between the coolness of its coolest and its hottest subarray.
Given an array A consisting of N elements, find its happiness.The first line of the input contains an integer N.
The next line contains N singly spaced integers A[1], A[2],...A[N]
Constraints
1 <= N <= 200000
-1000000000 <= A[i] <= 1000000000Output a single integer, the happiness of the array.
(The output may not fit into 32 bit integer datatype, use long long integer datatype instead).Sample Input
5
-1 2 -3 1 -5
Sample Output
9
Explanation: Coolest subarray of the array is [2], while the hottest subarray of the array is [-3, 1, -5]. The happiness of the array is 2-(-7)=9., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static long KadanesAlgoMax(int[] a,int n)
{
long maxSum=Integer.MIN_VALUE;
long currSum=0;
for(int i=0;i<n;i++)
{
currSum+=a[i];
if(currSum>maxSum)maxSum=currSum;
if(currSum<0)currSum=0;
}
return maxSum;
}
public static long KadanesAlgoMin(int[] a,int n)
{
long minSum=Integer.MAX_VALUE;
long currSum=0;
for(int i=0;i<n;i++)
{
currSum+=a[i];
if(currSum<minSum)minSum=currSum;
if(currSum>0)currSum=0;
}
return minSum;
}
public static void main (String[] args)throws IOException {
BufferedReader rd=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(rd.readLine());
String[] s=rd.readLine().split(" ");
int[] a=new int[n];
for(int i=0;i<n;i++)
{
a[i]=Integer.parseInt(s[i]);
}
System.out.print((long)Math.abs(KadanesAlgoMax(a,n)-KadanesAlgoMin(a,n)));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The coolness of the subarray of an array is the sum of elements in the subarray. The coolest subarray is the subarray having the maximum coolness, while the hottest subarray is the one having minimum coolness (obviously it can be negative). The coolest and the hottest subarrays are always <b>non-empty</b>.
Tono believes that cool and hot are happy together. The happiness of the array is the absolute difference between the coolness of its coolest and its hottest subarray.
Given an array A consisting of N elements, find its happiness.The first line of the input contains an integer N.
The next line contains N singly spaced integers A[1], A[2],...A[N]
Constraints
1 <= N <= 200000
-1000000000 <= A[i] <= 1000000000Output a single integer, the happiness of the array.
(The output may not fit into 32 bit integer datatype, use long long integer datatype instead).Sample Input
5
-1 2 -3 1 -5
Sample Output
9
Explanation: Coolest subarray of the array is [2], while the hottest subarray of the array is [-3, 1, -5]. The happiness of the array is 2-(-7)=9., I have written this Solution Code: l = int(input())
arr = list(map(int,input().split()))
maxSum = arr[0]
currSum =0
maxSumB = arr[0]
currSumB = 0
for j in range(0,l):
currSum = currSum + arr[j]
if(maxSum>currSum):
maxSum = currSum
if(currSum>0):
currSum = 0
currSumB = currSumB + arr[j]
if(maxSumB<currSumB):
maxSumB = currSumB
if(currSumB<0):
currSumB = 0
print(abs(maxSumB-maxSum)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The coolness of the subarray of an array is the sum of elements in the subarray. The coolest subarray is the subarray having the maximum coolness, while the hottest subarray is the one having minimum coolness (obviously it can be negative). The coolest and the hottest subarrays are always <b>non-empty</b>.
Tono believes that cool and hot are happy together. The happiness of the array is the absolute difference between the coolness of its coolest and its hottest subarray.
Given an array A consisting of N elements, find its happiness.The first line of the input contains an integer N.
The next line contains N singly spaced integers A[1], A[2],...A[N]
Constraints
1 <= N <= 200000
-1000000000 <= A[i] <= 1000000000Output a single integer, the happiness of the array.
(The output may not fit into 32 bit integer datatype, use long long integer datatype instead).Sample Input
5
-1 2 -3 1 -5
Sample Output
9
Explanation: Coolest subarray of the array is [2], while the hottest subarray of the array is [-3, 1, -5]. The happiness of the array is 2-(-7)=9., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define int long long
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define MOD 1000000007
#define INF 1000000000000000007LL
const int N = 200005;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
int arr[N];
int n;
int kadane(){
int sum = 0;
int mx = 0;
For(i, 0, n){
sum += arr[i];
if(sum < 0){
sum = 0;
}
mx = max(sum, mx);
}
if(mx > 0)
return mx;
// all elements negative
mx = -10000000000LL;
For(i, 0, n){
mx = max(mx, arr[i]);
}
return mx;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
cin>>n;
For(i, 0, n){
cin>>arr[i];
}
int v1 = kadane();
For(i, 0, n){
arr[i]=-1*arr[i];
}
int v2 = kadane();
cout<<v1+v2;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of size N and a number K, the task is to find the smallest number M such that the sum of the array becomes lesser than or equal to the number K when every element of that array is divided by the number M.
Note: Each result of the division is rounded to the nearest integer greater than or equal to that element. For example: 10/3 = 4 and 6/2 = 3The first line contains two integers N and K.
Next line contains N integers denoting the elements of arr[]
<b>Constraints:</b>
1 <= N <= 100000
1 <= arr[i] <= 100000
1 <= K <= 100000000Print a single integer the value of smallest number MSample Input:
4 6
2 3 4 9
Sample Output:
4
Explanation:
When every element is divided by 4 - 2/4 + 3/4 + 4/4 + 9/4 = 1 + 1 + 1 + 3 = 6
When every element is divided by 3 - 2/3 + 3/3 + 4/3 + 9/3 = 1 + 1 + 2 + 3 = 7 which is greater than K.
Hence the smallest integer which makes the sum less than or equal to K = 6 is 4., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
String s[] = br.readLine().split(" ");
int n = Integer.parseInt(s[0]);
int k = Integer.parseInt(s[1]);
String str[] = br.readLine().split(" ");
int[] arr = new int[n];
for(int i=0;i<n;i++){
arr[i] = Integer.parseInt(str[i]);
}
System.out.print(minDivisor(arr,n,k));
}
static int minDivisor(int arr[],int N, int limit) {
int low = 0, high = 1000000000;
while (low < high)
{
int mid = (low + high) / 2;
int sum = 0;
for(int i = 0; i < N; i++)
{
sum += Math.ceil((double) arr[i] / (double) mid);
}
if(sum <= limit){
high = mid;
}
else{
low = mid + 1;
}
}
return low;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of size N and a number K, the task is to find the smallest number M such that the sum of the array becomes lesser than or equal to the number K when every element of that array is divided by the number M.
Note: Each result of the division is rounded to the nearest integer greater than or equal to that element. For example: 10/3 = 4 and 6/2 = 3The first line contains two integers N and K.
Next line contains N integers denoting the elements of arr[]
<b>Constraints:</b>
1 <= N <= 100000
1 <= arr[i] <= 100000
1 <= K <= 100000000Print a single integer the value of smallest number MSample Input:
4 6
2 3 4 9
Sample Output:
4
Explanation:
When every element is divided by 4 - 2/4 + 3/4 + 4/4 + 9/4 = 1 + 1 + 1 + 3 = 6
When every element is divided by 3 - 2/3 + 3/3 + 4/3 + 9/3 = 1 + 1 + 2 + 3 = 7 which is greater than K.
Hence the smallest integer which makes the sum less than or equal to K = 6 is 4., I have written this Solution Code: from math import ceil
def kSum(arr, n, k):
low = 0
high = 10 ** 9
while (low < high):
mid = (low + high) // 2
sum = 0
for i in range(int(n)):
sum += ceil( int(arr[i]) / mid)
if (sum <= int(k)):
high = mid
else:
low = mid + 1
return low
n, k =input().split()
arr = input().split()
print( kSum(arr, n, k) ), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of size N and a number K, the task is to find the smallest number M such that the sum of the array becomes lesser than or equal to the number K when every element of that array is divided by the number M.
Note: Each result of the division is rounded to the nearest integer greater than or equal to that element. For example: 10/3 = 4 and 6/2 = 3The first line contains two integers N and K.
Next line contains N integers denoting the elements of arr[]
<b>Constraints:</b>
1 <= N <= 100000
1 <= arr[i] <= 100000
1 <= K <= 100000000Print a single integer the value of smallest number MSample Input:
4 6
2 3 4 9
Sample Output:
4
Explanation:
When every element is divided by 4 - 2/4 + 3/4 + 4/4 + 9/4 = 1 + 1 + 1 + 3 = 6
When every element is divided by 3 - 2/3 + 3/3 + 4/3 + 9/3 = 1 + 1 + 2 + 3 = 7 which is greater than K.
Hence the smallest integer which makes the sum less than or equal to K = 6 is 4., I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define ll long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N], n, k;
bool f(int x){
int sum = 0;
for(int i = 1; i <= n; i++)
sum += (a[i]-1)/x + 1;
return (sum <= k);
}
void solve(){
cin >> n >> k;
for(int i = 1; i <= n; i++)
cin >> a[i];
int l = 0, h = N;
while(l+1 < h){
int m = (l + h) >> 1;
if(f(m))
h = m;
else
l = m;
}
cout << h;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a postfix expression, your task is to evaluate given expression.
Infix expression: The expression of the form a operator b. When an operator is in-between every pair of operands.
Postfix expression: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.The first line denotes the size of the string(which contains number or operand in form of string) i.e. N.
