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For this Question: Doctor Strange stumbles upon a rather curious artefact while spending holidays in Egypt. An antique board game but it can only be unlocked by a magical die. The only clue on the box is to get a sum of B by throwing the die A times. Help Doctor Strange if it is possible to get the particular sum. Print “YES” if it is possible, otherwise print “NO”.
Note:- Consider the die to be a regular one with face {1, 2, 3, 4, 5, 6}.The input contains two integers A and B separated by spaces.
Constraints:-
• 1≤A≤100
• 1≤B≤1000
A and B are integers.Print "YES" if it is possible to get a Sum of B by throwing die A times else print "NO".Sample Input:-
10 100
Sample Output:-
NO
Sample Input:-
6 30
Sample Output:-
YES, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int a,b;
cin>>a>>b;
if(a*6>=b && a<=b){
cout<<"YES";
}
else{
cout<<"NO";
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Doctor Strange stumbles upon a rather curious artefact while spending holidays in Egypt. An antique board game but it can only be unlocked by a magical die. The only clue on the box is to get a sum of B by throwing the die A times. Help Doctor Strange if it is possible to get the particular sum. Print “YES” if it is possible, otherwise print “NO”.
Note:- Consider the die to be a regular one with face {1, 2, 3, 4, 5, 6}.The input contains two integers A and B separated by spaces.
Constraints:-
• 1≤A≤100
• 1≤B≤1000
A and B are integers.Print "YES" if it is possible to get a Sum of B by throwing die A times else print "NO".Sample Input:-
10 100
Sample Output:-
NO
Sample Input:-
6 30
Sample Output:-
YES, 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 A = sc.nextInt();
int B = sc.nextInt();
if (B<A || B>A*6) {
System.out.print("NO");
}
else {
System.out.print("YES");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Doctor Strange stumbles upon a rather curious artefact while spending holidays in Egypt. An antique board game but it can only be unlocked by a magical die. The only clue on the box is to get a sum of B by throwing the die A times. Help Doctor Strange if it is possible to get the particular sum. Print “YES” if it is possible, otherwise print “NO”.
Note:- Consider the die to be a regular one with face {1, 2, 3, 4, 5, 6}.The input contains two integers A and B separated by spaces.
Constraints:-
• 1≤A≤100
• 1≤B≤1000
A and B are integers.Print "YES" if it is possible to get a Sum of B by throwing die A times else print "NO".Sample Input:-
10 100
Sample Output:-
NO
Sample Input:-
6 30
Sample Output:-
YES, I have written this Solution Code: A,B=map(int, input().split())
if 1*A<=B and B<=A*6:
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 the length, breadth, and height of a cuboid. Your task is to calculate its Perimeter.
Note:- Formula for the perimeter of the cuboid is 4(Length + Breadth + height)<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>Perimeter()</b> that takes integers L, B, and H as parameters.
Constraints:-
1 <= L, B, H <= 100Return the length of the Cuboid.Sample Input:-
L = 3, B = 5, H = 1
Sample Output:-
36
Sample Input:-
L = 1, B = 1, H = 1
Sample Output:-
12, I have written this Solution Code:
L,B,H=input().split()
a=4*(int(L)+int(B)+int(H))
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length, breadth, and height of a cuboid. Your task is to calculate its Perimeter.
Note:- Formula for the perimeter of the cuboid is 4(Length + Breadth + height)<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>Perimeter()</b> that takes integers L, B, and H as parameters.
Constraints:-
1 <= L, B, H <= 100Return the length of the Cuboid.Sample Input:-
L = 3, B = 5, H = 1
Sample Output:-
36
Sample Input:-
L = 1, B = 1, H = 1
Sample Output:-
12, I have written this Solution Code: static int Perimeter(int L, int B, int H){
return 4*(L+B+H);
}
, 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: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: static int equationSum(int n)
{
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: def equationSum(N) :
re=(N*(N+1))//2
re=re*re
re=re+9*((N*(N+1))//2)
re=re+4*N
return re
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: int equationSum(int n){
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: int equationSum(int n){
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton loves EVEN numbers.
You are given two integers N and M. Generate 5 unique even numbers for Newton between N and M (excluding both).The first and only line of input contains integer N and integer M.
Constraints
-10<sup>3</sup> ≤ N ≤ M ≤ 10<sup>3</sup>
10 ≤ M-NThe only line of output contains 5 singly spaced integers satisfying the constraints.Sample Input
0 20
Sample Output
2 6 8 18 14, I have written this Solution Code: N, M = map(int, input().split())
start = N
if N % 2 != 0:
start = N+1
else:
start = N+2
print(f'{start} {start+2} {start+4} {start+6} {start+8}'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list, the task is to move all 0’s to the front of the linked list. The order of all another element except 0 should be same after rearrangement.
Note: Avoid use of any type of Java Collection frameworks.
Note: For custom input/output, enter the list in reverse order, and the output will come in right order.<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>moveZeroes()</b> that takes head node as parameter.
Constraints:
1 <= T <= 100
1 <= N <= 100000
0<=Node.data<=100000
Note:- Sum of all test cases doesn't exceed 10^5
Return the head of the modified linked list.Input:
2
10
0 4 0 5 0 2 1 0 1 0
7
1 1 2 3 0 0 0
Output:
0 0 0 0 0 4 5 2 1 1
0 0 0 1 1 2 3
Explanation:
Testcase 1:
Original list was 0->4->0->5->0->2->1->0->1->0->NULL.
After processing list becomes 0->0->0->0->0->4->5->2->1->1->NULL.
Testcase 2:
Original list was 1->1->2->3->0->0->0->NULL.
After processing list becomes 0->0->0->1->1->2->3->NULL., I have written this Solution Code: static public Node moveZeroes(Node head){
ArrayList<Integer> a=new ArrayList<>();
int c=0;
while(head!=null){
if(head.data==0){
c++;
}
else{
a.add(head.data);
}
head=head.next;
}
head=null;
for(int i=a.size()-1;i>=0;i--){
if(head==null){
head=new Node(a.get(i));
}
else{
Node temp=new Node(a.get(i));
temp.next=head;
head=temp;
}
}
while(c-->0){
Node temp=new Node(0);
temp.next=head;
head=temp;
}
return head;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In a new plot by Hydra to kill the avengers, Doctor Zola comes up with a new form of poison composed of chromosomes A, C, T and G. The poison is a string of length n (where n is divisible by 4) with each of the four letters occurring exactly n/4 times. For example, GACT and AAGTGCCT are both fatal poisons.Help Doctor Zola’s plan by choosing one (maybe empty) substring of minimal length in given poision and replace it with the new string (containing A,T,G,C) to make that poison fatal quickly.The first line of input contains a single integer N. The next line of input contains a string S of size N.
Constraints:-
1 <= N <= 100000
Note:- N is divisible by 4 and S will only contain {A, G, C, T}Print the length of the minimum length substring that can be replaced to detect the poison as fatal.Sample Input:-
8
GAAATAAA
Sample Output:-
5
Explanation:-
ATAAA can be replaced to TTCCG to get a fatal poison
Sample Input:-
4
AAAA
Sample Output:-
3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
FastReader sc = new FastReader();
int n= sc.nextInt();
String str = sc.next();
int x= n/4;
int l = 0;
int r = n;
while (l!=r){
int a = (l+r)/2;
if(checkL(str,a)) r=a;
else l=a+1;
}
System.out.print(l);
}
static boolean checkL(String str,int L){
int n = str.length();
int m = n/4;
int g=0,a=0,t=0,c=0;
for (int i = L; i < n; i++) {
char x = str.charAt(i);
if(x=='G') g+=1;
else if (x=='A') a+=1;
else if (x=='T') t+=1;
else c+=1;
}
if (g<=m && a<=m && t<=m && c<=m) return true;
for (int i = 0; i < n-L; i++) {
char x = str.charAt(i);
char y = str.charAt(i+L);
if(x=='G') g+=1;
else if (x=='A') a+=1;
else if (x=='T') t+=1;
else if(x=='C') c+=1;
if(y=='G') g-=1;
else if (y=='A') a-=1;
else if (y=='T') t-=1;
else if(y=='C') c-=1;
if (g<=m && a<=m && t<=m && c<=m) return true;
}
return false;
}
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;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In a new plot by Hydra to kill the avengers, Doctor Zola comes up with a new form of poison composed of chromosomes A, C, T and G. The poison is a string of length n (where n is divisible by 4) with each of the four letters occurring exactly n/4 times. For example, GACT and AAGTGCCT are both fatal poisons.Help Doctor Zola’s plan by choosing one (maybe empty) substring of minimal length in given poision and replace it with the new string (containing A,T,G,C) to make that poison fatal quickly.The first line of input contains a single integer N. The next line of input contains a string S of size N.
Constraints:-
1 <= N <= 100000
Note:- N is divisible by 4 and S will only contain {A, G, C, T}Print the length of the minimum length substring that can be replaced to detect the poison as fatal.Sample Input:-
8
GAAATAAA
Sample Output:-
5
Explanation:-
ATAAA can be replaced to TTCCG to get a fatal poison
Sample Input:-
4
AAAA
Sample Output:-
3, I have written this Solution Code: n = int(input(""))
st = input("")
fxmp = {'A':0,'C':0,'G':0,'T':0}
mp = {'A':0,'C':0,'G':0,'T':0}
for i in st:
fxmp[i] +=1
l = 0
r = len(st)
while l<r:
for k,m in fxmp.items():
mp[k] = m
md = (l+r+1)//2
for i in range(md):
mp[st[i]]-=1
if (mp['A'] <= n/4) and (mp['C'] <= n/4) and (mp['G'] <= n/4) and (mp['T'] <= n/4):
r = md - 1
k = 0
for i in range(md,n):
mp[st[k]] += 1
mp[st[i]] -= 1
k+=1
if mp['A'] <= n/4 and mp['C'] <= n/4 and mp['G'] <= n/4 and mp['T'] <= n/4:
r = md - 1
break
if r != md-1:
l = md
print(r+1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In a new plot by Hydra to kill the avengers, Doctor Zola comes up with a new form of poison composed of chromosomes A, C, T and G. The poison is a string of length n (where n is divisible by 4) with each of the four letters occurring exactly n/4 times. For example, GACT and AAGTGCCT are both fatal poisons.Help Doctor Zola’s plan by choosing one (maybe empty) substring of minimal length in given poision and replace it with the new string (containing A,T,G,C) to make that poison fatal quickly.The first line of input contains a single integer N. The next line of input contains a string S of size N.