The next line contains string. The string contains an integer called operand or any of the four operators (+ - * /)
Constraints:-
1 <= n <= 40
1<=number<=500
Output the value of arithmetic expression formed using reverse Polish Notation.Input 1:
5
2 1 + 3 *
Output 1:
9
Explaination 1:
starting from backside:
*: ( )*( )
3: ()*(3)
+: ( () + () )*(3)
1: ( () + (1) )*(3)
2: ( (2) + (1) )*(3)
((2)+(1))*(3) = 9
Input 2:
5
4 13 5 / +
Output 2:
6
Explanation 2:
+: ()+()
/: ()+(() / ())
5: ()+(() / (5))
1: ()+((13) / (5))
4: (4)+((13) / (5))
(4)+((13) / (5)) = 6, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
String str[] = br.readLine().split(" ");
Stack <Integer> st = new Stack<>();
for(int i=0;i<t;i++){
char ch = str[i].charAt(0);
if(Character.isDigit(ch)){
st.push(Integer.parseInt(str[i]));
}
else{
int str1 = st.pop();
int str2 = st.pop();
switch(ch) {
case '+':
st.push(str2+str1);
break;
case '-':
st.push(str2- str1);
break;
case '/':
st.push(str2/str1);
break;
case '*':
st.push(str2*str1);
break;
}
}
}
System.out.println(st.peek());
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a postfix expression, your task is to evaluate given expression.
Infix expression: The expression of the form a operator b. When an operator is in-between every pair of operands.
Postfix expression: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.The first line denotes the size of the string(which contains number or operand in form of string) i.e. N.
The next line contains string. The string contains an integer called operand or any of the four operators (+ - * /)
Constraints:-
1 <= n <= 40
1<=number<=500
Output the value of arithmetic expression formed using reverse Polish Notation.Input 1:
5
2 1 + 3 *
Output 1:
9
Explaination 1:
starting from backside:
*: ( )*( )
3: ()*(3)
+: ( () + () )*(3)
1: ( () + (1) )*(3)
2: ( (2) + (1) )*(3)
((2)+(1))*(3) = 9
Input 2:
5
4 13 5 / +
Output 2:
6
Explanation 2:
+: ()+()
/: ()+(() / ())
5: ()+(() / (5))
1: ()+((13) / (5))
4: (4)+((13) / (5))
(4)+((13) / (5)) = 6, I have written this Solution Code: stack = []
n = int(input())
exp = [i for i in input().split()]
for i in exp:
try:
stack.append(int(i))
except:
a1 = stack.pop()
a2 = stack.pop()
if i == '+':
stack.append(a1+a2)
if i == '-':
stack.append(a2-a1)
if i == '/':
stack.append(a2//a1)
if i == '*':
stack.append(a1*a2)
print(*stack), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a postfix expression, your task is to evaluate given expression.
Infix expression: The expression of the form a operator b. When an operator is in-between every pair of operands.
Postfix expression: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.The first line denotes the size of the string(which contains number or operand in form of string) i.e. N.
The next line contains string. The string contains an integer called operand or any of the four operators (+ - * /)
Constraints:-
1 <= n <= 40
1<=number<=500
Output the value of arithmetic expression formed using reverse Polish Notation.Input 1:
5
2 1 + 3 *
Output 1:
9
Explaination 1:
starting from backside:
*: ( )*( )
3: ()*(3)
+: ( () + () )*(3)
1: ( () + (1) )*(3)
2: ( (2) + (1) )*(3)
((2)+(1))*(3) = 9
Input 2:
5
4 13 5 / +
Output 2:
6
Explanation 2:
+: ()+()
/: ()+(() / ())
5: ()+(() / (5))
1: ()+((13) / (5))
4: (4)+((13) / (5))
(4)+((13) / (5)) = 6, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
class Solution {
public:
int evalRPN(vector<string> &tokens) {
int size = tokens.size();
if (size == 0)
return 0;
std::stack<int> operands;
for (int i = 0; i < size; ++i)
{
std::string cur_token = tokens[i];
if ((cur_token == "*") || (cur_token == "/") || (cur_token == "+") || (cur_token == "-"))
{
int opr2 = operands.top();
operands.pop();
int opr1 = operands.top();
operands.pop();
int result = this->eval(opr1, opr2, cur_token);
operands.push(result);
}
else{
operands.push(std::atoi(cur_token.c_str()));
}
}
return operands.top();
}
int eval(int opr1, int opr2, string opt)
{
if (opt == "*")
{
return opr1 * opr2;
}
else if (opt == "+")
{
return opr1 + opr2;
}
else if (opt == "-")
{
return opr1 - opr2;
}
else if (opt == "/")
{
return opr1 / opr2;
}
return 0;
}
};
signed main() {
IOS;
int n; cin >> n;
vector<string> v(n, "");
for(int i = 0; i < n; i++)
cin >> v[i];
Solution s;
cout << s.evalRPN(v);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean Matrix of size N*M, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, I have written this Solution Code: // mat is the matrix/ 2d array
// n,m are dimensions
function goodCell(mat, n, m) {
// write code here
// do not console.log
// return the answer as a number
let cnt = 0;
for (let i = 1; i < n - 1; i++) {
for (let j = 1; j < m - 1; j++) {
if (mat[i - 1][j] == 1 && mat[i + 1][j] == 1 && mat[i][j - 1] == 1 && mat[i][j + 1] == 1) {
cnt++;
}
}
}
return cnt
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean Matrix of size N*M, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, I have written this Solution Code: N, M= list(map(int,input().split()))
mat =[]
for i in range(N):
List =list(map(int,input().split()))[:M]
mat.append(List)
count =0
for i in range(1,N-1):
for j in range(1,M-1):
if (mat[i][j-1] == 1 and mat[i][j+1] == 1 and mat[i-1][j] == 1 and mat[i+1][j] == 1):
count +=1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean Matrix of size N*M, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1000001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
signed main(){
int n,m;
cin>>n>>m;
int a[n][m];
FOR(i,n){
FOR(j,m){
cin>>a[i][j];}}
int sum=0,sum1=0;;
FOR1(i,1,n-1){
FOR1(j,1,m-1){
if(a[i-1][j]==1 && a[i+1][j]==1 && a[i][j-1]==1 && a[i][j+1]==1){
sum++;
}
}
}
out1(sum);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean Matrix of size N*M, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m= sc.nextInt();
int a[][]= new int[n][m];
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
a[i][j]=sc.nextInt();}}
int cnt=0;
for(int i=1;i<n-1;i++){
for(int j=1;j<m-1;j++){
if(a[i-1][j]==1 && a[i+1][j]==1 && a[i][j-1]==1 && a[i][j+1]==1){
cnt++;
}
}
}
System.out.print(cnt);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Walter white is considered very intelligent person. He has a problem to solve. As he is suffering from cancer, can you help him solve it?
Given two integer arrays C and S of length c and s respectively. Index i of array S can be considered good if a subarray of length c can be formed starting from index i which is complimentary to array C.
Two arrays A, B of same length are considered complimentary if any cyclic permutation of A satisfies the property (A[i]- A[i-1]=B[i]-B[i-1]) for all i from 2 to length of A (1 indexing).
Calculate number of good positions in S .
<a href="https://mathworld.wolfram.com/CyclicPermutation.html">Cyclic Permutation</a>
1 2 3 4 has 4 cyclic permutations 2 3 4 1, 3 4 1 2, 4 1 2 3,1 2 3 4First line contains integer s (length of array S).
Second line contains s space separated integers of array S.
Third line contain integer c (length of array C).
Forth line contains c space separated integers of array C.
Constraints:
1 <= s <=1000000
1 <= c <=1000000
1 <= S[i], C[i] <= 10^9
Print the answer.
Input :
9
1 2 3 1 2 4 1 2 3
3
1 2 3
Output :
4
Explanation :
index 1- 1 2 3 matches with 1 2 3
index 2- 2 3 1 matches with 2 3 1(2 3 1 is cyclic permutation of 1 2 3)
index 3- 3 1 2 matches with 3 1 2(3 1 2 is cyclic permutation of 1 2 3)
index 7- 1 2 3 matches with 1 2 3
Input :
4
3 4 3 4
2
1 2
Output :
3
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
// #define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 2000015
#define qw1 freopen("input1.txt", "r", stdin); freopen("output1.txt", "w", stdout);
#define qw2 freopen("input2.txt", "r", stdin); freopen("output2.txt", "w", stdout);
#define qw3 freopen("input3.txt", "r", stdin); freopen("output3.txt", "w", stdout);
#define qw4 freopen("input4.txt", "r", stdin); freopen("output4.txt", "w", stdout);
#define qw5 freopen("input5.txt", "r", stdin); freopen("output5.txt", "w", stdout);
#define qw6 freopen("input6.txt", "r", stdin); freopen("output6.txt", "w", stdout);
#define qw freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
int A[sz],B[sz],C[sz],D[sz],E[sz],F[sz],G[sz];
int n,m;
signed main()
{
cin>>n;
for(int i=0;i<n;i++)
{
cin>>A[i];
F[i]=A[i];
}
cin>>m;
for(int i=0;i<m;i++)
{
cin>>B[i];
G[i]=B[i];
C[m-i-1]=B[i];
}
C[m]=-500000000;
for(int i=0;i<n;i++)
{
C[i+m+1]=A[n-i-1];
}
int l=0,r=0;
for(int i=1;i<=n+m;i++)
{
if(i<=r)
{
E[i]=min(r-i+1,E[i-l]);
}
while(i+E[i]<=n+m && C[E[i]]-C[0]==C[i+E[i]]-C[i])
E[i]++;
if(i+E[i]-1>r) {
l=i;r=i+E[i]-1;
}
}
for(int i=0;i<m;i++)
{
C[i]=B[i];
}
for(int i=0;i<n;i++)
{
C[i+m+1]=A[i];
}
for(int i=0;i<n;i++)
{
A[i]=E[n+m-i];
}
l=0;
r=0;
for(int i=1;i<=n+m;i++)
{
if(i<=r)
{
D[i]=min(r-i+1,D[i-l]);
}
while(i+D[i]<=n+m && C[D[i]]-C[0]==C[i+D[i]]-C[i])
D[i]++;
if(i+D[i]-1>r) {
l=i;r=i+D[i]-1;
}
}
// cout<<0<<" ";
for(int i=0;i<n;i++)
{
B[i]=D[i+m+1];
// cout<<A[i]<<" ";
}
// cout<<endl;
// for(int i=0;i<n;i++)
// {
// cout<<B[i]<<" ";
// }cout<<endl;
// for(int i=0;i<=n;i++)
// {
// cout<<i<<" ";
// }
// cout<<endl;
int cnt=0;
vector<pii> xx,yy;
for(int i=0;i<=n;i++){
int a=0;
int b=0;
if(i>0) a=A[i-1];
if(i<n) b=B[i];
// cout<<i<<" "<<a<<" "<<b<<endl;
if(a+b>=m && (a==0 || b==0 ||(F[i]-F[i-1]==G[0]-G[m-1])))
{xx.pu(mp(i-a,i+b-m)); }
if(a==m) xx.pu(mp(i-a,i-a));
if(b==m ) xx.pu(mp(i,i));
}
sort(xx.begin(),xx.end());
for(int i=0;i<xx.size();i++)
{ // cout<<xx[i].fi<<" "<<xx[i].se<<endl;
if(yy.size()==0) yy.pu(mp(xx[i].fi,xx[i].se));
else{
int p=yy.size()-1;
// cout<<i<<" "<<xx[i].fi<<" "<<xx[i].se<<" " <<yy[p].se<<endl;
if(yy[p].se>=xx[i].se) continue;
if(yy[p].se>=xx[i].fi) yy[p].se=xx[i].se;
else yy.pu(mp(xx[i].fi,xx[i].se));
}
}
for(int i=0;i<yy.size();i++)
{ // cout<<yy[i].fi<<" "<<yy[i].se<<endl;
cnt+=yy[i].se-yy[i].fi+1;
}
cout<<cnt<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer Q representing the number of queries and an integer K. There are two types of queries:
(i) 1 x : Add the number x to the stream
(ii) 2 : Print the sum of last K numbers of the stream. If there are less than K numbers in the stream, print the sum of current stream.
Process all the queries.First line contains two integers Q and K.
Next Q lines contains the queries.
Constraints
1 <= Q <= 10^5
1 <= x <= 10^5
1 <= K <= Q
There is atleast one query of 2nd type. For each query of type 2, print the sum of last K numbers of the stream on a new line.Sample Input 1:
5 2
1 4
2
1 1
1 3
2
Output
4
4
Explanation:
Initial Stream = {}
Add 4. Stream = {4}
Sum of last two elements = 4
Add 1. Stream = {4, 1}
Add 3. Stream = {4, 1, 3}
Sum of last two elements = 4
Sample Input 2:
3 1
1 1
2
2
Output
1
1
Explanation
Initial Stream = {}
Add 1. Stream = {1}
Sum of last element = 1
Sum of last element = 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main{
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n,k;
String line = br.readLine();
String[] strs = line.trim().split("\\s+");
n=Integer.parseInt(strs[0]);
k=Integer.parseInt(strs[1]);
Long sum=0L;
Queue<Integer>x=new LinkedList<Integer>();
while(n>0){
n--;
line = br.readLine();
strs = line.trim().split("\\s+");
int y=Integer.parseInt(strs[0]);
if(y==1){
y=Integer.parseInt(strs[1]);
x.add(y);
sum += y;
if (x.size() > k) {
sum -= x.peek();
x.remove();
}
}
else{
System.out.print(sum+"\n");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer Q representing the number of queries and an integer K. There are two types of queries:
(i) 1 x : Add the number x to the stream
(ii) 2 : Print the sum of last K numbers of the stream. If there are less than K numbers in the stream, print the sum of current stream.
Process all the queries.First line contains two integers Q and K.
Next Q lines contains the queries.
Constraints
1 <= Q <= 10^5
1 <= x <= 10^5
1 <= K <= Q
There is atleast one query of 2nd type. For each query of type 2, print the sum of last K numbers of the stream on a new line.Sample Input 1:
5 2
1 4
2
1 1
1 3
2
Output
4
4
Explanation:
Initial Stream = {}
Add 4. Stream = {4}
Sum of last two elements = 4
Add 1. Stream = {4, 1}
Add 3. Stream = {4, 1, 3}
Sum of last two elements = 4
Sample Input 2:
3 1
1 1
2
2
Output
1
1
Explanation
Initial Stream = {}
Add 1. Stream = {1}
Sum of last element = 1
Sum of last element = 1, I have written this Solution Code: /**
* author: tourist1256
* created: 2022-06-14 14:26:47
**/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k) returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
#define int long long
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int64_t random_long(long long l = LLONG_MIN, long long r = LLONG_MAX) {
uniform_int_distribution<int64_t> generator(l, r);
return generator(rng);
}
int32_t main() {
int Q, K;
cin >> Q >> K;
deque<int> st;
int sum = 0;
while (Q--) {
int x;
cin >> x;
if (x == 1) {
int y;
cin >> y;
if (st.size() == K) {
sum -= st.back();
st.pop_back();
st.push_front(y);
sum += y;
} else {
st.push_front(y);
sum += y;
}
} else {
cout << sum << "\n";
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For two strings A and B, let A + B denote the concatenation of A and B in this order.
You are given N strings Si,. , SN. Modify and print them as follows, in the order i = 1,. , N:
Let X be the number of strings S1,. , Si-1 are equal to Si.
if X = 0 print Si.
if X (X > 0) print Si+(+X+), treating X as a string.The first line of the input contains a single integer N.
The next N lines contains N strings S1, S2,. SN.
Constraints:
1 <= N <= 2 x 10^5
Si is a string of length between 1 and 10 (inclusive) consisting of lowercase English letters.Print N lines as specified in the Problem Statement.Sample Input 1:
5
newfile
newfile
newfolder
newfile
newfolder
Sample Output 1:
newfile
newfile(1)
newfolder
newfile(2)
newfolder(1)
Sample Input 2:
11
a
a
a
a
a
a
a
a
a
a
a
Sample Output 2:
a
a(1)
a(2)
a(3)
a(4)
a(5)
a(6)
a(7)
a(8)
a(9)
a(10), I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
int n =Integer.parseInt(br.readLine());
String [] inp = new String [n];
Map <String,Integer> mp = new HashMap<>();
for(int i= 0 ;i<n;i++){
inp[i]=br.readLine();
}
for(String ele : inp ){
mp.put(ele,mp.getOrDefault(ele,0)+1);
if(mp.get(ele)==1) System.out.println(ele);
else System.out.println(ele+'('+(mp.get(ele)-1)+')');
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For two strings A and B, let A + B denote the concatenation of A and B in this order.
You are given N strings Si,. , SN. Modify and print them as follows, in the order i = 1,. , N:
Let X be the number of strings S1,. , Si-1 are equal to Si.
if X = 0 print Si.
if X (X > 0) print Si+(+X+), treating X as a string.The first line of the input contains a single integer N.
The next N lines contains N strings S1, S2,. SN.
Constraints:
1 <= N <= 2 x 10^5
Si is a string of length between 1 and 10 (inclusive) consisting of lowercase English letters.Print N lines as specified in the Problem Statement.Sample Input 1:
5
newfile
newfile
newfolder
newfile
newfolder
Sample Output 1:
newfile
newfile(1)
newfolder
newfile(2)
newfolder(1)
Sample Input 2:
11
a
a
a
a
a
a
a
a
a
a
a
Sample Output 2:
a
a(1)
a(2)
a(3)
a(4)
a(5)
a(6)
a(7)
a(8)
a(9)
a(10), I have written this Solution Code: N = int(input())
d = dict()
for ind in range(N):
s = input().strip()
if(s not in d):
print(s)
d[s] = 0
else:
print("%s(%s)"%(s,d[s]+1))
d[s] += 1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For two strings A and B, let A + B denote the concatenation of A and B in this order.
You are given N strings Si,. , SN. Modify and print them as follows, in the order i = 1,. , N:
Let X be the number of strings S1,. , Si-1 are equal to Si.
if X = 0 print Si.
if X (X > 0) print Si+(+X+), treating X as a string.The first line of the input contains a single integer N.
The next N lines contains N strings S1, S2,. SN.
Constraints:
1 <= N <= 2 x 10^5
Si is a string of length between 1 and 10 (inclusive) consisting of lowercase English letters.Print N lines as specified in the Problem Statement.Sample Input 1:
5
newfile
newfile
newfolder
newfile
newfolder
Sample Output 1:
newfile
newfile(1)
newfolder
newfile(2)
newfolder(1)
Sample Input 2:
11
a
a
a
a
a
a
a
a
a
a
a
Sample Output 2:
a
a(1)
a(2)
a(3)
a(4)
a(5)
a(6)
a(7)
a(8)
a(9)
a(10), I have written this Solution Code: #include<iostream>
#include<cstring>
#include<map>
using namespace std;
const int N=1e3+10;
int main(){
int n;
scanf("%d",&n);
map<string,int>ans;
for(int i=1;i<=n;i++){
string res;
cin>>res;
if(ans[res]){
cout<<res<<"("<<ans[res]<<")"<<endl;
}
else{
cout<<res<<endl;
}
ans[res]++;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is <b>Palindrome</b> or not. A number is said to be Palindrome when it reads the same from backward as forward.User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isPalindrome()</b> which contains N as a parameter.
<b>Constraints:</b>
1 <= N <= 9999You need to return "true" is the number is palindrome otherwise "false".Sample Input:
5
Sample Output:
true
Sample Input:
121
Sample Output:
true, I have written this Solution Code: static boolean isPalindrome(int N)
{
int sum = 0;
int rev = N;
while(N > 0)
{
int digit = N%10;
sum = sum*10+digit;
N = N/10;
}
if(rev == sum)
return true;
else return false;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is <b>Palindrome</b> or not. A number is said to be Palindrome when it reads the same from backward as forward.User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isPalindrome()</b> which contains N as a parameter.