Constraints:-
1 <= N <= 100000
Note:- N is divisible by 4 and S will only contain {A, G, C, T}Print the length of the minimum length substring that can be replaced to detect the poison as fatal.Sample Input:-
8
GAAATAAA
Sample Output:-
5
Explanation:-
ATAAA can be replaced to TTCCG to get a fatal poison
Sample Input:-
4
AAAA
Sample Output:-
3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define MAX 500005
typedef long longll;
int ara[MAX],art[MAX];
int arc[MAX],argg[MAX];
int main()
{
//freopen("ou.txt", "r", stdin);
int n; cin>>n;
string s; cin>>s;
int count_a=0,count_t=0,count_c=0,count_g=0,xox;
for(int i=0;i<n;i++)
{
char c = s[i];
if(c=='A')
{
count_a++; ara[i]=count_a;
}
else if(c=='T')
count_t++;
else if(c=='C')
count_c++;
else
count_g++;
ara[i]=count_a;
art[i]=count_t;
arc[i]=count_c;
argg[i]=count_g;
}
if(count_a==count_t&&count_a==count_c&&count_a==count_g&&count_t==count_c&&count_t==count_g&&count_c==count_g)
{
cout<< 0; return 0;
}
xox = n/4;
count_a=xox-count_a;
if(count_a>= 0)
count_a = 0;
else
count_a = -count_a;
count_t=xox-count_t;
if(count_t>=0)
count_t=0;
else
count_t = -count_t;
count_c = xox-count_c;
if(count_c>=0)
count_c=0;
else
count_c = -count_c;
count_g = xox-count_g;
if(count_g>=0)
count_g=0;
else
count_g = -count_g;
int ans=n;
for(int i=0;i<n;i++)
{
int index1,index2,index3,index4;
if(i==0)
{
index1 = lower_bound(ara+i,ara+n,count_a)-ara;
if(index1==n)
continue;
index2 = lower_bound(art+i,art+n,count_t)-art;
if(index2==n)
continue;
index3 = lower_bound(arc+i,arc+n,count_c)-arc;
if(index3==n)
continue;
index4 = lower_bound(argg+i,argg+n,count_g)-argg;
if(index4==n)
continue;
}
else{
index1 = lower_bound(ara+i,ara+n,ara[i-1]+count_a)-ara;
if(index1==n)
continue;
index2 = lower_bound(art+i,art+n,art[i-1]+count_t)-art;
if(index2==n)
continue;
index3 = lower_bound(arc+i,arc+n,arc[i-1]+count_c)-arc;
if(index3==n)
continue;
index4 = lower_bound(argg+i,argg+n,argg[i-1]+count_g)-argg;
if(index4==n)
continue;
}
ans = min(ans,max(index1-i+1,max(index2-i+1,max(index3-i+1,index4-i+1))));
}
printf("%d\n",ans);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Andrew is now bored with usual prime numbers and decides to find prime numbers which has exactly three factors. He sits back and keeps on thinking random numbers, and asks you to tell him if the number he has chosen had exactly 3 factors. Now he has given <b>N</b> random numbers, you need to tell him "<b>Yes</b>" if those numbers had exactly 3 factors otherwise "<b>No</b>".
<b>Note:</b> You can consider the number to be factor of its own.First-line contains N, the number of random numbers he thinks of. Second-line contains the N numbers separated by space.
<b>Constraints:</b>
1 <= N <= 10^5
1 <= numbers <= 10^12Print N lines containing "Yes" or "No".Sample Input 1
3
2 4 6
Sample Output 1
No
Yes
No
Explanation:
2 has 2 factors: 1, 2
4 has 3 factors: 1,2,4
6 has 4 factors: 1,2,3,6
Sample Input 2
3
3 5 7
Sample Output 2
No
No
No, I have written this Solution Code: from math import sqrt
def isPrime(n):
if (n <= 1):
return False
if (n <= 3):
return True
if (n % 2 == 0 or n % 3 == 0):
return False
k= int(sqrt(n))+1
for i in range(5,k,6):
if (n % i == 0 or n % (i + 2) == 0):
return False
return True
def isThreeDisctFactors(n):
sq = int(sqrt(n))
if (1 * sq * sq != n):
return False
if (isPrime(sq)):
return True
else:
return False
N = int(input())
numbers = input().split()
for x in numbers:
if isThreeDisctFactors(int(x)):
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Andrew is now bored with usual prime numbers and decides to find prime numbers which has exactly three factors. He sits back and keeps on thinking random numbers, and asks you to tell him if the number he has chosen had exactly 3 factors. Now he has given <b>N</b> random numbers, you need to tell him "<b>Yes</b>" if those numbers had exactly 3 factors otherwise "<b>No</b>".
<b>Note:</b> You can consider the number to be factor of its own.First-line contains N, the number of random numbers he thinks of. Second-line contains the N numbers separated by space.
<b>Constraints:</b>
1 <= N <= 10^5
1 <= numbers <= 10^12Print N lines containing "Yes" or "No".Sample Input 1
3
2 4 6
Sample Output 1
No
Yes
No
Explanation:
2 has 2 factors: 1, 2
4 has 3 factors: 1,2,4
6 has 4 factors: 1,2,3,6
Sample Input 2
3
3 5 7
Sample Output 2
No
No
No, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
static int MAX = 1000005;
static long spf[] = new long[MAX+5];
public static void main (String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int N = Integer.parseInt(read.readLine());
SOF();
String str[] = read.readLine().trim().split(" ");
for(int i = 0; i < N; i++)
{
long x = Long.parseLong(str[i]);
long g = (long)Math.sqrt(x);
if((g*g)!=x)
{
System.out.println("No");
}
else
{
if(spf[(int)g]==g)
{
System.out.println("Yes");
}
else
System.out.println("No");
}
}
}
public static void SOF()
{
spf[1] = 0;
for (int i=2; i<MAX; i++)
// marking smallest prime factor for every
// number to be itself.
spf[i] = i;
// separately marking spf for every even
// number as 2
for (int i=4; i<MAX; i+=2)
spf[i] = 2;
for (int i=3; i*i<MAX; i++)
{
// checking if i is prime
if (spf[i] == i)
{
// marking SPF for all numbers divisitedible by i
for (int j=i*i; j<MAX; j+=i)
// marking spf[j] if it is not
// previously marked
if (spf[j]==j)
spf[j] = i;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Andrew is now bored with usual prime numbers and decides to find prime numbers which has exactly three factors. He sits back and keeps on thinking random numbers, and asks you to tell him if the number he has chosen had exactly 3 factors. Now he has given <b>N</b> random numbers, you need to tell him "<b>Yes</b>" if those numbers had exactly 3 factors otherwise "<b>No</b>".
<b>Note:</b> You can consider the number to be factor of its own.First-line contains N, the number of random numbers he thinks of. Second-line contains the N numbers separated by space.
<b>Constraints:</b>
1 <= N <= 10^5
1 <= numbers <= 10^12Print N lines containing "Yes" or "No".Sample Input 1
3
2 4 6
Sample Output 1
No
Yes
No
Explanation:
2 has 2 factors: 1, 2
4 has 3 factors: 1,2,4
6 has 4 factors: 1,2,3,6
Sample Input 2
3
3 5 7
Sample Output 2
No
No
No, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 1000001
bool a[max1];
signed main()
{
string s="abcdefghijklmnopqrstuvwxyz";
for(int i=0;i<=1000000;i++){
a[i]=false;
}
for(int i=2;i<1000000;i++){
if(a[i]==false){
for(int j=i+i;j<=1000000;j+=i){
a[j]=true;
}
}
}
int n;
cin>>n;
long long x;
long long y;
for(int i=0;i<n;i++){
cin>>x;
long long y=sqrt(x);
if(y*y==x){
if(a[y]==false){cout<<"Yes"<<endl;}
else{cout<<"No"<<endl;}
}
else{
cout<<"No"<<endl;
}
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a rope of length L which you have to cut at 11 positions to divide it into 12 ropes. Here, each of the 12 resulting ropes must have a positive integer length.
Find the number of ways to do this division.
Note: Two ways to do the division are considered different if and only if there is a position cut in only one of those ways.Input contains a single integer L.
Constraints:
12 <= L <= 200Print the number of ways to do the division.Sample Input 1
12
Sample Output 1
1
Sample Input 2
13
Sample Output 2
12, I have written this Solution Code: def ncr(n, r):
ans = 1
for i in range(1,r+1):
ans *= (n - r + i)
ans //= i
return ans
def NoOfDistributions(N, R):
return ncr(N - 1, R - 1)
N = int(input())
print(NoOfDistributions(N, 12)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a rope of length L which you have to cut at 11 positions to divide it into 12 ropes. Here, each of the 12 resulting ropes must have a positive integer length.
Find the number of ways to do this division.
Note: Two ways to do the division are considered different if and only if there is a position cut in only one of those ways.Input contains a single integer L.
Constraints:
12 <= L <= 200Print the number of ways to do the division.Sample Input 1
12
Sample Output 1
1
Sample Input 2
13
Sample Output 2
12, 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;
cin>>n;
int ans=1;
for(int i=1;i<12;i++){
ans=ans*(n-i);
ans=ans/i;
}
cout<<ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a rope of length L which you have to cut at 11 positions to divide it into 12 ropes. Here, each of the 12 resulting ropes must have a positive integer length.
Find the number of ways to do this division.
Note: Two ways to do the division are considered different if and only if there is a position cut in only one of those ways.Input contains a single integer L.
Constraints:
12 <= L <= 200Print the number of ways to do the division.Sample Input 1
12
Sample Output 1
1
Sample Input 2
13
Sample Output 2
12, I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
// don't change the name of this class
// you can add inner classes if needed
class Main {
public static void main (String[] args) {
// Your code here
Scanner sc = new Scanner(System.in);
int rope = sc.nextInt();
System.out.println(possibleValues(rope));
}
static long possibleValues(int rope)
{
long ans = 1;
for(int i = 1; i < 12; i++)
{
ans *= (rope-i);
ans /= i;
}
return ans;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Padding is a process of adding layers of zeros to the input image. If we want to have p layers of padding then we have to add p layers of zeroes on borders.
Given two integers N and p. How many zeroes are needed to be added to N X N image to give p layers of zero padding?First line contains N and p.
<b>Constraints</b>
1 ≤ N, p ≤ 10<sup>8</sup>Output a single integer denoting the required number of zeroes.Input:
3 1
Output:
16
Explanation :
0 0 0 0 0
0 1 2 3 0
0 4 5 6 0
0 7 8 9 0
0 0 0 0 0, I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
long n=Long.parseLong(in.next());
long p=Long.parseLong(in.next());
long ans=(n+2*p) * (n+2*p) - (n*n);
out.print(ans);
out.close();
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array "a" of non-Integers, check if any integer in the array is less than k.The first line contains two Integers n and k, which is the size of the array and k.
The second line contains n space- separated Integers, that is array elements.
<b>Constraints</b>
1 ≤ n ≤ 100
1 ≤ k ≤ 100
1 ≤ a[i] ≤ 100Print "true" if any element of array is less than k, otherwise "false".Sample 1:
Input:
4 3
1 1 4 8
Output:
true
Explanation:
Both Integers at Indices 0 and 1 are less than k=3., I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
import java.util.*;
public class Main {
public static boolean checkIfAnyLessThanK(int[] a, int k) {
for (int i = 0; i < a.length; i++) {
if (a[i] < k) {
return true;
}
}
return false;
}
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
// String s=sc.nextLine();
int n=sc.nextInt();
int k=sc.nextInt();
sc.nextLine();
assert n>=1&&n<=100 : "Invalid Input";
assert k>=1&&k<=100 : "Invalid Input";
int a[]=new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
assert a[i]>=1&&a[i]<=100 : "Invalid Input";
}
System.out.print(checkIfAnyLessThanK(a,k));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number, you task is to reverse its digits. If the reversed number contains 0s in the beginning, you must remove them as well.Functional problem, Number is the input
Constraint:-
1<=size of n<=100000Output is a single line containing reversed number n.Sample Input
123445
Sample Output
544321
Sample Input
16724368
Sample Output
86342761, 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();
System.out.println(reverseDigits(n));
}
static int reverseDigits(int num)
{
int rev_num = 0;
while (num > 0) {
rev_num = rev_num * 10 + num % 10;
num = num / 10;
}
return rev_num;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number, you task is to reverse its digits. If the reversed number contains 0s in the beginning, you must remove them as well.Functional problem, Number is the input
Constraint:-
1<=size of n<=100000Output is a single line containing reversed number n.Sample Input
123445
Sample Output
544321
Sample Input
16724368
Sample Output
86342761, I have written this Solution Code: // str is input string
function reverseNumber(num) {
// write code here
// do not console.log
// return the sum
const numStr = `${num}`
return Number(numStr.split('').reverse().join(''))
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string of length N, count the number of vowels present at even places.First line of input contains a single integer N. Second line contains a string of size N.