<b>Constraints:</b>
1 <= N <= 9999You need to return "true" is the number is palindrome otherwise "false".Sample Input:
5
Sample Output:
true
Sample Input:
121
Sample Output:
true, I have written this Solution Code: def isPalindrome(N):
sum1 = 0
rev = N
while(N > 0):
digit = N%10
sum1 = sum1*10+digit
N = N//10
if(rev == sum1):
return True
return False, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given a <code>list of employees</code> with their information in a <code>JSON format as an argument</code>. Your task is to implement a JavaScript function, <code>performEmployeeOperations</code>, that performs the following tasks:
<ol>
<li>Find the <code>employee with the highest salary</code> and store it in the <code>highestSalaryEmployee</code>. Note, the <code>highestSalaryEmployee</code> logic has already been provided in the boilerplate. For the rest of the operations, you might have to look into array loop methods, like reduce, filter, map, etc.</li>
<li>The <code>employeesByDepartment</code> returns an <code>object</code> of employees grouped by department. The <code>department should be the key</code> and an <code>array of employees with their information as value</code>.</li>
<li>The <code>averageAgeByDepartment</code> returns an <code>object</code>, with the <code>department as key</code> and the <code>average age of that department as value</code>.</li>
<li><code>employeesWithLongestName</code> returns an <code>array</code> with the <code>employee(s) information having the longest name(s)</code>.</li>
</ol>
<code>Note:</code>
**Remove the extra spaces from the list while inserting the list of employees in the input section.**The <code>performEmployeeOperations</code> function will take in an <code>array of objects in JSON format</code>. Each Object will have a <code>name</code>, <code>age</code>, <code>department</code>, and <code>salary</code> property.The <code>performEmployeeOperations</code> function should return an <code>object</code> of <code>highestSalaryEmployee</code>, <code>employeesByDepartment</code>, <code>averageAgeByDepartment</code>, <code>employeesWithLongestName</code>.
<code>highestSalaryEmployee</code> should store the <code>object of the employee having the highest salary</code>.
<code>employeesByDepartment</code> should return an <code>object</code>.
<code>averageAgeByDepartment</code> should return an <code>object</code>.
<code>employeesWithLongestName</code> should return an <code>array of object</code>.const employees = [{"name":"John","age":30,"department":"HR","salary":50000},{"name":"Jane","age":28,"department":"IT","salary":60000},{"name":"Mark","age":35,"department":"HR","salary":55000},{"name":"Alice","age":32,"department":"Finance","salary":65000},{"name":"Charlie","age":40,"department":"IT","salary":70000}]
const operations = performEmployeeOperations(employees);
console.log(operations.highestSalaryEmployee); // Output: { name: 'Charlie', age: 40, department: 'IT', salary: 70000 }
console.log(operations.employeesByDepartment);
// Output:
{
HR: [
{ name: 'John', age: 30, department: 'HR', salary: 50000 },
{ name: 'Mark', age: 35, department: 'HR', salary: 55000 }
],
IT: [
{ name: 'Jane', age: 28, department: 'IT', salary: 60000 },
{ name: 'Charlie', age: 40, department: 'IT', salary: 70000 }
],
Finance: [ { name: 'Alice', age: 32, department: 'Finance', salary: 65000 } ]
}
console.log(operations.averageAgeByDepartment); //Output: { HR: 32.5, IT: 34, Finance: 32 }
console.log(operations.employeesWithLongestName); //Output: [ { name: 'Charlie', age: 40, department: 'IT', salary: 70000 } ], I have written this Solution Code: function performEmployeeOperations(employees) {
const highestSalaryEmployee = employees.reduce((acc, emp) => emp.salary > acc.salary ? emp : acc, employees[0]);
const employeesByDepartment = employees.reduce((acc, emp) => {
if (!acc[emp.department]) {
acc[emp.department] = [];
}
acc[emp.department].push(emp);
return acc;
}, {});
const averageAgeByDepartment = Object.keys(employeesByDepartment).reduce((acc, department) => {
const employeesInDepartment = employeesByDepartment[department];
const totalAge = employeesInDepartment.reduce((sum, emp) => sum + emp.age, 0);
const averageAge = totalAge / employeesInDepartment.length;
acc[department] = averageAge;
return acc;
}, {});
const longestNameLength = Math.max(...employees.map(emp => emp.name.length));
const employeesWithLongestName = employees.filter(emp => emp.name.length === longestNameLength);
return {
highestSalaryEmployee,
employeesByDepartment,
averageAgeByDepartment,
employeesWithLongestName
};
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find number of ways an integer N can be represented as a sum of unique natural numbers.First line contain an integer T denoting number of test cases. Each test case contains a single integer N.
Constraint:-
1 <= T <= 100
0 <= N <= 120Print a single integer containing number of ways.Sample input
4
6
1
4
2
Sample output:-
4
1
2
1
Explanation:-
TestCase1:-
6 can be represented as (1, 2, 3), (1, 5), (2, 4), (6), I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
for(int i=0;i<t;i++){
int n = Integer.parseInt(br.readLine());
System.out.println(Ways(n,1));
}
}
static int Ways(int x, int num)
{
int val =(x - num);
if (val == 0)
return 1;
if (val < 0)
return 0;
return Ways(val, num + 1) + Ways(x, num + 1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find number of ways an integer N can be represented as a sum of unique natural numbers.First line contain an integer T denoting number of test cases. Each test case contains a single integer N.
Constraint:-
1 <= T <= 100
0 <= N <= 120Print a single integer containing number of ways.Sample input
4
6
1
4
2
Sample output:-
4
1
2
1
Explanation:-
TestCase1:-
6 can be represented as (1, 2, 3), (1, 5), (2, 4), (6), I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
long cnt=0;
int j;
void sum(int X,int j) {
if(X==0){cnt++;}
if(X<0)
{return;}
else {
for(int i=j;i<=(X);i++){
X=X-i;
sum(X,i+1);
X=X+i;
}
}
}
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
if(n<1){cout<<0<<endl;continue;}
sum(n,1);
cout<<cnt<<endl;
cnt=0;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find number of ways an integer N can be represented as a sum of unique natural numbers.First line contain an integer T denoting number of test cases. Each test case contains a single integer N.
Constraint:-
1 <= T <= 100
0 <= N <= 120Print a single integer containing number of ways.Sample input
4
6
1
4
2
Sample output:-
4
1
2
1
Explanation:-
TestCase1:-
6 can be represented as (1, 2, 3), (1, 5), (2, 4), (6), I have written this Solution Code: // ignore number of testcases
// n is the number as provided in input
function numberOfWays(n) {
// write code here
// do not console.log the answer
// return answer as a number
if(n < 1) return 0;
let cnt = 0;
let j;
function sum(X, j) {
if (X == 0) { cnt++; }
if (X < 0) { return; }
else {
for (let i = j; i <= X; i++) {
X = X - i;
sum(X, i + 1);
X = X + i;
}
}
}
sum(n,1);
return cnt
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find number of ways an integer N can be represented as a sum of unique natural numbers.First line contain an integer T denoting number of test cases. Each test case contains a single integer N.
Constraint:-
1 <= T <= 100
0 <= N <= 120Print a single integer containing number of ways.Sample input
4
6
1
4
2
Sample output:-
4
1
2
1
Explanation:-
TestCase1:-
6 can be represented as (1, 2, 3), (1, 5), (2, 4), (6), I have written this Solution Code: for _ in range(int(input())):
x = int(input())
n = 1
dp = [1] + [0] * x
for i in range(1, x + 1):
u = i ** n
for j in range(x, u - 1, -1):
dp[j] += dp[j - u]
if x==0:
print(0)
else:
print(dp[-1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, find the first non-repeating character in the string and return its index. If it does not exist, return -1.First line of the input contains the string s.
<b>Constraints</b>
1<= s. length <= 100000Print the index of the first non- repeating character in a stringInput
s = "newtonschool"
Output
1
Explanation
"e" is the first non- repeating character in a string, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine();
System.out.print(nonRepeatChar(s));
}
static int nonRepeatChar(String s){
char count[] = new char[256];
for(int i=0; i< s.length(); i++){
count[s.charAt(i)]++;
}
for (int i=0; i<s.length(); i++) {
if (count[s.charAt(i)]==1){
return i;
}
}
return -1;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, find the first non-repeating character in the string and return its index. If it does not exist, return -1.First line of the input contains the string s.
<b>Constraints</b>
1<= s. length <= 100000Print the index of the first non- repeating character in a stringInput
s = "newtonschool"
Output
1
Explanation
"e" is the first non- repeating character in a string, I have written this Solution Code: from collections import defaultdict
s=input()
d=defaultdict(int)
for i in s:
d[i]+=1
ans=-1
for i in range(len(s)):
if(d[s[i]]==1):
ans=i
break
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, find the first non-repeating character in the string and return its index. If it does not exist, return -1.First line of the input contains the string s.
<b>Constraints</b>
1<= s. length <= 100000Print the index of the first non- repeating character in a stringInput
s = "newtonschool"
Output
1
Explanation
"e" is the first non- repeating character in a string, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-01-10 12:51:16
**/
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
int firstUniqChar(string s) {
map<char, int> charCount;
int len = s.length();
for (int i = 0; i < len; i++) {
charCount[s[i]]++;
}
for (int i = 0; i < len; i++) {
if (charCount[s[i]] == 1)
return i;
}
return -1;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
string str;
cin>>str;
cout<<firstUniqChar(str)<<"\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let's define P[i] as the ith Prime Number. Therefore, P[1]=2, P[2]=3, P[3]=5, so on.
Given two integers L, R (L<=R), find the value of P[L]+P[L+1]+P[L+2]...+P[R].The first line of the input contains an integer T denoting the number of test cases.
The next T lines contain two integers L and R.
Constraints
1 <= T <= 50000
1 <= L <= R <= 50000For each test case, print one line corresponding to the required valueSample Input
4
1 3
2 4
5 5
1 5
Sample Output
10
15
11
28, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t= Integer.parseInt(br.readLine());
int max = 1000001;
boolean isNotPrime[] = new boolean[max];
ArrayList<Integer> arr = new ArrayList<Integer>();
isNotPrime[0] = true; isNotPrime[1] = true;
for (int i=2; i*i <max; i++) {
if (!isNotPrime[i]) {
for (int j=i*i; j<max; j+= i) {
isNotPrime[j] = true;
}
}
}
for(int i=2; i<max; i++) {
if(!isNotPrime[i]) {
arr.add(i);
}
}
while(t-- > 0) {
String str[] = br.readLine().trim().split(" ");
int l = Integer.parseInt(str[0]);
int r = Integer.parseInt(str[1]);
System.out.println(primeRangeSum(l,r,arr));
}
}
static long primeRangeSum(int l , int r, ArrayList<Integer> arr) {
long sum = 0;
for(int i=l; i<=r;i++) {
sum += arr.get(i-1);
}
return sum;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let's define P[i] as the ith Prime Number. Therefore, P[1]=2, P[2]=3, P[3]=5, so on.
Given two integers L, R (L<=R), find the value of P[L]+P[L+1]+P[L+2]...+P[R].The first line of the input contains an integer T denoting the number of test cases.
The next T lines contain two integers L and R.
Constraints
1 <= T <= 50000
1 <= L <= R <= 50000For each test case, print one line corresponding to the required valueSample Input
4
1 3
2 4
5 5
1 5
Sample Output
10
15
11
28, I have written this Solution Code: def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
pri = []
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
for p in range(2, n+1):
if prime[p]:
pri.append(p)
return pri
N = int(input())
X = []
prim = SieveOfEratosthenes(1000000)
for i in range(1,len(prim)):
prim[i] = prim[i]+prim[i-1]
for i in range(N):
nnn = input()
X.append((int(nnn.split()[0]),int(nnn.split()[1])))
for xx,yy in X:
if xx==1:
print(prim[yy-1])
else:
print(prim[yy-1]-prim[xx-2])
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let's define P[i] as the ith Prime Number. Therefore, P[1]=2, P[2]=3, P[3]=5, so on.
Given two integers L, R (L<=R), find the value of P[L]+P[L+1]+P[L+2]...+P[R].The first line of the input contains an integer T denoting the number of test cases.
The next T lines contain two integers L and R.
Constraints
1 <= T <= 50000
1 <= L <= R <= 50000For each test case, print one line corresponding to the required valueSample Input
4
1 3
2 4
5 5
1 5
Sample Output
10
15
11
28, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
vector<int> v;
v.push_back(0);
for(int i = 2; i < N; i++){
if(a[i]) continue;
v.push_back(i);
for(int j = i*i; j < N; j += i)
a[j] = 1;
}
int p = 0;
for(auto &i: v){
i += p;
p = i;
}
int t; cin >> t;
while(t--){
int l, r;
cin >> l >> r;
cout << v[r] - v[l-1] << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string is called square if it is some string written twice in a row. For example, the strings "aa", "abcabc", "abab" and "baabaa" are square. But the strings "aaa", "abaaab" and "abcdabc" are not square. For a given string s determine if it is square.The input consists of a string S consisting of lowercase English alphabets.
<b>Constraints</b>
The length of the string is between 1 to 100.Print Yes if the string in the corresponding test case is square, No otherwise.<b>Sample Input 1</b>
aaa
<b>Sample Output 1</b>
No
<b>Sample Input 2</b>
xyxy
<b>Sample Output 2</b>
Yes, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t=1;
// cin >> t;
for (int i = 0; i < t; i++){
string s;
cin >> s;
int N = s.size();
if (N % 2 == 1){
cout << "No" << endl;
} else {
if (s.substr(0, N / 2) == s.substr(N / 2)){
cout << "Yes" << endl;
} else {
cout << "No" << endl;
}
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a reference to the head of a sorted doubly-linked list and an integer k, your task is to insert the integer k in your doubly linked-list while maintaining the sort.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>insertnew()</b> that takes head node of the linked list and the integer k as parameter.
Constraints:
1 <= T <= 100
1 <= Node.data <= 1000Return the head of the modified linked list.Sample Input
1
4
1
3
4
10
5
Sample Output
1 3 4 5 10, I have written this Solution Code: public static Node insertnew(Node head,int k) {
if(head == null){
Node temp = new Node(k);
return temp;
}
Node temp=head;
while(temp != null){
if (temp.val >= k){
Node x=new Node(k);
x.prev = temp.prev;
x.next = temp;
temp.prev = x;
if (x.prev == null){
return x;
}
else {
x.prev.next =x;
return head;
}
}
if (temp.next == null){
Node x=new Node(k);
x.prev = temp;
x.next = null;
temp.next = x;
break;
}
temp = temp.next;
}
return head;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N strings, find the longest common prefix among all strings present in the array.The fist line of each test contains a single integer N. Next line has space-separated N strings.
Constraints:-
1 <= N <= 10^3
1 <= |S| <= 10^3
Print the longest common prefix as a string in the given array. If no such prefix exists print "-1"(without quotes).Sample Input:
4
geeksforgeeks geeks geek geezer
Sample Output:
gee
Sampel Input:-
3
a b c
Sample Output:-
-1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
String[] str = br.readLine().split(" ");
String ref = str[0];
for(String s: str){
if(s.length()<ref.length()){
ref = s;
}
}
for(int i=0; i<n; i++){
if(str[i].contains(ref) == false){
ref = ref.substring(0, ref.length()-1);
}
}
if(ref.length()>0)
System.out.println(ref);
else
System.out.println(-1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N strings, find the longest common prefix among all strings present in the array.The fist line of each test contains a single integer N. Next line has space-separated N strings.
Constraints:-
1 <= N <= 10^3
1 <= |S| <= 10^3
Print the longest common prefix as a string in the given array. If no such prefix exists print "-1"(without quotes).Sample Input:
4
geeksforgeeks geeks geek geezer
Sample Output:
gee
Sampel Input:-
3
a b c
Sample Output:-
-1, I have written this Solution Code: def longestPrefix( strings ):
size = len(strings)
if (size == 0):
return -1
if (size == 1):
return strings[0]
strings.sort()
end = min(len(strings[0]), len(strings[size - 1]))
i = 0
while (i < end and
strings[0][i] == strings[size - 1][i]):
i += 1
pre = strings[0][0: i]
if len(pre) > 1:
return pre
else:
return -1
N=int(input())
strings=input().split()
print( longestPrefix(strings) ), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N strings, find the longest common prefix among all strings present in the array.The fist line of each test contains a single integer N. Next line has space-separated N strings.
Constraints:-
1 <= N <= 10^3
1 <= |S| <= 10^3
Print the longest common prefix as a string in the given array. If no such prefix exists print "-1"(without quotes).Sample Input:
4
geeksforgeeks geeks geek geezer
Sample Output:
gee
Sampel Input:-
3
a b c
Sample Output:-
-1, I have written this Solution Code: function commonPrefixUtil(str1,str2)
{
let result = "";
let n1 = str1.length, n2 = str2.length;
// Compare str1 and str2
for (let i = 0, j = 0; i <= n1 - 1 && j <= n2 - 1; i++, j++) {
if (str1[i] != str2[j]) {
break;
}
result += str1[i];
}
return (result);
}
// n is number of individual space seperated strings inside strings variable,
// strings is the string which contains space seperated words.
function longestCommonPrefix(strings,n){
// write code here
// do not console,log answer
// return the answer as string
if(n===1) return strings;
const arr= strings.split(" ")
let prefix = arr[0];
for (let i = 1; i <= n - 1; i++) {
prefix = commonPrefixUtil(prefix, arr[i]);
}
if(!prefix) return -1;
return (prefix);
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N strings, find the longest common prefix among all strings present in the array.The fist line of each test contains a single integer N. Next line has space-separated N strings.
Constraints:-
1 <= N <= 10^3
1 <= |S| <= 10^3
Print the longest common prefix as a string in the given array. If no such prefix exists print "-1"(without quotes).Sample Input:
4
geeksforgeeks geeks geek geezer
Sample Output:
gee
Sampel Input:-
3
a b c
Sample Output:-
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
string a[n];
int x=INT_MAX;
for(int i=0;i<n;i++){
cin>>a[i];
x=min(x,(int)a[i].length());
}
string ans="";
for(int i=0;i<x;i++){
for(int j=1;j<n;j++){
if(a[j][i]==a[0][i]){continue;}
goto f;
}
ans+=a[0][i];
}
f:;
if(ans==""){cout<<-1;return 0;}
cout<<(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function
<b>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code:
char Race(int A, int B, int C){
if(abs(C-A)==abs(C-B)){return 'D';}
if(abs(C-A)>abs(C-B)){return 'S';}
else{
return 'N';}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function
<b>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code: def Race(A,B,C):
if abs(C-A) ==abs(C-B):
return 'D'
if abs(C-A)>abs(C-B):
return 'S'
return 'N'
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function
<b>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code:
char Race(int A, int B, int C){
if(abs(C-A)==abs(C-B)){return 'D';}
if(abs(C-A)>abs(C-B)){return 'S';}
else{
return 'N';}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function
<b>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code: static char Race(int A,int B,int C){
if(Math.abs(C-A)==Math.abs(C-B)){return 'D';}
if(Math.abs(C-A)>Math.abs(C-B)){return 'S';}
else{
return 'N';}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
Sample Output:-
NO, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
long n = sc.nextLong();
int p=(int)Math.sqrt(n);
for(int i=2;i<=p;i++){
if(n%i==0){System.out.print("NO");return;}
}
System.out.print("YES");
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
Sample Output:-
NO, I have written this Solution Code: import math
def isprime(A):
if A == 1:
return False
sqrt = int(math.sqrt(A))
for i in range(2,sqrt+1):
if A%i == 0:
return False
return True
inp = int(input())
if isprime(inp):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
Sample Output:-
NO, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
long n;
cin>>n;
long x = sqrt(n);
for(int i=2;i<=x;i++){
if(n%i==0){cout<<"NO";return 0;}
}
cout<<"YES";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Remove duplicates of an array and return an array of only unique elements.An array containing numbers.Space separated unique elements from the array.Sample Input:-
1 2 3 5 1 5 9 1 2 8
Sample Output:-
1 2 3 5 9 8
<b>Explanation:-</b>
Extra 1, 2, and 5 were removed since they were occurring multiple times.