Constraints:-
1 <= N <= 100000
Note:- String will contain only lowercase english letterPrint the number of vowels present at the even placesSample Input:-
12
newtonschool
Sample Output:-
2
Sample Input:-
5
basid
Sample Output:-
2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader reader =new BufferedReader(new InputStreamReader(System.in));
String name = reader.readLine();
String s=reader.readLine();
long c=0;
for(int i=1;i<s.length();i=i+2){
if(s.charAt(i)=='a'||s.charAt(i)=='e'||s.charAt(i)=='i'||s.charAt(i)=='o'||s.charAt(i)=='u')
c++;
}
System.out.println(c);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string of length N, count the number of vowels present at even places.First line of input contains a single integer N. Second line contains a string of size N.
Constraints:-
1 <= N <= 100000
Note:- String will contain only lowercase english letterPrint the number of vowels present at the even placesSample Input:-
12
newtonschool
Sample Output:-
2
Sample Input:-
5
basid
Sample Output:-
2, I have written this Solution Code: def check(c):
if c=='a' or c=='e' or c=='i' or c=='o' or c=='u':
return True
else:
return False
n=int(input())
s=input()
j=0
cnt=0
ans=0
for i in range(0,n):
j=i+1
if j%2==0:
if check(s[i]):
#=='a' or s[i]=='e' or s[i]=='i' or s[i]=='o' or s[i]=='u':
cnt+=1
print(cnt)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string of length N, count the number of vowels present at even places.First line of input contains a single integer N. Second line contains a string of size N.
Constraints:-
1 <= N <= 100000
Note:- String will contain only lowercase english letterPrint the number of vowels present at the even placesSample Input:-
12
newtonschool
Sample Output:-
2
Sample Input:-
5
basid
Sample Output:-
2, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
bool check(char c){
if(c=='a' || c=='e' || c=='i' || c=='o' || c=='u'){return true;}
else{
return false;
}
}
int main() {
int n; cin>>n;
string s;
cin>>s;
int j=0;
int cnt=0;
int ans=0;
for(int i=0;i<n;i++){
if(i&1){
if(s[i]=='a' || s[i]=='e' || s[i]=='i' || s[i]=='o' || s[i]=='u'){
cnt++;
}}
}
cout<<cnt;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: static int equationSum(int n)
{
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: def equationSum(N) :
re=(N*(N+1))//2
re=re*re
re=re+9*((N*(N+1))//2)
re=re+4*N
return re
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: int equationSum(int n){
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: int equationSum(int n){
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string, the task is to remove duplicates from it. Expected time complexity O(n) where n is length of the input string and extra space O(1) under the assumption that there are total of 256 possible characters in a string.
Note: that original order of characters must be kept same.The first line of the input is the number of test cases T. And the first line of a test case contains a string.
Constraints:
1 <= T <= 15
1 <= |string|<= 15000Modified string without duplicates and the same order of characters.Input:
2
hello
helloworld
Output:
helo
helowrd, 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 test = Integer.parseInt(br.readLine());
while(test-->0)
{
String line=br.readLine();
char str[] = line.toCharArray();
int n = str.length;
System.out.println(duplicates(str, n));
}
}
static String duplicates(char str[], int n)
{
int index = 0;
for (int i = 0; i < n; i++)
{
int j;
for (j = 0; j < i; j++)
if (str[i] == str[j])
break;
if (j == i)
str[index++] = str[i];
}
return String.valueOf(Arrays.copyOf(str, index));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string, the task is to remove duplicates from it. Expected time complexity O(n) where n is length of the input string and extra space O(1) under the assumption that there are total of 256 possible characters in a string.
Note: that original order of characters must be kept same.The first line of the input is the number of test cases T. And the first line of a test case contains a string.
Constraints:
1 <= T <= 15
1 <= |string|<= 15000Modified string without duplicates and the same order of characters.Input:
2
hello
helloworld
Output:
helo
helowrd, I have written this Solution Code: for _ in range(int(input())):
s=input()
a=[]
for i in s:
if (i not in a):
print(i,end="")
a.append(i)
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string, the task is to remove duplicates from it. Expected time complexity O(n) where n is length of the input string and extra space O(1) under the assumption that there are total of 256 possible characters in a string.
Note: that original order of characters must be kept same.The first line of the input is the number of test cases T. And the first line of a test case contains a string.
Constraints:
1 <= T <= 15
1 <= |string|<= 15000Modified string without duplicates and the same order of characters.Input:
2
hello
helloworld
Output:
helo
helowrd, I have written this Solution Code: // C++ implementation of above approach
#include <bits/stdc++.h>
#include <string>
using namespace std;
// Function to remove duplicates
string removeDuplicatesFromString(string str)
{
// keeps track of visited characters
int counter = 0;
int i = 0;
int size = str.size();
// gets character value
int x;
// keeps track of length of resultant string
int length = 0;
int table[500];
memset(table,0,sizeof(table));
while (i < size) {
x = str[i]-'a'+97;
// check if Xth bit of counter is unset
if (table[x] == 0) {
str[length] = 'a' + x -97;
// mark current character as visited
table[x]++;
length++;
}
i++;
}
return str.substr(0, length);
}
// Driver code
int main()
{
int t;
cin>>t;
while(t>0)
{ t--;
string str;
cin>>str;
cout << removeDuplicatesFromString(str)<<"\n";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<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>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code: public static long SumOfDivisors(long N){
long sum=0;
long c=(long)Math.sqrt(N);
for(long i=1;i<=c;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){sum+=N/i;}
}
}
return sum;
}
, 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 return the sum of all of its divisors.<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>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
long long SumOfDivisors(long long N){
long long sum=0;
long sq=sqrt(N);
for(long i=1;i<=sq;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){
sum+=N/i;
}
}
}
return sum;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<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>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
long long SumOfDivisors(long long N){
long long sum=0;
long sq=sqrt(N);
for(long i=1;i<=sq;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){
sum+=N/i;
}
}
}
return sum;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<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>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
def SumOfDivisors(num) :
# Final result of summation of divisors
result = 0
# find all divisors which divides 'num'
i = 1
while i<= (math.sqrt(num)) :
# if 'i' is divisor of 'num'
if (num % i == 0) :
# if both divisors are same then
# add it only once else add both
if (i == (num / i)) :
result = result + i;
else :
result = result + (i + num/i);
i = i + 1
# Add 1 to the result as 1 is also
# a divisor
return (result); , In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers N and D, your task is to find the numbers between 1 to N which contains the digit D at least 1 time.Input contains only two integers N and D.
Constraints:-
1 < = N < = 100000
1 < = D < = 9Print all the numbers from 1 to N which contains the digit D in it separated by space in non decreasing order.Sample Input:-
20 5
Sample Output:-
5 15
Sample Input:-
15 1
Sample Output:-
1 10 11 12 13 14 15, I have written this Solution Code: /**
* author: tourist1256
* created: 2022-07-08 04:16:54
**/
#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(int l = LLONG_MIN, int r = LLONG_MAX) {
uniform_int_distribution<int64_t> generator(l, r);
return generator(rng);
}
int32_t main() {
auto start = std::chrono::high_resolution_clock::now();
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, d;
cin >> n >> d;
int cnt = 0;
function<int(int, int)> f = [&](int d, int m) {
while (d) {
int x = d % 10;
d /= 10;
if (x == m) {
return 1;
}
}
return 0;
};
for (int i = 1; i <= n; i++) {
if (f(i, d)) {
cout << i << " ";
}
}
auto stop = std::chrono::high_resolution_clock::now();
auto duration = std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start);
cerr << "Time taken : " << ((long double)duration.count()) / ((long double)1e9) << "s\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers N and D, your task is to find the numbers between 1 to N which contains the digit D at least 1 time.Input contains only two integers N and D.
Constraints:-
1 < = N < = 100000
1 < = D < = 9Print all the numbers from 1 to N which contains the digit D in it separated by space in non decreasing order.Sample Input:-
20 5
Sample Output:-
5 15
Sample Input:-
15 1
Sample Output:-
1 10 11 12 13 14 15, I have written this Solution Code: def index(st, ch):
for i in range(len(st)):
if (st[i] == ch):
return i;
return -1
def printNumbers(n, d):
# Converting d to character
st = "" + str(d)
ch = st[0]
l = []
# Loop to check each digit one by one.
for i in range(0, n + 1, 1):
# initialize the string
st = ""
st = st + str(i)
# checking for digit
if (i == d or index(st, ch) >= 0):
# print(i, end = " ")
l.append(i)
for k in range(0, len(l) - 1):
print(l[k], end=" ")
print(l[len(l) - 1])
# Driver code
if __name__ == '__main__':
li = list(map(int, input().strip().split()))
n = li[0]
d = li[1]
printNumbers(n, d), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers N and D, your task is to find the numbers between 1 to N which contains the digit D at least 1 time.Input contains only two integers N and D.