Note: You only have to remove the extra occurrences i.e. each element in the final array should have a frequency equal to one., I have written this Solution Code: nan, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Remove duplicates of an array and return an array of only unique elements.An array containing numbers.Space separated unique elements from the array.Sample Input:-
1 2 3 5 1 5 9 1 2 8
Sample Output:-
1 2 3 5 9 8
<b>Explanation:-</b>
Extra 1, 2, and 5 were removed since they were occurring multiple times.
Note: You only have to remove the extra occurrences i.e. each element in the final array should have a frequency equal to one., I have written this Solution Code:
inp = eval(input(""))
new_set = []
for i in inp:
if(str(i) not in new_set):
new_set.append(str(i))
print(" ".join(new_set)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary tree of N nodes, with root 1 and Q queries with nodes u and v. For each query find the sum of nodes on the shortest path from u to v, where node v is the ancestor of node u.First line contains the integer N and Q, denoting the number of nodes in the binary tree and the number of queries respectively.
Next N lines contains two integers denoting the left and right child of the i'th node respectively.
If the node doesn't have a left or right child, it is denoted by '-1'
Next Q lines contain two integers u and v
1 <= N <= 10000
1 <= Q <= 100000
1 <= u, v <= NPrint Q lines denoting the sum of nodes on the shortest path from u to vSample Input 1:
6 3
2 4
5 3
-1 -1
-1 -1
6 -1
-1 -1
6 2
3 1
5 5
Sample output 1:
13
6
5
Explanation: Given binary tree
1
/ \
2 4
/ \
5 3
/
6
Query 1: 6+5+2 = 13
Query 2: 3+2+1 = 6
Query 3: 5 , I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc= new Scanner(System.in);
int n=sc.nextInt();
int q=sc.nextInt();
int arr[]=new int[n+1];
arr[1]=1;
for(int i=1;i<=n;i++){
int l=sc.nextInt();
int r=sc.nextInt();
if(l!=-1){
arr[l]=l+arr[i];
}
if(r!=-1){
arr[r]=r+arr[i];
}
}
for(int i=0;i<q;i++){
int u=sc.nextInt();
int v=sc.nextInt();
System.out.println(arr[u]-arr[v]+v);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary tree of N nodes, with root 1 and Q queries with nodes u and v. For each query find the sum of nodes on the shortest path from u to v, where node v is the ancestor of node u.First line contains the integer N and Q, denoting the number of nodes in the binary tree and the number of queries respectively.
Next N lines contains two integers denoting the left and right child of the i'th node respectively.
If the node doesn't have a left or right child, it is denoted by '-1'
Next Q lines contain two integers u and v
1 <= N <= 10000
1 <= Q <= 100000
1 <= u, v <= NPrint Q lines denoting the sum of nodes on the shortest path from u to vSample Input 1:
6 3
2 4
5 3
-1 -1
-1 -1
6 -1
-1 -1
6 2
3 1
5 5
Sample output 1:
13
6
5
Explanation: Given binary tree
1
/ \
2 4
/ \
5 3
/
6
Query 1: 6+5+2 = 13
Query 2: 3+2+1 = 6
Query 3: 5 , I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define ll long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int l[N], r[N], p[N], s[N];
void dfs(int u, int p = 0){
if(u == -1) return;
s[u] = s[p] + u;
dfs(l[u], u);
dfs(r[u], u);
}
signed main() {
IOS;
clock_t start = clock();
int n, q; cin >> n >> q;
for(int i = 1; i <= n; i++){
cin >> l[i] >> r[i];
p[l[i]] = p[r[i]] = i;
}
dfs(1);
while(q--){
int u, v;
cin >> u >> v;
cout << s[u] - s[p[v]] << endl;
}
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary matrix, your task is to calculate the maximum area of a rectangle formed by only 1's in it.First line contains two integers number of rows N and number of columns M.
Next N lines contain M integers with each integer being either 0 or 1
Constraints:-
1 <= N, M <=1000Print the maximum area of rectangle.Sample Input 1:
4 4
0 1 1 0
1 1 1 1
1 1 1 1
1 1 0 0
Sample Output 1:
8
Explanation
The max size rectangle is
1 1 1 1
1 1 1 1
and its area is 4*2 = 8
Sample Input 2:-
1 4
1 1 1 1
Sample Output 2:-
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main { static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
public static void main (String[] args) {
FastReader sc = new FastReader();
int n = sc.nextInt();int m = sc.nextInt();
char a[][] = new char[n][m];
for(int i=0;i<n;i++){for(int j=0;j<m;j++){
a[i][j] = sc.next().charAt(0);
}}
System.out.println(findRectangleArea2Ds(a));
}
private static int findRectangleArea2Ds(char[][] matrix) {
if (matrix == null || matrix.length == 0) {
return 0;
}
int maxArea = 0;
int rows = matrix.length;
int cols = matrix[0].length;
int[][] dp = new int[rows][cols];
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (matrix[i][j] == '1') {
dp[i][j] = j == 0 ? 1 : dp[i][j - 1] + 1;
int length = dp[i][j];
for (int k = i; k >= 0; k--) {
length = Math.min(length, dp[k][j]);
int width = i - k + 1;
maxArea = Math.max(maxArea, length * width);
}
}
}
}
return maxArea;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary matrix, your task is to calculate the maximum area of a rectangle formed by only 1's in it.First line contains two integers number of rows N and number of columns M.
Next N lines contain M integers with each integer being either 0 or 1
Constraints:-
1 <= N, M <=1000Print the maximum area of rectangle.Sample Input 1:
4 4
0 1 1 0
1 1 1 1
1 1 1 1
1 1 0 0
Sample Output 1:
8
Explanation
The max size rectangle is
1 1 1 1
1 1 1 1
and its area is 4*2 = 8
Sample Input 2:-
1 4
1 1 1 1
Sample Output 2:-
4, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e3 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int p[N][N], l[N], r[N];
signed main() {
IOS;
int n, m;
cin >> n >> m;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
cin >> p[i][j];
if(p[i][j])
p[i][j] += p[i-1][j];
}
}
int ans = 0;
for(int i = 1; i <= n; i++){
memset(l, 0, sizeof l);
memset(r, 0, sizeof r);
stack<int> s;
p[i][0] = p[i][m+1] = -1;
s.push(0);
for(int j = 1; j <= m; j++){
while(!s.empty() && p[i][s.top()] >= p[i][j])
s.pop();
l[j] = s.top();
s.push(j);
}
while(!s.empty())
s.pop();
s.push(m+1);
for(int j = m; j >= 1; j--){
while(!s.empty() && p[i][s.top()] >= p[i][j])
s.pop();
r[j] = s.top();
s.push(j);
}
for(int j = 1; j <= m; j++)
ans = max(ans, p[i][j]*(r[j]-l[j]-1));
}
cout << ans;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary sorted non-increasing array arr of size <b>N</b>. You need to print the count of <b>1's</b> in the binary array.
Try to solve the problem using binary searchThe input line contains T, denotes the number of testcases.
Each test case contains two lines. The first line contains N (size of binary array). The second line contains N elements of binary array separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^6
arr[i] = 0,1
<b>Sum of N over all testcases does not exceed 10^6</b>For each testcase in new line, print the count 1's in binary array.Input:
2
8
1 1 1 1 1 0 0 0
8
1 1 0 0 0 0 0 0
Output:
5
2
Explanation:
Testcase 1: Number of 1's in given binary array : 1 1 1 1 1 0 0 0 is 5.
Testcase 2: Number of 1's in given binary array : 1 1 0 0 0 0 0 0 is 2., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main (String[] args) throws IOException {
Reader sc=new Reader();
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
int[] a=new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int count=search(a,0,n-1);
System.out.println(count);
}
}
public static int search(int[] a,int l,int h){
while(l<=h){
int mid=l+(h-l)/2;
if ((mid==h||a[mid+1]==0)&&(a[mid]==1))
return mid+1;
if (a[mid]==1)
l=mid+1;
else h=mid-1;
}
return 0;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary sorted non-increasing array arr of size <b>N</b>. You need to print the count of <b>1's</b> in the binary array.
Try to solve the problem using binary searchThe input line contains T, denotes the number of testcases.
Each test case contains two lines. The first line contains N (size of binary array). The second line contains N elements of binary array separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^6
arr[i] = 0,1
<b>Sum of N over all testcases does not exceed 10^6</b>For each testcase in new line, print the count 1's in binary array.Input:
2
8
1 1 1 1 1 0 0 0
8
1 1 0 0 0 0 0 0
Output:
5
2
Explanation:
Testcase 1: Number of 1's in given binary array : 1 1 1 1 1 0 0 0 is 5.
Testcase 2: Number of 1's in given binary array : 1 1 0 0 0 0 0 0 is 2., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
int l = 0, h = n+1;
while(l+1 < h){
int m = (l + h) >> 1;
if(a[m] == 1)
l = m;
else
h = m;
}
cout << l << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary sorted non-increasing array arr of size <b>N</b>. You need to print the count of <b>1's</b> in the binary array.
Try to solve the problem using binary searchThe input line contains T, denotes the number of testcases.
Each test case contains two lines. The first line contains N (size of binary array). The second line contains N elements of binary array separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^6
arr[i] = 0,1
<b>Sum of N over all testcases does not exceed 10^6</b>For each testcase in new line, print the count 1's in binary array.Input:
2
8
1 1 1 1 1 0 0 0
8
1 1 0 0 0 0 0 0
Output:
5
2
Explanation:
Testcase 1: Number of 1's in given binary array : 1 1 1 1 1 0 0 0 is 5.
Testcase 2: Number of 1's in given binary array : 1 1 0 0 0 0 0 0 is 2., I have written this Solution Code: c=int(input())
for x in range(c):
size=int(input())
s=input()
print(s.count('1')), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, your task is to find if it is a perfect square or not.The first line of input contains the number of test cases T, the next T lines contains a single integer N.
<b>Constraints:</b>
1 < = T < = 1000
1 < = N < = 10<sup>12</sup>For each test case, print "YES" if the number N is a perfect square else print "NO".Sample Input:-
2
625
624
Sample Output:-
YES
NO, I have written this Solution Code: import math
cases=int(input(""))
for case in range(cases):
n=int(input())
if math.sqrt(n) % 1 == 0:print("YES")
else:print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, your task is to find if it is a perfect square or not.The first line of input contains the number of test cases T, the next T lines contains a single integer N.
<b>Constraints:</b>
1 < = T < = 1000
1 < = N < = 10<sup>12</sup>For each test case, print "YES" if the number N is a perfect square else print "NO".Sample Input:-
2
625
624
Sample Output:-
YES
NO, I have written this Solution Code: /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
long n = sc.nextLong();
if(check(n)==1){
System.out.println("YES");
}
else{
System.out.println("NO");
}
}
}
static int check(long n){
long l=1;
long h = 10000000;
long m=0;
long p=0;
long ans=0;
while(l<=h){
m=l+h;
m/=2;
p=m*m;
if(p > n){
h=m-1;
}
else{
l=m+1;
ans=m;
}
//System.out.println(l+" "+h);
}
if(ans*ans==n){return 1;}
return 0;
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, your task is to find if it is a perfect square or not.The first line of input contains the number of test cases T, the next T lines contains a single integer N.
<b>Constraints:</b>
1 < = T < = 1000
1 < = N < = 10<sup>12</sup>For each test case, print "YES" if the number N is a perfect square else print "NO".Sample Input:-
2
625
624
Sample Output:-
YES
NO, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
long long x=sqrt(n);
long long p = x*x;
if(p==n){cout<<"YES"<<endl;;}
else{
cout<<"NO"<<endl;
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in a queue in which A of them like apple and B of them like oranges. How many people like both apple and oranges.
<b>Note</b>:- It is guaranteed that each and every person likes at least one of the given two.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LikesBoth()</b> that takes integers N, A, and B as arguments.
Constraints:-
1 <= N <= 10000
1 <= A <= N
1 <= B <= NReturn the number of people that like both of the fruit.Sample Input:-
5 3 4
Sample Output:-
2
Sample Input:-
5 5 5
Sample Output:-
5, I have written this Solution Code: static int LikesBoth(int N, int A, int B){
return (A+B-N);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in a queue in which A of them like apple and B of them like oranges. How many people like both apple and oranges.
<b>Note</b>:- It is guaranteed that each and every person likes at least one of the given two.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LikesBoth()</b> that takes integers N, A, and B as arguments.
Constraints:-
1 <= N <= 10000
1 <= A <= N
1 <= B <= NReturn the number of people that like both of the fruit.Sample Input:-
5 3 4
Sample Output:-
2
Sample Input:-
5 5 5
Sample Output:-
5, I have written this Solution Code: def LikesBoth(N,A,B):
return (A+B-N)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in a queue in which A of them like apple and B of them like oranges. How many people like both apple and oranges.
<b>Note</b>:- It is guaranteed that each and every person likes at least one of the given two.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LikesBoth()</b> that takes integers N, A, and B as arguments.
Constraints:-
1 <= N <= 10000
1 <= A <= N
1 <= B <= NReturn the number of people that like both of the fruit.Sample Input:-
5 3 4
Sample Output:-
2
Sample Input:-
5 5 5
Sample Output:-
5, I have written this Solution Code: int LikesBoth(int N,int A, int B){
return (A+B-N);
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in a queue in which A of them like apple and B of them like oranges. How many people like both apple and oranges.
<b>Note</b>:- It is guaranteed that each and every person likes at least one of the given two.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LikesBoth()</b> that takes integers N, A, and B as arguments.
Constraints:-
1 <= N <= 10000
1 <= A <= N
1 <= B <= NReturn the number of people that like both of the fruit.Sample Input:-
5 3 4
Sample Output:-
2
Sample Input:-
5 5 5
Sample Output:-
5, I have written this Solution Code: int LikesBoth(int N,int A, int B){
return (A+B-N);
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements. In the array, each element is present twice except for 1 element whose frequency in the array is 1. Hence the length of the array will always be odd.
Find the unique number.An integer, N, representing the size of the array. In the next line, N space-separated integers follow.
<b>Constraints:</b>
1 <= N <=10<sup>5</sup>
1 <= A[i] <=10<sup>8</sup>Output the element with frequency 1.Input :
5
1 1 2 2 3
Output:
3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int noofterm=Integer.parseInt(br.readLine());
int arr[] = new int[noofterm];
String s[] = br.readLine().split(" ");
for(int i=0; i<noofterm;i++){
arr[i]= Integer.parseInt(s[i]);
}
System.out.println(unique(arr));
}
public static int unique(int[] inputArray)
{
int result = 0;
for(int i=0;i<inputArray.length;i++)
{
result ^= inputArray[i];
}
return (result>0 ? result : -1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements. In the array, each element is present twice except for 1 element whose frequency in the array is 1. Hence the length of the array will always be odd.
Find the unique number.An integer, N, representing the size of the array. In the next line, N space-separated integers follow.
<b>Constraints:</b>
1 <= N <=10<sup>5</sup>
1 <= A[i] <=10<sup>8</sup>Output the element with frequency 1.Input :
5
1 1 2 2 3
Output:
3, I have written this Solution Code: n = int(input())
a = [int (x) for x in input().split()]
mapp={}
for index,val in enumerate(a):
if val in mapp:
del mapp[val]
else:
mapp[val]=1
for key, value in mapp.items():
print(key), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements. In the array, each element is present twice except for 1 element whose frequency in the array is 1. Hence the length of the array will always be odd.
Find the unique number.An integer, N, representing the size of the array. In the next line, N space-separated integers follow.
<b>Constraints:</b>
1 <= N <=10<sup>5</sup>
1 <= A[i] <=10<sup>8</sup>Output the element with frequency 1.Input :
5
1 1 2 2 3
Output:
3, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
signed main()
{
int n;
cin>>n;
int p=0;
for(int i=0;i<n;i++)
{
int a;
cin>>a;
p^=a;
}
cout<<p<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Remove duplicates of an array and return an array of only unique elements.An array containing numbers.Space separated unique elements from the array.Sample Input:-
1 2 3 5 1 5 9 1 2 8
Sample Output:-
1 2 3 5 9 8
<b>Explanation:-</b>
Extra 1, 2, and 5 were removed since they were occurring multiple times.
Note: You only have to remove the extra occurrences i.e. each element in the final array should have a frequency equal to one., I have written this Solution Code: nan, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Remove duplicates of an array and return an array of only unique elements.An array containing numbers.Space separated unique elements from the array.Sample Input:-
1 2 3 5 1 5 9 1 2 8
Sample Output:-
1 2 3 5 9 8
<b>Explanation:-</b>
Extra 1, 2, and 5 were removed since they were occurring multiple times.
Note: You only have to remove the extra occurrences i.e. each element in the final array should have a frequency equal to one., I have written this Solution Code:
inp = eval(input(""))
new_set = []
for i in inp:
if(str(i) not in new_set):
new_set.append(str(i))
print(" ".join(new_set)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a positive integer A. You have to apply the following operation on it N times:
<ul><li>Let the maximum and minimum digits in the decimal representation of A be <i>mx</i> and <i>mn</i> respectively. </li><li>Add mx*mn to A, i. e, convert A to A + mx*mn. </li></ul>Find the value of A after applying this operation N times.First and the only line of the input contains two integers A and N.
Constraints:
1 <= A <= 10<sup>18</sup>
1 <= N <= 10<sup>16</sup>Print the value of A after applying the above operation N times.Sample Input
8 3
Sample Output
134
Explaination:
During the first operation, mx = 8, mn = 8.
A = 8 + (8*8) = 8 + 64 = 72
During the second operation, mx = 7, mn = 2
A = 72 + (7*2) = 72 + 14 = 86
During the final operation, mx = 8, mn = 6
A = 86 + (8*6) = 86 + 48 = 134, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define int long long
#define pb push_back
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
using T = pair<int, int>;
typedef long double ld;
const int mod = 1e9 + 7;
const int INF = 2e9;
int chk(int n){
int mi = 100, ma = -1;
while(n){
mi = min(mi, n%10);
ma = max(ma, n%10);
n /= 10;
}
return ma*mi;
}
void solve(){
int x, y, f=0;
cin >> x >> y;
while(y--){
int a = chk(x);
x += a;
if(a == 0){
break;
}
}
cout << x;
}
signed main(){
fast
int t = 1;
// cin >> t;
for(int i = 1; i <= t; i++){
solve();
if(i != t) cout << endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is <b>Palindrome</b> or not. A number is said to be Palindrome when it reads the same from backward as forward.User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isPalindrome()</b> which contains N as a parameter.