Constraints:-
1 < = N < = 100000
1 < = D < = 9Print all the numbers from 1 to N which contains the digit D in it separated by space in non decreasing order.Sample Input:-
20 5
Sample Output:-
5 15
Sample Input:-
15 1
Sample Output:-
1 10 11 12 13 14 15, 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 d=sc.nextInt();
printnumber(n,d);
}
static boolean ispresent(int n,int d){
while(n>0){
if(n%10==d)
break;
n=n / 10;
}
return n>0;
}
static void printnumber(int n,int d){
for(int i=0;i<=n;i++){
if(i==d||ispresent(i,d)){
System.out.print(i+" ");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to a given number S.Each test case consists of two lines. The first line of each test case is N and S, where N is the size of the array and S is the sum. The second line of each test case contains N space-separated integers denoting the array elements.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ Ai ≤ 10<sup>5</sup>Print the starting and ending positions (1 indexing) of first such occurring subarray from the left if sum equals to subarray, else print -1.Sample Input
5 12
1 2 3 7 5
Sample Output
2 4
Explanation:
subarray starting from index 2 and ending at index 4 => {2 , 3 , 7}
sum = 2 + 3 + 7 = 12
Sample Input
10 15
1 2 3 4 5 6 7 8 9 10
Sample Output
1 5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException
{
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String StrInput[] = br.readLine().trim().split(" ");
int n = Integer.parseInt(StrInput[0]);
int s = Integer.parseInt(StrInput[1]);
int arr[] = new int[n];
String StrInput2[] = br.readLine().trim().split(" ");
for(int i=0;i<n;i++)
{
arr[i] = Integer.parseInt(StrInput2[i]);
}
int sum = arr[0];
int startingindex = 0;
int endingindex = 1;
int j = 0;
int i;
for(i=1;i<=n;i++)
{
if(sum < s && arr[i] != 0)
{
sum += arr[i];
}
while(sum > s && startingindex < i-1)
{
sum -= arr[startingindex];
startingindex++;
}
if(sum == s)
{
endingindex = i+1;
if(arr[0] == 0)
{
System.out.print(startingindex+2 + " " + endingindex);
}
else
{
System.out.print(startingindex+1 + " "+ endingindex);
}
break;
}
if(i == n && sum < s)
{
System.out.print(-1);
break;
}
}
}
catch(Exception e)
{
System.out.print(-1);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to a given number S.Each test case consists of two lines. The first line of each test case is N and S, where N is the size of the array and S is the sum. The second line of each test case contains N space-separated integers denoting the array elements.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ Ai ≤ 10<sup>5</sup>Print the starting and ending positions (1 indexing) of first such occurring subarray from the left if sum equals to subarray, else print -1.Sample Input
5 12
1 2 3 7 5
Sample Output
2 4
Explanation:
subarray starting from index 2 and ending at index 4 => {2 , 3 , 7}
sum = 2 + 3 + 7 = 12
Sample Input
10 15
1 2 3 4 5 6 7 8 9 10
Sample Output
1 5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n,k;
cin>>n>>k;
int a[n];
for(int i=0;i<n;i++)
{
cin>>a[i];
}
int sum=0;
unordered_map<int,int> m;
for(int i=0;i<n;i++){
sum+=a[i];
if(sum==k){cout<<1<<" "<<i+1;return 0;}
if(m.find(sum-k)!=m.end()){
cout<<m[sum-k]+2<<" "<<i+1;
return 0;
}
m[sum]=i;
}
cout<<-1;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to a given number S.Each test case consists of two lines. The first line of each test case is N and S, where N is the size of the array and S is the sum. The second line of each test case contains N space-separated integers denoting the array elements.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ Ai ≤ 10<sup>5</sup>Print the starting and ending positions (1 indexing) of first such occurring subarray from the left if sum equals to subarray, else print -1.Sample Input
5 12
1 2 3 7 5
Sample Output
2 4
Explanation:
subarray starting from index 2 and ending at index 4 => {2 , 3 , 7}
sum = 2 + 3 + 7 = 12
Sample Input
10 15
1 2 3 4 5 6 7 8 9 10
Sample Output
1 5, I have written this Solution Code: def sumFinder(N,S,a):
currentSum = a[0]
start = 0
i = 1
while i <= N:
while currentSum > S and start < i-1:
currentSum = currentSum - a[start]
start += 1
if currentSum == S:
return (start+1,i)
if i < N:
currentSum = currentSum + a[i]
i += 1
return(-1)
N, S = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
ans = sumFinder(N,S,a)
if(ans==-1):
print(ans)
else:
print(ans[0],ans[1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of N integers arranged in a circular fashion. Your task is to find the maximum contiguous subarray sum containing at least 1 element.The first line of input contains a single integer T which denotes the number of test cases. The first line of each test case contains a single integer N which denotes the total number of elements. The second line of each test case contains N space-separated integers denoting the elements of the array.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^5
-10^6 <= Arr[i] <= 10^6
The Sum of N over all test cases is less than equal to 10^6.For each test case print the maximum sum obtained by adding the consecutive elements.<b>Input:</b>
4
7
8 -8 9 -9 10 -11 12
8
10 -3 -4 7 6 5 -4 -1
8
-1 40 -14 7 6 5 -4 -1
4
-1 -2 -3 -4
<b>Output:</b>
22
23
52
-1
Explanation:
Testcase 1: Starting from the last element of the array, i.e, 12, and moving in a circular fashion, we have max subarray as 12, 8, -8, 9, -9, 10, which gives the maximum sum as 22., 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 long maxSubarraySumCircular(long[] array) {
long currentSum = 0, maxSum = Long.MIN_VALUE;
for(int i = 0; i < array.length; i++) {
currentSum = Math.max(currentSum + array[i], array[i]);
maxSum = Math.max(maxSum, currentSum);
}
if(maxSum < 0) return maxSum;
currentSum = 0;
long minSum = Long.MAX_VALUE;
for(int i = 0; i < array.length; i++) {
currentSum = Math.min(currentSum + array[i], array[i]);
minSum = Math.min(minSum, currentSum);
}
long totalSum = 0;
for(long element : array) totalSum += element;
return Math.max(maxSum, totalSum - minSum);
}
public static void main (String[] args) {
FastReader sc = new FastReader();
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
long a[] = new long[n];
long sum = 0l;
for(int i=0;i<n;i++){
a[i] = sc.nextLong();
}
System.out.println(maxSubarraySumCircular(a));
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of N integers arranged in a circular fashion. Your task is to find the maximum contiguous subarray sum containing at least 1 element.The first line of input contains a single integer T which denotes the number of test cases. The first line of each test case contains a single integer N which denotes the total number of elements. The second line of each test case contains N space-separated integers denoting the elements of the array.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^5
-10^6 <= Arr[i] <= 10^6
The Sum of N over all test cases is less than equal to 10^6.For each test case print the maximum sum obtained by adding the consecutive elements.<b>Input:</b>
4
7
8 -8 9 -9 10 -11 12
8
10 -3 -4 7 6 5 -4 -1
8
-1 40 -14 7 6 5 -4 -1
4
-1 -2 -3 -4
<b>Output:</b>
22
23
52
-1
Explanation:
Testcase 1: Starting from the last element of the array, i.e, 12, and moving in a circular fashion, we have max subarray as 12, 8, -8, 9, -9, 10, which gives the maximum sum as 22., I have written this Solution Code: def kadane(a):
Max = a[0]
temp = Max
for i in range(1,len(a)):
temp += a[i]
if temp < a[i]:
temp = a[i]
Max = max(Max,temp)
return Max
def maxCircularSum(a):
n = len(a)
max_kadane = kadane(a)
neg_a = [-1*x for x in a]
max_neg_kadane = kadane(neg_a)
max_wrap = -(sum(neg_a)-max_neg_kadane)
res = max(max_wrap,max_kadane)
return res if res != 0 else max_kadane
for _ in range(int(input())):
s=int(input())
a=list(map(int,input().split()))
print(maxCircularSum(a)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of N integers arranged in a circular fashion. Your task is to find the maximum contiguous subarray sum containing at least 1 element.The first line of input contains a single integer T which denotes the number of test cases. The first line of each test case contains a single integer N which denotes the total number of elements. The second line of each test case contains N space-separated integers denoting the elements of the array.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^5
-10^6 <= Arr[i] <= 10^6
The Sum of N over all test cases is less than equal to 10^6.For each test case print the maximum sum obtained by adding the consecutive elements.<b>Input:</b>
4
7
8 -8 9 -9 10 -11 12
8
10 -3 -4 7 6 5 -4 -1
8
-1 40 -14 7 6 5 -4 -1
4
-1 -2 -3 -4
<b>Output:</b>
22
23
52
-1
Explanation:
Testcase 1: Starting from the last element of the array, i.e, 12, and moving in a circular fashion, we have max subarray as 12, 8, -8, 9, -9, 10, which gives the maximum sum as 22., 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 EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 10001
#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
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int INF = 4557430888798830399ll;
signed main()
{
fast();
int t;
cin>>t;
while(t--){
int n;
cin>>n;
vector<int> A(n);
FOR(i,n){
cin>>A[i];
}
int maxTillNow = -INF;
int maxEndingHere = -INF;
int start = 0;
while (start < A.size()) {
maxEndingHere = max(maxEndingHere + A[start], (int)A[start]);
maxTillNow = max(maxTillNow, maxEndingHere);
start++;
}
vector<int> prefix(n, 0), suffix(n, 0), maxSuffTill(n, 0);
for (int i = 0; i < n; ++i) {
prefix[i] = A[i];
if (i != 0) prefix[i] += prefix[i - 1];
}
for (int i = n - 1; i >= 0; --i) {
suffix[i] = A[i];
maxSuffTill[i] = max(A[i],(int) 0);
if (i != n - 1) {
suffix[i] += suffix[i + 1];
maxSuffTill[i] = max(suffix[i], maxSuffTill[i + 1]);
}
}
for (int i = 0; i < n; ++i) {
int sum = prefix[i];
if (i != n - 1) sum += maxSuffTill[i + 1];
maxTillNow = max(maxTillNow, sum);
}
out(maxTillNow);
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two elements A and B, your task is to swap the given two elements.<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>Swap(int A, int B)</b>.Print one line which contains the output of swapped elements.Sample Input:-
5 7
Sample Output:-
7 5
Sample Input:-
3 6
Sample Output:-
6 3, I have written this Solution Code: class Solution {
public static void Swap(int A, int B){
int C = A;
A = B;
B = C;
System.out.println(A + " " + B);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two elements A and B, your task is to swap the given two elements.<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>Swap(int A, int B)</b>.Print one line which contains the output of swapped elements.Sample Input:-
5 7
Sample Output:-
7 5
Sample Input:-
3 6
Sample Output:-
6 3, I have written this Solution Code: # Python program to swap two variables
li= list(map(int,input().strip().split()))
x=li[0]
y=li[1]
# create a temporary variable and swap the values
temp = x
x = y
y = temp
print(x,end=" ")
print(y,end="")
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, give the number with maximum frequency. If multiple numbers have maximum frequency print the maximum number among them.The first line of the input contains an integer N, and the Second line contains N space-separated integers of the array.
<b>Constraints:</b>
3 <= N <= 1000
1 <= Arr[i] <= 100The output should contain single integer, the number with maximum frequency.If multiple numbers have maximum frequency print the maximum number among them.Sample Input
5
1 4 2 4 5
Sample Output
4
<b>Explanation:-</b>
4 has max frequency=2, 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().trim().split(" ");
int a[]=new int[n];
for(int i=0;i<n;i++){
a[i]=Integer.parseInt(str[i]);
}
Arrays.sort(a);
int size=a[n-1]+1;
int c[]=new int[size];
for(int i=0;i<size;i++) c[i]=0;
for(int i=0;i<n;i++) c[a[i]]++;
int max=0,freq=c[1];
for(int i=2;i<size;i++){
if(freq<=c[i]){
freq=c[i];
max=i;
}
}
System.out.println(max);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, give the number with maximum frequency. If multiple numbers have maximum frequency print the maximum number among them.The first line of the input contains an integer N, and the Second line contains N space-separated integers of the array.
<b>Constraints:</b>
3 <= N <= 1000
1 <= Arr[i] <= 100The output should contain single integer, the number with maximum frequency.If multiple numbers have maximum frequency print the maximum number among them.Sample Input
5
1 4 2 4 5
Sample Output
4
<b>Explanation:-</b>
4 has max frequency=2, I have written this Solution Code: n = int(input())
a = [int(x) for x in input().split()]
freq = {}
for x in a:
if x not in freq:
freq[x] = 1
else:
freq[x] += 1
mx = max(freq.values())
rf = sorted(freq)
for i in range(len(rf) - 1, -1, -1):
if freq[rf[i]] == mx:
print(rf[i])
break, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, give the number with maximum frequency. If multiple numbers have maximum frequency print the maximum number among them.The first line of the input contains an integer N, and the Second line contains N space-separated integers of the array.
<b>Constraints:</b>
3 <= N <= 1000
1 <= Arr[i] <= 100The output should contain single integer, the number with maximum frequency.If multiple numbers have maximum frequency print the maximum number among them.Sample Input
5
1 4 2 4 5
Sample Output
4
<b>Explanation:-</b>
4 has max frequency=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 = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int n; cin >> n;
for(int i = 1; i <= n; i++){
int p; cin >> p;
a[p]++;
}
int mx = 0, id = -1;
for(int i = 1; i <= 100; i++){
if(a[i] >= mx)
mx = a[i], id = i;
}
cout << id;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size N*N, your task is to find the sum of the primary and secondary diagonal of the matrix.
For Matrix:-
M<sub>00</sub> M<sub>01</sub> M<sub>02</sub>
M<sub>10</sub> M<sub>11</sub> M<sub>12</sub>
M<sub>20</sub> M<sub>21</sub> M<sub>22</sub>
Primary diagonal:- M<sub>00</sub> M<sub>11</sub> M<sub>22</sub>
Secondary diagonal:- M<sub>02</sub> M<sub>11</sub> M<sub>20</sub>The first line of input contains a single integer N, The next N lines of input contains N space-separated integers depicting the values of the matrix.
Constraints:-
1 <= N <= 500
1 <= Matrix[][] <= 100000Print the sum of primary and secondary diagonal separated by a space.Sample Input:-
2
1 4
2 6
Sample Output:-
7 6
Sample Input:-
3
1 4 2
1 5 7
3 8 1
Sample Output:-
7 10, 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;
cin>>n;
int a[n][n];
FOR(i,n){
FOR(j,n){
cin>>a[i][j];}}
int sum=0,sum1=0;;
FOR(i,n){
sum+=a[i][i];
sum1+=a[n-i-1][i];
}
out1(sum);out(sum1);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size N*N, your task is to find the sum of the primary and secondary diagonal of the matrix.
For Matrix:-
M<sub>00</sub> M<sub>01</sub> M<sub>02</sub>
M<sub>10</sub> M<sub>11</sub> M<sub>12</sub>
M<sub>20</sub> M<sub>21</sub> M<sub>22</sub>
Primary diagonal:- M<sub>00</sub> M<sub>11</sub> M<sub>22</sub>
Secondary diagonal:- M<sub>02</sub> M<sub>11</sub> M<sub>20</sub>The first line of input contains a single integer N, The next N lines of input contains N space-separated integers depicting the values of the matrix.
Constraints:-
1 <= N <= 500
1 <= Matrix[][] <= 100000Print the sum of primary and secondary diagonal separated by a space.Sample Input:-
2
1 4
2 6
Sample Output:-
7 6
Sample Input:-
3
1 4 2
1 5 7
3 8 1
Sample Output:-
7 10, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String args[])throws Exception {
InputStreamReader inr= new InputStreamReader(System.in);
BufferedReader br= new BufferedReader(inr);
String str=br.readLine();
int row = Integer.parseInt(str);
int col=row;
int [][] arr=new int [row][col];
for(int i=0;i<row;i++){
String line =br.readLine();
String[] elements = line.split(" ");
for(int j=0;j<col;j++){
arr[i][j]= Integer.parseInt(elements[j]);
}
}
int sumPrimary=0;
int sumSecondary=0;
for(int i=0;i<row;i++){
sumPrimary=sumPrimary + arr[i][i];
sumSecondary= sumSecondary + arr[i][row-1-i];
}
System.out.println(sumPrimary+ " " +sumSecondary);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size N*N, your task is to find the sum of the primary and secondary diagonal of the matrix.
For Matrix:-
M<sub>00</sub> M<sub>01</sub> M<sub>02</sub>
M<sub>10</sub> M<sub>11</sub> M<sub>12</sub>
M<sub>20</sub> M<sub>21</sub> M<sub>22</sub>
Primary diagonal:- M<sub>00</sub> M<sub>11</sub> M<sub>22</sub>
Secondary diagonal:- M<sub>02</sub> M<sub>11</sub> M<sub>20</sub>The first line of input contains a single integer N, The next N lines of input contains N space-separated integers depicting the values of the matrix.
Constraints:-
1 <= N <= 500
1 <= Matrix[][] <= 100000Print the sum of primary and secondary diagonal separated by a space.Sample Input:-
2
1 4
2 6
Sample Output:-
7 6
Sample Input:-
3
1 4 2
1 5 7
3 8 1
Sample Output:-
7 10, I have written this Solution Code: // mat is the matrix/ 2d array
// the dimensions of array are n * n
function diagonalSum(mat, n) {
// write code here
// console.log the answer as in example
let principal = 0, secondary = 0;
for (let i = 0; i < n; i++) {
principal += mat[i][i];
secondary += mat[i][n - i - 1];
}
console.log(`${principal} ${secondary}`);
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size N*N, your task is to find the sum of the primary and secondary diagonal of the matrix.
For Matrix:-
M<sub>00</sub> M<sub>01</sub> M<sub>02</sub>
M<sub>10</sub> M<sub>11</sub> M<sub>12</sub>
M<sub>20</sub> M<sub>21</sub> M<sub>22</sub>
Primary diagonal:- M<sub>00</sub> M<sub>11</sub> M<sub>22</sub>
Secondary diagonal:- M<sub>02</sub> M<sub>11</sub> M<sub>20</sub>The first line of input contains a single integer N, The next N lines of input contains N space-separated integers depicting the values of the matrix.
Constraints:-
1 <= N <= 500
1 <= Matrix[][] <= 100000Print the sum of primary and secondary diagonal separated by a space.Sample Input:-
2
1 4
2 6
Sample Output:-
7 6
Sample Input:-
3
1 4 2
1 5 7
3 8 1
Sample Output:-
7 10, I have written this Solution Code: n = int(input())
sum1 = 0
sum2 = 0
for i in range(n):
a = [int(j) for j in input().split()]
sum1 = sum1+a[i]
sum2 = sum2+a[n-1-i]
print(sum1,sum2), 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: 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: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
bool EqualOrNot(int h1, int h2, int v1,int v2){
if (v2>v1&&(h1-h2)%(v2-v1)==0){
return true;
}
return false;
}
int main(){
int n1,n2,v1,v2;
cin>>n1>>n2>>v1>>v2;
if(EqualOrNot(n1,n2,v1,v2)){
cout<<"Yes";}
else{
cout<<"No";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code: def EqualOrNot(h1,h2,v1,v2):
if (v2>v1 and (h1-h2)%(v2-v1)==0):
return True
else:
return False
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code: static boolean EqualOrNot(int h1, int h2, int v1,int v2){
if (v2>v1&&(h1-h2)%(v2-v1)==0){
return true;
}
return false;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
bool EqualOrNot(int h1, int h2, int v1,int v2){
if (v2>v1&&(h1-h2)%(v2-v1)==0){
return true;
}
return false;
}
int main(){
int n1,n2,v1,v2;
cin>>n1>>n2>>v1>>v2;
if(EqualOrNot(n1,n2,v1,v2)){
cout<<"Yes";}
else{
cout<<"No";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code: def EqualOrNot(h1,h2,v1,v2):
if (v2>v1 and (h1-h2)%(v2-v1)==0):
return True
else:
return False
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code: static boolean EqualOrNot(int h1, int h2, int v1,int v2){
if (v2>v1&&(h1-h2)%(v2-v1)==0){
return true;
}
return false;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a chessboard of size N x N, where the top left square is black. Each square contains a value. Find the sum of the values of all black squares and all white squares.
Remember that in a chessboard black and white squares are alternate.The first line of input will be the N size of the matrix. Then next N lines will consist of elements of the matrix. Each row will contain N elements since it is a square matrix.
<b>Constraints:-</b>
1 ≤ N ≤ 800
1 ≤ Matrix[i][j] ≤ 100000
Print two lines, the first line containing the sum of black squares and the second line containing the sum of white squares.Input 1:
3
1 2 3
4 5 6
7 8 9
Output 1:
25
20
Sample Input 2:
4
1 2 3 4
6 8 9 10
11 12 13 14
15 16 17 18
Sample Output 2:
80
79
<b>Explanation 1</b>
The black square contains 1, 3, 5, 7, 9; sum = 25
The white square contains 2, 4, 6, 8; sum = 20, I have written this Solution Code: n=int(input())
bSum=0
wSum=0
for i in range(n):
c=(input().split())
for j in range(len(c)):
if(i%2==0):
if(j%2==0):
bSum+=int(c[j])
else:
wSum+=int(c[j])
else:
if(j%2==0):
wSum+=int(c[j])
else:
bSum+=int(c[j])
print(bSum)
print(wSum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a chessboard of size N x N, where the top left square is black. Each square contains a value. Find the sum of the values of all black squares and all white squares.
Remember that in a chessboard black and white squares are alternate.The first line of input will be the N size of the matrix. Then next N lines will consist of elements of the matrix. Each row will contain N elements since it is a square matrix.
<b>Constraints:-</b>
1 ≤ N ≤ 800
1 ≤ Matrix[i][j] ≤ 100000
Print two lines, the first line containing the sum of black squares and the second line containing the sum of white squares.Input 1:
3
1 2 3
4 5 6
7 8 9
Output 1:
25
20
Sample Input 2:
4
1 2 3 4
6 8 9 10
11 12 13 14
15 16 17 18
Sample Output 2:
80
79
<b>Explanation 1</b>
The black square contains 1, 3, 5, 7, 9; sum = 25
The white square contains 2, 4, 6, 8; sum = 20, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main (String[] args)throws IOException {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int mat[][] = new int[N][N];
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
mat[i][j] = sc.nextInt();
}
alternate_Matrix_Sum(mat,N);
}
static void alternate_Matrix_Sum(int mat[][], int N)
{
long sum =0, sum1 = 0;
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
if((i+j)%2 == 0)
sum += mat[i][j];
else sum1 += mat[i][j];
}
}
System.out.println(sum);
System.out.print(sum1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a 2*N matrix in which each cell contains some candies in it. You are at the top left corner of the matrix and want to reach the bottom right corner of the matrix i. e from (1, 1) to (2, N). You can only move right or down. You have to find the maximum number of candies you can collect in your journey.The first line of input contains a single integer N. The second line of input contains N spaces separated integers. The last line of input contains N space-separated integers.
<b>Constraints:-</b>
2 <= N <= 10000
1 <= Matrix[i][j] <= 100000Print the maximum amount of candies you can have at the end of your journey.Sample Input 1:-
5
1 3 5 6 2
2 4 8 4 2
Sample Output 1:-
23
Sample Input 2:-
4
1 1 1 1
1 1 1 1
Sample Output 2:-
5, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n;
cin>>n;
int a[n][2];
for(int i=0;i<n;i++){
cin>>a[i][0];
}
for(int i=0;i<n;i++){
cin>>a[i][1];
}
for(int i=1;i<n;i++){
a[i][0]+=a[i-1][0];
}
for(int i = n-2;i>=0;i--){
a[i][1]+=a[i+1][1];
}
int ans=0;
for(int i=0;i<n;i++){
ans=max(a[i][0]+a[i][1],ans);
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S. Two position i and j of the string are friends if they have the same character. The distance between two friends at positions i and j is defined as |i- j|. Find the sum of distances of all the pairs of friends in the given strings.First line of input contains a single string S.
Constraints:
1 <= |S| <= 1000000
String contains lowercase english letters.Output a single integer which is the sum of distance of all the pair of friends in the given strings.Sample Input
ababa
Sample Output
10
Explanation: Friend pairs - (1, 3) (1, 5) (2, 4) (3, 5), I have written this Solution Code: import java.io.*;
import java.io.IOException;
import java.util.*;
class Main {
public static long mod = (long)Math.pow(10,9)+7 ;
public static double epsilon=0.00000000008854;
public static InputReader sc = new InputReader(System.in);
public static PrintWriter pw = new PrintWriter(System.out);
public static void main(String[] args) {
String s=sc.nextLine();
int n=s.length();
int hnum[]=new int[26];
int hpast[]=new int[26];
Arrays.fill(hpast,-1);
long hsum[]=new long[26];
long ans=0;
for(int i=0;i<n;i++){
int k=s.charAt(i)-'a';
if(hpast[k]!=-1)
hsum[k]=hsum[k]+(i-hpast[k])*hnum[k];
ans+=hsum[k];
hnum[k]++;
hpast[k]=i;
}
pw.println(ans);
pw.flush();
pw.close();
}
public static Comparator<Long[]> column(int i){
return
new Comparator<Long[]>() {
@Override
public int compare(Long[] o1, Long[] o2) {
return o1[i].compareTo(o2[i]);
}
};
}
public static Comparator<Integer[]> col(int i){
return
new Comparator<Integer[]>() {
@Override
public int compare(Integer[] o1, Integer[] o2) {
return o1[i].compareTo(o2[i]);
}
};
}
public static String reverseString(String s){
StringBuilder input1 = new StringBuilder();
input1.append(s);
input1 = input1.reverse();
return input1.toString();
}
public static int[] scanArray(int n){
int a[]=new int [n];
for(int i=0;i<n;i++)
a[i]=sc.nextInt();
return a;
}
public static long[] scanLongArray(int n){
long a[]=new long [n];
for(int i=0;i<n;i++)
a[i]=sc.nextLong();
return a;
}
public static String [] scanStrings(int n){
String a[]=new String [n];
for(int i=0;i<n;i++)
a[i]=sc.nextLine();
return a;
}
public static class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[8192];
private int curChar, snumChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int snext() {
if (snumChars == -1)
throw new InputMismatchException();
if (curChar >= snumChars) {
curChar = 0;
try {
snumChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (snumChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public int[] nextIntArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
return a;
}
public String readString() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isSpaceChar(c));
return res.toString();
}
public String nextLine() {
int c = snext();
while (isSpaceChar(c))
c = snext();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isEndOfLine(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public 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: Given a string S. Two position i and j of the string are friends if they have the same character. The distance between two friends at positions i and j is defined as |i- j|. Find the sum of distances of all the pairs of friends in the given strings.First line of input contains a single string S.
Constraints:
1 <= |S| <= 1000000
String contains lowercase english letters.Output a single integer which is the sum of distance of all the pair of friends in the given strings.Sample Input
ababa
Sample Output
10
Explanation: Friend pairs - (1, 3) (1, 5) (2, 4) (3, 5), I have written this Solution Code: def findS(s):
visited= [ 0 for i in range(256)];
distance =[0 for i in range (256)];
for i in range(256):
visited[i]=0;
distance[i]=0;
sum=0;
for i in range(len(s)):
sum+=visited[ord(s[i])] * i - distance[ord(s[i])];
visited[ord(s[i])] +=1;
distance[ord(s[i])] +=i;
return sum;
if __name__ == '__main__':
s=input("");
print(findS(s));, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S. Two position i and j of the string are friends if they have the same character. The distance between two friends at positions i and j is defined as |i- j|. Find the sum of distances of all the pairs of friends in the given strings.First line of input contains a single string S.
Constraints:
1 <= |S| <= 1000000
String contains lowercase english letters.Output a single integer which is the sum of distance of all the pair of friends in the given strings.Sample Input
ababa
Sample Output
10
Explanation: Friend pairs - (1, 3) (1, 5) (2, 4) (3, 5), I have written this Solution Code: #pragma GCC optimize ("O3")
#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;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// 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
string s;
cin>>s;
int c[26]={};
int f[26]={};
int ans=0;
int n=s.length();
for(int i=0;i<n;++i){
ans+=f[s[i]-'a']*i-c[s[i]-'a'];
f[s[i]-'a']++;
c[s[i]-'a']+=i;
}
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: A permutation is simply a name for a reordering. So the permutations of the string
‘abc’ are ‘abc’, ‘acb’, ‘bac’, ‘bca’, ‘cab’, and ‘cba’. Note that a sequence is a
permutation of itself (the trivial permutation). For this problem, you’ll need
to write a recursive function that takes a string and returns a
list of all its permutations.
A couple of notes on the requirements:
1. The order of the returned permutations must be lexicographically.
2. Avoid returning duplicates in your final list.Input contains a single string S.
Constraints:-
1<=|S|<=8Print all the permutations of string S in lexicographical order.Sample Input:
ABC
Sample Output :
ABC ACB BAC BCA CAB CBA
Explanation:
all permutation are arranged in lexicographical order .
Sample Input:
(T(
Sample Output:-
((T (T( T((, I have written this Solution Code: def sol(arr):
dict = {}
for i in arr:
if i in dict.keys():
dict[i] = dict[i] + 1
else:
dict[i] = 1
keys = sorted(dict)
str = []
c = []
s=0
for key in keys:
str.append(key)
c.append(dict[key])
total = [0]*len(arr)
sol2(str, c, total, s)
def sol2(str, c, total, s):
if s == len(total):
str1=''
k=str1.join(total)
print(k,end=' ')
return
for i in range(len(str)):
if c[i] == 0:
continue
total[s] = str[i]
c[i] -= 1
sol2(str, c, total, s + 1)
c[i] += 1
str2=input()
n = list(str2)
sol(n), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A permutation is simply a name for a reordering. So the permutations of the string
‘abc’ are ‘abc’, ‘acb’, ‘bac’, ‘bca’, ‘cab’, and ‘cba’. Note that a sequence is a
permutation of itself (the trivial permutation). For this problem, you’ll need
to write a recursive function that takes a string and returns a
list of all its permutations.
A couple of notes on the requirements:
1. The order of the returned permutations must be lexicographically.
2. Avoid returning duplicates in your final list.Input contains a single string S.
Constraints:-
1<=|S|<=8Print all the permutations of string S in lexicographical order.Sample Input:
ABC
Sample Output :
ABC ACB BAC BCA CAB CBA
Explanation:
all permutation are arranged in lexicographical order .
Sample Input:
(T(
Sample Output:-
((T (T( T((, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.util.Arrays;
class Main
{
public static void main(String[] args)
{
Scanner sc= new Scanner(System.in);
String str = sc.next();
if(str.length()==1)
System.out.print(str);
else
permutations(str);
}
public static void permutations(String str)
{
char[] charstr = str.toCharArray();
Arrays.sort(charstr);
while (true)
{
System.out.print(new String(charstr) + " ");
if (!next_String(charstr)) {
break;
}
}
}
static void swap(char[] charstr, int i, int j)
{
char ch = charstr[i];
charstr[i] = charstr[j];
charstr[j] = ch;
}
static void reverse(char[] charstr, int start)
{
for (int i = start, j = charstr.length - 1; i < j; i++, j--) {
swap(charstr, i, j);
}
}
public static boolean next_String(char[] charstr)
{
int i = charstr.length - 1;
while (charstr[i - 1] >= charstr[i])
{
if (--i == 0) {
return false;
}
}
int j = charstr.length - 1;
while (j > i && charstr[j] <= charstr[i - 1]) {
j--;
}
swap(charstr, i - 1, j);
reverse(charstr, i);
return true;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A permutation is simply a name for a reordering. So the permutations of the string
‘abc’ are ‘abc’, ‘acb’, ‘bac’, ‘bca’, ‘cab’, and ‘cba’. Note that a sequence is a
permutation of itself (the trivial permutation). For this problem, you’ll need
to write a recursive function that takes a string and returns a
list of all its permutations.
A couple of notes on the requirements:
1. The order of the returned permutations must be lexicographically.
2. Avoid returning duplicates in your final list.Input contains a single string S.
Constraints:-
1<=|S|<=8Print all the permutations of string S in lexicographical order.Sample Input:
ABC
Sample Output :
ABC ACB BAC BCA CAB CBA
Explanation:
all permutation are arranged in lexicographical order .
Sample Input:
(T(
Sample Output:-
((T (T( T((, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
set<string> se;
// Function to find all Permutations of a given string
// containing all distinct characters
void permutations(string str, int n, string res)
{
// base condition (only one character is left in the string)
if (n == 1)
{
string s= res + str;
se.insert(s);
return;
}
// process each character of the remaining string
for (int i = 0; i < n; i++)
{
// push current character to the output string and recur
// for the remaining characters
permutations(str.substr(1), n - 1, res + str[0]);
// left rotate the string by 1 unit for next iteration
// to right rotate the string use reverse iterator
rotate(str.begin(), str.begin() + 1, str.end());
}
}
// Find all Permutations of a string
int main()
{
string s;
cin>>s;
string res="";
permutations(s, s.size(), res);
for(auto it=se.begin();it!=se.end();it++){
cout<<*it<<" ";
}
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: You are given an array of integers arr of size n, a window of size k. Your task is to find the sum of the maximum element from each window.The first line contains the input of n, k ie the size of the array, and the size of the window.
The next line contains the input of the array.
<b>Constraints</b>
1 <= k < n <= 10<sup>5</sup>
1 <= arr[i] <= 10<sup>5</sup>Print the single line containing the maximum sum.Sample Input 1:
5 3
1 2 3 4 5
Sample Output 1:
12
Sample Input 2:
6 2
2 3 1 7 8 3
Sample Output 2:
29, 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 srr[] = br.readLine().trim().split(" ");
int n = Integer.parseInt(srr[0]);
int k = Integer.parseInt(srr[1]);
String ksrr[] = br.readLine().trim().split(" ");
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] =Integer.parseInt(ksrr[i]);
}
long sum=0,ans=0;
for(int i=0;i<=n-k;i++){
int maxval=0;
for(int j=i;j<i+k;j++){
if(j>=n) break;
maxval=Math.max(maxval,arr[j]);
}
ans+=maxval;
}
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of integers arr of size n, a window of size k. Your task is to find the sum of the maximum element from each window.The first line contains the input of n, k ie the size of the array, and the size of the window.
The next line contains the input of the array.
<b>Constraints</b>
1 <= k < n <= 10<sup>5</sup>
1 <= arr[i] <= 10<sup>5</sup>Print the single line containing the maximum sum.Sample Input 1:
5 3
1 2 3 4 5
Sample Output 1:
12
Sample Input 2:
6 2
2 3 1 7 8 3
Sample Output 2:
29, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-01-18 01:39:40
**/
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
#define int long long
int sumWindow(vector<int> a, int n, int k) {
int result = 0;
for (int i = 0; i < n - k + 1; i++) {
int windowMax = INT_MIN;
for (int j = i; j < i + k; j++) {
windowMax = max(windowMax, a[j]);
}
debug(windowMax);
result += windowMax;
}
return result;
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n, k, sum;
cin >> n >> k;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int result = sumWindow(a, n, k);
cout << result << "\n";
debug(result);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Create a <code>Product</code> constructor function which should take three arguments: <code>name</code>, <code>price</code>, and <code>manufacturer</code>. These arguments should be used to set the initial values of the name, price, and manufacturer properties respectively.
<ol>
<li>Add a <code>getProductDetails</code> method that returns the string<code>"Name: {name}, Price: {price}, Manufacturer: {manufacturer}"</code> where {name},{price},{manufacturer} are the corresponding properties of the object</li>
<li> Now, create a private property, <code>tax</code> which will store the tax percentage. This property should not be directly accessible by the objects created using the <code>Product</code> constructor.
<li> Create a method <code>setTax</code>, which will take an argument <code>amount</code>, and assign its value to the <code>tax</code>. Create another method <code>getTax</code> method, which should return the <code>tax</code></li>
<li>Create a private method <code>calculateTotalPrice</code> which would return the total price which can be calculated as <code>price + price*(tax/100)</code></li>
<li>Create a method <code>getTotalPrice</code> which would call <code>calculateTotalPrice</code> function, and return the value which <code>calculateTotalPrice</code> returned</li>
<li>Create a method <code>deleteMethod</code> which would take <code>methodName</code> as argument and should delete the <code>methodName</code> function of the constructor.</li>
<li>All functions other than <code>calculateTotalPrice</code> should be accessible to the objects created using the <code>Product</code> constructor</li>
</ol>The <code>Product</code> constructor should take three arguments: <code>name</code> , <code>price</code>, and <code>manufacturer</code>.
The <code>setTax()</code> method within the constructor should take <code>amount</code> as an argument.
The <code>deleteMethod()</code> function within the constructor should take <code>methodName</code> as an argument.
Providing the functions with the input is handled internally.
Example:
Bulb,1000,Philips,10<code>getProductDetails</code> method should return string <code>"Name: name, Price: price, Manufacturer: manufacturer"</code>
<code>getTax</code> method should return the <code>tax</code>
<code>setTax</code> method should set the value of <code>tax</code> and need not return any value
<code>getTotalPrice</code> should return the total price which can be calculated by calling the <code>calculateTotalPrice</code> function
<code>calculateTotalPrice</code> function should return the total price calculated as <code>price + price*(tax/100)</code>
<code>deleteMethod(methodName)</code> should delete <code>methodName</code> function and need not return any valueconst arg1 = "Bulb";
const arg2 = "1000";
const arg3 = "Philips";
const arg4 = "5";
const myProduct = new Product(arg1, parseInt(arg2), arg3);
myProduct.setTax(parseInt(arg4));
console.log(myProduct.getProductDetails()); //prints "Name: Bulb, Price: 1000, Manufacturer: Philips"
console.log(myProduct.getTax()); //prints 5
console.log(myProduct.tax); //prints "undefined" as <code>tax</code> is private variable
console.log(myProduct.getTotalPrice()); //prints 1050
console.log(myProduct.calculateTotalPrice); //prints "undefined" as <code>calculateTotalPrice</code> is a private function
myProduct.deleteMethod("getProductDetails")
console.log(myProduct.getProductDetails); //prints "undefined" as <code>getProductDetails</code> function is deleted, I have written this Solution Code: function Product(name, price, manufacturer) {
let tax;
this.name = name;
this.price = price;
this.manufacturer = manufacturer;
this.getProductDetails = function() {
return `Name: ${this.name}, Price: ${this.price}, Manufacturer: ${this.manufacturer}`;
};
this.setTax = function(amount) {
tax = amount;
};
this.getTax = function() {
return tax;
};
const calculateTotalPrice = function() {
return this.price + (this.price * (tax / 100));
};
this.getTotalPrice = function() {
return calculateTotalPrice.bind(this)();
};
this.deleteMethod = function(methodName){
delete this[methodName];
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We have five cards with integers A, B, C, D, and E written on them, one on each card.
This set of five cards is called a <b>Full house</b> if and only if the following below condition is satisfied:
1. Any three cards with the same number written on them
2. Rest two cards with the same numbers written on them.
Determine whether the set is a <b>Full house</b>.The input consists of 5 space separated integers.
A B C D E
<b>Constraints</b>
1≤A, B, C, D, E≤13
Not all of A, B, C, D, and E are the same.
All values in input are integers.If the set is a Full house, print Yes; otherwise, print No.<b>Sample Input 1</b>
1 2 1 2 1
<b>Sample Output 1</b>
Yes
<b>Sample Input 2</b>
12 12 11 1 2
<b>Sample Output 2</b>
No, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
vector<int> a(5);
for(int i = 0; i < 5; i++) cin >> a[i];
sort(a.begin(), a.end());
if((a[0] == a[2] and a[3] == a[4]) or (a[0] == a[1] and a[2] == a[4])) 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: You are given a square matrix of size N*N. Initially all elements of this matrix are equal to 0. You are given Q queries. Each query consists of two integers, i and j (1 <= i, j <= N) wherein you increase the value of all elements in the i<sup>th</sup> row and j<sup>th</sup> column by 1. After doing this, for each query print the number of zeroes left in the matrix.The first line of the input consists of two integers N and Q.
The next Q lines each contains two integers i and j.
Constraints:
1 <= N, Q <= 10<sup>5</sup>
1 <= i, j <= NFor each query print the number of zeroes left in the matrix.Sample Input:
3 3
1 1
1 2
3 2
Sample Output:
4 2 1
Explaination:
Initially, the matrix will look like:
0 0 0
0 0 0
0 0 0
After the first query, the matrix will look something like this:
1 1 1
1 0 0
1 0 0
<b>Number of zeroes = 4</b>
After the second query, the matrix will look something like this:
1 1 1
1 1 0
1 1 0
<b>Number of zeroes = 2</b>
After the third query, the matrix will look something like this:
1 1 1
1 1 0
1 1 1
<b>Number of zeroes = 1</b>
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
signed main() {
int n, r = 0, c = 0;
cin >> n;
int k;
cin >> k;
int ans = n*n;
vector<int> row(n + 1), col(n + 1);
while(k--){
int i, j;
cin >> i >> j;
if(row[i] == 0) ans -= n - c, row[i] = 1, r++;
if(col[j] == 0) ans -= n - r, col[j] = 1, c++;
cout << ans << ' ';
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A and an integer K. Find the maximum for each and every contiguous subarray of size K.
Problem asked in Amazon, Flipkart.The first line of each test case contains a single integer N denoting the size of array and the size of subarray K. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.
Constraints:
1 ≤ N ≤ 10^5
1 ≤ K ≤ N
0 ≤ A[i] <= 10^5Print the maximum for every subarray of size K.Sample Input:
9 3
1 2 3 1 4 5 2 3 6
Sample Output:
3 3 4 5 5 5 6
Explanation:
Starting from the first subarray of size k = 3, we have 3 as maximum. Moving the window forward, maximum element are as 3, 4, 5, 5, 5 and 6., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void printMax(int arr[], int n, int k) {
int j, max;
for(int i = 0; i <= n - k; i++) {
max = arr[i];
for(j = 1; j < k; j++) {
if(arr[i + j] > max) {
max = arr[i + j];
}
}
System.out.print(max + " ");
}
}
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str1[] = br.readLine().trim().split(" ");
int n = Integer.parseInt(str1[0]);
int k = Integer.parseInt(str1[1]);
String str2[] = br.readLine().trim().split(" ");
int arr[] = new int[n];
for(int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(str2[i]);
}
printMax(arr, n ,k);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A and an integer K. Find the maximum for each and every contiguous subarray of size K.
Problem asked in Amazon, Flipkart.The first line of each test case contains a single integer N denoting the size of array and the size of subarray K. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.
Constraints:
1 ≤ N ≤ 10^5
1 ≤ K ≤ N
0 ≤ A[i] <= 10^5Print the maximum for every subarray of size K.Sample Input:
9 3
1 2 3 1 4 5 2 3 6
Sample Output:
3 3 4 5 5 5 6
Explanation:
Starting from the first subarray of size k = 3, we have 3 as maximum. Moving the window forward, maximum element are as 3, 4, 5, 5, 5 and 6., I have written this Solution Code: n,k=input().split()
n=int(n)
k=int(k)
arr=input().split()
for i in range(0,n):
arr[i]=int(arr[i])
m=max(arr[0:k])
for i in range(k-1,n):
if(arr[i] > m):
m=arr[i]
if(arr[i-k]==m):
m=max(arr[i-k+1:i+1])
print (m, end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A and an integer K. Find the maximum for each and every contiguous subarray of size K.
Problem asked in Amazon, Flipkart.The first line of each test case contains a single integer N denoting the size of array and the size of subarray K. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.
Constraints:
1 ≤ N ≤ 10^5
1 ≤ K ≤ N
0 ≤ A[i] <= 10^5Print the maximum for every subarray of size K.Sample Input:
9 3
1 2 3 1 4 5 2 3 6
Sample Output:
3 3 4 5 5 5 6
Explanation:
Starting from the first subarray of size k = 3, we have 3 as maximum. Moving the window forward, maximum element are as 3, 4, 5, 5, 5 and 6., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
// A Dequeue (Double ended queue) based method for printing maximum element of
// all subarrays of size k
void printKMax(int arr[], int n, int k)
{
// Create a Double Ended Queue, Qi that will store indexes of array elements
// The queue will store indexes of useful elements in every window and it will
// maintain decreasing order of values from front to rear in Qi, i.e.,
// arr[Qi.front[]] to arr[Qi.rear()] are sorted in decreasing order
std::deque<int> Qi(k);
/* Process first k (or first window) elements of array */
int i;
for (i = 0; i < k; ++i) {
// For every element, the previous smaller elements are useless so
// remove them from Qi
while ((!Qi.empty()) && arr[i] >= arr[Qi.back()])
Qi.pop_back(); // Remove from rear
// Add new element at rear of queue
Qi.push_back(i);
}
// Process rest of the elements, i.e., from arr[k] to arr[n-1]
for (; i < n; ++i) {
// The element at the front of the queue is the largest element of
// previous window, so print it
cout << arr[Qi.front()] << " ";
// Remove the elements which are out of this window
while ((!Qi.empty()) && Qi.front() <= i - k)
Qi.pop_front(); // Remove from front of queue
// Remove all elements smaller than the currently
// being added element (remove useless elements)
while ((!Qi.empty()) && arr[i] >= arr[Qi.back()])
Qi.pop_back();
// Add current element at the rear of Qi
Qi.push_back(i);
}
// Print the maximum element of last window
cout << arr[Qi.front()];
}
// Driver program to test above functions
int main()
{
int n,k;
cin>>n>>k;
int arr[n];
for(int i=0;i<n;i++){
cin>>arr[i];
}
printKMax(arr, n, k);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array <strong>arr[]</strong> of size <strong>N</strong> and an element <strong>X</strong>. The task is to find and print the indices of the given element if it is present in array if not then print “<strong>Not found</strong>” without quotes.
<strong>Note</strong>: The elements may be present more than once.The first line of input contains T, denoting the number of test cases.
The first line of each test case contains N and X, N is the size of array and X is an element. Second line contains elements of array space separated. If not present then print "Not found" without quotes
<strong>Constraints</strong>:
1 <= T <= 100
1 <= N, X <= 10000
1 <= arr[i] <= 100000For each test case in new line you need to print all the positions where you find the X separated by space.
Assume 0-indexingInput:
2
5 6
2 3 6 5 6
4 3
2 4 6 5
Output:
2 4
Not found, 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 t; cin >> t;
while(t--){
vector<int> v;
int n, x; cin >> n >> x;
for(int i = 1; i <= n; i++){
int p; cin >> p;
if(p == x)
v.push_back(i-1);
}
if(v.size() == 0)
cout << "Not found\n";
else{
for(auto i: v)
cout << i << " ";
cout << endl;
}
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array <strong>arr[]</strong> of size <strong>N</strong> and an element <strong>X</strong>. The task is to find and print the indices of the given element if it is present in array if not then print “<strong>Not found</strong>” without quotes.
<strong>Note</strong>: The elements may be present more than once.The first line of input contains T, denoting the number of test cases.
The first line of each test case contains N and X, N is the size of array and X is an element. Second line contains elements of array space separated. If not present then print "Not found" without quotes
<strong>Constraints</strong>:
1 <= T <= 100
1 <= N, X <= 10000
1 <= arr[i] <= 100000For each test case in new line you need to print all the positions where you find the X separated by space.
Assume 0-indexingInput:
2
5 6
2 3 6 5 6
4 3
2 4 6 5
Output:
2 4
Not found, I have written this Solution Code: def position(n,arr,x):
res = []
cnt = 0
for i in arr:
if(i == x):
res.append(cnt)
cnt += 1
return res
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array <strong>arr[]</strong> of size <strong>N</strong> and an element <strong>X</strong>. The task is to find and print the indices of the given element if it is present in array if not then print “<strong>Not found</strong>” without quotes.
<strong>Note</strong>: The elements may be present more than once.The first line of input contains T, denoting the number of test cases.
The first line of each test case contains N and X, N is the size of array and X is an element. Second line contains elements of array space separated. If not present then print "Not found" without quotes
<strong>Constraints</strong>:
1 <= T <= 100
1 <= N, X <= 10000
1 <= arr[i] <= 100000For each test case in new line you need to print all the positions where you find the X separated by space.
Assume 0-indexingInput:
2
5 6
2 3 6 5 6
4 3
2 4 6 5
Output:
2 4
Not found, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main (String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t =Integer.parseInt(read.readLine());
while(t-- > 0)
{
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
int x = Integer.parseInt(str[1]);
int arr[] = new int[n];
str = read.readLine().trim().split(" ");
for(int i = 0; i < n; i++)
arr[i] = Integer.parseInt(str[i]);
findPositions(arr, n, x);
}
}
static void findPositions(int arr[], int n, int x)
{
boolean flag = false;
StringBuffer sb = new StringBuffer();
for(int i = 0; i < n; i++)
{
if(arr[i] == x)
{
sb.append(i + " ");
flag = true;
}
}
if(flag ==true)
System.out.println(sb.toString());
else System.out.println("Not found");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given positive integer N, determine if there exist three <b>positive</b> integers a, b and c such that the following two conditions hold:
1. a + b + c = N
2. a ^ b ^ c = 0 where ^ denotes the bitwise XOR operation.
If there exist such a triple (a, b, c), print the lexicographically smallest one.
Else, print -1.The first line of input contains a single integer, T.
T lines follow, each containing a single integer, N.
<b>Constraints:</b>
1 <= T <= 10<sup>3</sup>
3 <= N <= 10<sup>18</sup>For each test, in a new line, print the lexicographically smallest triple (a, b, c) if it exists, else print -1.Sample Input:
3
3
6
12
Sample Output:
-1
1 2 3
2 4 6, I have written this Solution Code: for _ in range(int(input())):
n = int(input())
if n%2 or n<6:
print(-1)
continue
m = n//2
sl = 1
while m % 2 == 0:
m >>= 1
sl <<= 1
if n == sl*2:
print(-1)
continue
print(sl,n//2-sl,n//2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given positive integer N, determine if there exist three <b>positive</b> integers a, b and c such that the following two conditions hold:
1. a + b + c = N
2. a ^ b ^ c = 0 where ^ denotes the bitwise XOR operation.
If there exist such a triple (a, b, c), print the lexicographically smallest one.
Else, print -1.The first line of input contains a single integer, T.
T lines follow, each containing a single integer, N.
<b>Constraints:</b>
1 <= T <= 10<sup>3</sup>
3 <= N <= 10<sup>18</sup>For each test, in a new line, print the lexicographically smallest triple (a, b, c) if it exists, else print -1.Sample Input:
3
3
6
12
Sample Output:
-1
1 2 3
2 4 6, I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// #include <sys/resource.h>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// PRAGMAS (do these even work?)
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define f first
#define s second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
#define start_clock() auto start_time = std::chrono::high_resolution_clock::now();
#define measure() auto end_time = std::chrono::high_resolution_clock::now(); cerr << (end_time - start_time)/std::chrono::milliseconds(1) << "ms" << endl;
typedef long long ll;
typedef long double ld;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifdef LOCALY
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
// DEBUG FUNCTIONS END
// CUSTOM HASH TO SPEED UP UNORDERED MAP AND TO AVOID FORCED CLASHES
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count()); // FOR RANDOM NUMBER GENERATION
ll mod_exp(ll a, ll b, ll c)
{
ll res=1; a=a%c;
while(b>0)
{
if(b%2==1)
res=(res*a)%c;
b/=2;
a=(a*a)%c;
}
return res;
}
ll mymod(ll a,ll b)
{
return (((a = a%b) < 0) ? a + b : a);
}
ll gcdExtended(ll,ll,ll *,ll *);
ll modInverse(ll a, ll m)
{
ll x, y;
ll g = gcdExtended(a, m, &x, &y);
g++; //this line was added just to remove compiler warning
ll res = (x%m + m) % m;
return res;
}
ll gcdExtended(ll a, ll b, ll *x, ll *y)
{
if (a == 0)
{
*x = 0, *y = 1;
return b;
}
ll x1, y1;
ll gcd = gcdExtended(b%a, a, &x1, &y1);
*x = y1 - (b/a) * x1;
*y = x1;
return gcd;
}
struct Graph
{
vector<vector<int>> adj;
Graph(int n)
{
adj.resize(n+1);
}
void add_edge(int a, int b, bool directed = false)
{
adj[a].pb(b);
if(!directed) adj[b].pb(a);
}
};
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll t;
cin >> t;
REP(i, 0, t)
{
ll n;
cin >> n;
if(n&1) cout << "-1\n";
else
{
ll x = n/2;
vll bits;
REP(i, 0, 60)
{
if((x >> i)&1)
{
bits.pb(i);
}
}
if(bits.size() == 1)
{
cout << "-1\n";
}
else
{
ll a = (1ll << bits[0]);
cout << a << " " << x - a << " " << x << "\n";
}
}
}
return 0;
}
/*
1. Check borderline constraints. Can a variable you are dividing by be 0?
2. Use ll while using bitshifts
3. Do not erase from set while iterating it
4. Initialise everything
5. Read the task carefully, is something unique, sorted, adjacent, guaranteed??
6. DO NOT use if(!mp[x]) if you want to iterate the map later
7. Are you using i in all loops? Are the i's conflicting?
*/
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given positive integer N, determine if there exist three <b>positive</b> integers a, b and c such that the following two conditions hold:
1. a + b + c = N
2. a ^ b ^ c = 0 where ^ denotes the bitwise XOR operation.
If there exist such a triple (a, b, c), print the lexicographically smallest one.
Else, print -1.The first line of input contains a single integer, T.
T lines follow, each containing a single integer, N.
<b>Constraints:</b>
1 <= T <= 10<sup>3</sup>
3 <= N <= 10<sup>18</sup>For each test, in a new line, print the lexicographically smallest triple (a, b, c) if it exists, else print -1.Sample Input:
3
3
6
12
Sample Output:
-1
1 2 3
2 4 6, I have written this Solution Code:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
import static java.lang.Math.*;
import static java.lang.System.out;
public class Main {
void solve(){
long n= in.nextLong();
if(n%2==1){
sb.append(-1).append("\n");
}
else{
long cnt=0;
n=n/2;
while(n%2==0){
cnt++;
n=n>>1;
}
long x=(1L<<cnt);
long y=(1L<<cnt);
long z=0;
while(n!=0){
n=n>>1;
cnt++;
if((n&1)==1){
y+=(1L<<cnt);
z+=(1L<<cnt);
}
}
long[] a= new long[3];
a[0]=x;
a[1]=y;
a[2]=z;
sort(a);
if(a[0]!=0){
sb.append(a[0]).append(" ");
sb.append(a[1]).append(" ");
sb.append(a[2]).append("\n");
}
else{
sb.append(-1).append("\n");
}
}
}
FastReader in;
StringBuffer sb;
public static void main(String[] args) {
new Main().run();
}
void run(){
in= new FastReader();
start();
}
void start(){
sb= new StringBuffer();
for(int t=in.nextInt();t>0;t--) {
solve();
}
out.print(sb);
}
void swap( int i , int j) {
int tmp = i;
i = j;
j = tmp;
}
long power(long x, long y, long p) {
long res = 1;
x = x % p;
while (y > 0) {
if (y % 2 == 1)
res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
int lower_bound(long[] a, long x) {
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]>=x) r=m;
else l=m;
}
return r;
}
int upper_bound(long[] arr, int key) {
int i=0, j=arr.length-1;
if (arr[j]<=key) return j+1;
if(arr[i]>key) return i;
while (i<j){
int mid= (i+j)/2;
if(arr[mid]<=key){
i= mid+1;
}else{
j=mid;
}
}
return i;
}
void sort(long[] A){
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i){
long tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
int[] intArr(int n){
int[] res= new int[n];
for(int i=0;i<n;i++){
res[i]= in.nextInt();
}
return res;
}
long[] longArr(int n){
long[] res= new long[n];
for(int i=0;i<n;i++){
res[i]= in.nextLong();
}
return res;
}
boolean isDigitSumPalindrome(long N) {
long sum= sumOfDigits(String.valueOf(N));
long rev=0;
long org= sum;
while (sum!=0){
long d= sum%10;
rev = rev*10 +d;
sum /= 10;
}
return org == rev;
}
long sumOfDigits(String n){
long sum= 0;
for (char c: n.toCharArray()){
sum += Integer.parseInt(String.valueOf(c));
}
return sum;
}
long[] revArray(long[] arr) {
int n= arr.length;
int i=0, j=n-1;
while (i<j){
long temp= arr[i];
arr[i]= arr[j];
arr[j]= temp;
i++;
j--;
}
return arr;
}
long gcd(long a, long b){
if (b==0)
return a;
return gcd(b, a%b);
}
long lcm(long a,long b){
return (a*b)/gcd(a,b);
}
static class Pair implements Comparable<Pair>{
long first;
long second;
Pair(long x, long y){
this.first=x;
this.second=y;
}
@Override
public int compareTo(Pair o) {
return 0;
}
}
public 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 (Exception e){
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
float nextFloat(){
return Float.parseFloat(next());
}
String nextLine(){
String str="";
try{
str=br.readLine();
}catch (Exception e){
e.printStackTrace();
}
return str;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ketani is fond of super simple problems. This time he has come up with another simple problem. Given two non negative numbers A and B. Find the number of multiples of k in the range [A, B] (all integers between A and B, A and B inclusive).The first and the only line of input contains three integers: A B k.
Constraints
0 <= a <= b <= 1000000000000000000 (10^18)
1 <= x <= 1000000000000000000 (10^18)Output a single integer, the number of multiples of k in the given range.Sample Input
4 8 2
Sample Output
3
Explanation: The divisors of 2 in the range [4, 8] are 4, 6, and 8.
Sample Input:
0 5 1
Sample Output:
6, I have written this Solution Code: a,b,k = list(map(int,input().strip().split()))
if a == 0:
count = 1
a =1
else:
count = 0
if a%k == 0 :
count += b//k - a//k +1
else:
count += b//k - a//k
print(count)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ketani is fond of super simple problems. This time he has come up with another simple problem. Given two non negative numbers A and B. Find the number of multiples of k in the range [A, B] (all integers between A and B, A and B inclusive).The first and the only line of input contains three integers: A B k.
Constraints
0 <= a <= b <= 1000000000000000000 (10^18)
1 <= x <= 1000000000000000000 (10^18)Output a single integer, the number of multiples of k in the given range.Sample Input
4 8 2
Sample Output
3
Explanation: The divisors of 2 in the range [4, 8] are 4, 6, and 8.
Sample Input:
0 5 1
Sample Output:
6, 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[] inputs = br.readLine().split(" ");
long A = Long.parseLong(inputs[0]);
long B = Long.parseLong(inputs[1]);
long k = Long.parseLong(inputs[2]);
long minMultiple = A/k;
long maxMultiple = B/k;
long compensattion = 0;
if(A % k == 0){
compensattion = 1;
}
long answer = maxMultiple+compensattion-minMultiple;
System.out.println(answer);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ketani is fond of super simple problems. This time he has come up with another simple problem. Given two non negative numbers A and B. Find the number of multiples of k in the range [A, B] (all integers between A and B, A and B inclusive).The first and the only line of input contains three integers: A B k.
Constraints
0 <= a <= b <= 1000000000000000000 (10^18)
1 <= x <= 1000000000000000000 (10^18)Output a single integer, the number of multiples of k in the given range.Sample Input
4 8 2
Sample Output
3
Explanation: The divisors of 2 in the range [4, 8] are 4, 6, and 8.
Sample Input:
0 5 1
Sample Output:
6, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
// Driver code
int main()
{
unsigned long long a,b,k,p,q,s=0;
cin>>a>>b>>k;
p=a/k;
if(a%k==0){p--;}
q=b/k;
cout<<max(s,q-p);
}
, In this Programming Language: C++, 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 an array A[] and a range [a, b]. The task is to partition the array around the range such that array is divided into three parts.
1) All elements smaller than a come first.
2) All elements in range a to b come next.
3) All elements greater than b appear in the end.
The individual elements of three sets can appear in any order. You are required to return the modified arranged array.
<b>Note:-</b>
In the case of custom input, you will get 1 if your code is correct else get a 0.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>threeWayPartition()</b> which contains following