<b>Constraints:</b>
1 <= N <= 9999You need to return "true" is the number is palindrome otherwise "false".Sample Input:
5
Sample Output:
true
Sample Input:
121
Sample Output:
true, I have written this Solution Code: static boolean isPalindrome(int N)
{
int sum = 0;
int rev = N;
while(N > 0)
{
int digit = N%10;
sum = sum*10+digit;
N = N/10;
}
if(rev == sum)
return true;
else return false;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is <b>Palindrome</b> or not. A number is said to be Palindrome when it reads the same from backward as forward.User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isPalindrome()</b> which contains N as a parameter.
<b>Constraints:</b>
1 <= N <= 9999You need to return "true" is the number is palindrome otherwise "false".Sample Input:
5
Sample Output:
true
Sample Input:
121
Sample Output:
true, I have written this Solution Code: def isPalindrome(N):
sum1 = 0
rev = N
while(N > 0):
digit = N%10
sum1 = sum1*10+digit
N = N//10
if(rev == sum1):
return True
return False, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A child is running up a staircase with n steps and can hop either 1 step, 2 steps or 3 steps at a time. Implement a method to count how many possible ways the child can run-up to the stairs. You need to return all possible number of ways.
The time complexity of your code should be O(n).
Print answer modulo 10^9 + 7.The input line contains T, denoting the number of testcases. Each testcase contains single line containing N, i.e. stairs having number of steps.
Constraints
1 <= T <= 10
1 <= N <= 10^5Find the total number of ways. Since the answer can be very large, thus print them as modulo 10^9 + 7.Sample Input
2
4
3
Sample output
7
4
Explanation:
Testcase 1:
In this case we have stairs with 4 steps. So we can figure out how many ways are there to reach at the top.
1st: (1, 1, 1, 1)
2nd: (1, 1, 2)
3rd: (1, 2, 1)
4th: (1, 3)
5th: (2, 1, 1)
6th: (2, 2)
7th: (3, 1), I have written this Solution Code:
dp = [1,2,4]
n = int(input())
for i in range(n):
k=int(input())
one = 1
two = 2
three = 4
four = 0
if k<4:
print(dp[k-1])
continue
for i in range(3,k):
four = one + two + three
one = two
two = three
three = four
print(four%(10**9+7)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A child is running up a staircase with n steps and can hop either 1 step, 2 steps or 3 steps at a time. Implement a method to count how many possible ways the child can run-up to the stairs. You need to return all possible number of ways.
The time complexity of your code should be O(n).
Print answer modulo 10^9 + 7.The input line contains T, denoting the number of testcases. Each testcase contains single line containing N, i.e. stairs having number of steps.
Constraints
1 <= T <= 10
1 <= N <= 10^5Find the total number of ways. Since the answer can be very large, thus print them as modulo 10^9 + 7.Sample Input
2
4
3
Sample output
7
4
Explanation:
Testcase 1:
In this case we have stairs with 4 steps. So we can figure out how many ways are there to reach at the top.
1st: (1, 1, 1, 1)
2nd: (1, 1, 2)
3rd: (1, 2, 1)
4th: (1, 3)
5th: (2, 1, 1)
6th: (2, 2)
7th: (3, 1), I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static long waysBottomUp(int n, long dp[]){
for(int i=4; i<=n; i++){
dp[i]= ((dp[i-1]%1000000007) + (dp[i-2]%1000000007) + (dp[i-3]%1000000007))%1000000007;
}
return dp[n];
}
public static void main (String[] args)throws IOException {
BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
long dp[] = new long[100001];
int t = Integer.parseInt(rd.readLine());
while(t-->0){
int n = Integer.parseInt(rd.readLine());
dp[0] =1;
dp[1] = 1;
dp[2] = 2;
dp[3] = 4;
System.out.println(waysBottomUp(n, dp)%1000000007);}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A child is running up a staircase with n steps and can hop either 1 step, 2 steps or 3 steps at a time. Implement a method to count how many possible ways the child can run-up to the stairs. You need to return all possible number of ways.
The time complexity of your code should be O(n).
Print answer modulo 10^9 + 7.The input line contains T, denoting the number of testcases. Each testcase contains single line containing N, i.e. stairs having number of steps.
Constraints
1 <= T <= 10
1 <= N <= 10^5Find the total number of ways. Since the answer can be very large, thus print them as modulo 10^9 + 7.Sample Input
2
4
3
Sample output
7
4
Explanation:
Testcase 1:
In this case we have stairs with 4 steps. So we can figure out how many ways are there to reach at the top.
1st: (1, 1, 1, 1)
2nd: (1, 1, 2)
3rd: (1, 2, 1)
4th: (1, 3)
5th: (2, 1, 1)
6th: (2, 2)
7th: (3, 1), I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
a[0] = 1;
for(int i = 1; i < N; i++){
if(i >= 1) a[i] += a[i-1];
if(i >= 2) a[i] += a[i-2];
if(i >= 3) a[i] += a[i-3];
a[i] %= mod;
}
while(t--){
int n; cin >> n;
cout << a[n] << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Take input from standard input while you do not get 0 as an input. Print all the inputs separated by space. It is guaranteed that the number of integers are less than 100000.The input will contain a series of integers in one line each. Input should be taken while you have not get a 0 as an input.
0 <= input <= 10Print the input integers seperated by space.Sample Input
6
5
5
0
Sample Output
6 5 5 0
Sample Input
9
3
5
7
6
9
8
3
2
7
7
3
5
0
Sample Output
9 3 5 7 6 9 8 3 2 7 7 3 5 0, I have written this Solution Code: n=1
index=0
li=[]
while n!=0:
n=int(input())
li.append(n)
index=index+1
#li = list(map(int,input().strip().split()))
for i in range(0,len(li)-1):
print(li[i],end=" ")
print(li[len(li)-1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Take input from standard input while you do not get 0 as an input. Print all the inputs separated by space. It is guaranteed that the number of integers are less than 100000.The input will contain a series of integers in one line each. Input should be taken while you have not get a 0 as an input.
0 <= input <= 10Print the input integers seperated by space.Sample Input
6
5
5
0
Sample Output
6 5 5 0
Sample Input
9
3
5
7
6
9
8
3
2
7
7
3
5
0
Sample Output
9 3 5 7 6 9 8 3 2 7 7 3 5 0, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n=100001;
int a;
for(int i=0;i<n;i++){
a=sc.nextInt();
System.out.print(a+" ");
if(a==0){break;}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Take input from standard input while you do not get 0 as an input. Print all the inputs separated by space. It is guaranteed that the number of integers are less than 100000.The input will contain a series of integers in one line each. Input should be taken while you have not get a 0 as an input.
0 <= input <= 10Print the input integers seperated by space.Sample Input
6
5
5
0
Sample Output
6 5 5 0
Sample Input
9
3
5
7
6
9
8
3
2
7
7
3
5
0
Sample Output
9 3 5 7 6 9 8 3 2 7 7 3 5 0, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int n;
while(cin >> n){
cout << n << " ";
if(n == 0) break;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
int[] arr=new int[n];
String[] s=br.readLine().trim().split(" ");
for(int i=0;i<arr.length;i++)
arr[i]=Integer.parseInt(s[i]);
int max=Integer.MIN_VALUE,c=0;
for(int i=0;i<arr.length;i++){
if(max==arr[i])
c++;
if(max<arr[i]){
max=arr[i];
c=1;
}
}
System.out.println(c);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: n=int(input())
a=list(map(int,input().split()))
print(a.count(max(a))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
vector<int> a(101, 0);
For(i, 0, n){
int x; cin>>x;
a[x]++;
}
for(int i=100; i>=1; i--){
if(a[i]){
cout<<a[i];
return;
}
}
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given resistance value of N resistors. Find the net resistance of the system when all of these N resistors are connected in parallel.
If there are three resistors A1, A2, A3, when they are connected in parallel, the net resistance will be 1/((1/A1) + (1/A2) + (1/A3))
Since this number can also have a fraction part, you only have to print the floor of the result obtained.
For example, if value of 1/((1/A1) + (1/A2) + (1/A3)) if 7.54567, you only have to print 7.First line contains a single integer N denoting the number of resistors.
Next line contains N space separated integers containing the value of different resistors.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^9Print the integral part or floor of the value obtained from the formula 1/((1/A1) + (1/A2) + ..... + (1/AN)).Input
2
10 30
Output
7
1/((1/10) + (1/30)) = 30/4 = 7.5 and floor of 7.5 is 7, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int N = Integer.parseInt(br.readLine());
double arr[] = new double[N];
String str[] = br.readLine().trim().split(" ");
for(int i=0;i<N;i++)
arr[i]=Integer.parseInt(str[i]);
double resistance=0;
int equResistance=0;
for(int i=0;i<N;i++)
arr[i]=Integer.parseInt(str[i]);
for(int i=0;i<N;i++)
{
resistance=resistance+(1/arr[i]);
}
equResistance = (int)Math.floor((1/resistance));
System.out.println(equResistance);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given resistance value of N resistors. Find the net resistance of the system when all of these N resistors are connected in parallel.
If there are three resistors A1, A2, A3, when they are connected in parallel, the net resistance will be 1/((1/A1) + (1/A2) + (1/A3))
Since this number can also have a fraction part, you only have to print the floor of the result obtained.
For example, if value of 1/((1/A1) + (1/A2) + (1/A3)) if 7.54567, you only have to print 7.First line contains a single integer N denoting the number of resistors.
Next line contains N space separated integers containing the value of different resistors.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^9Print the integral part or floor of the value obtained from the formula 1/((1/A1) + (1/A2) + ..... + (1/AN)).Input
2
10 30
Output
7
1/((1/10) + (1/30)) = 30/4 = 7.5 and floor of 7.5 is 7, I have written this Solution Code: r = input("")
r = int(r)
n = input("").split()
resistance=0.0
for i in range(0,r):
resistor = float(n[i])
resistance = resistance + (1/resistor)
print(int(1/resistance)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |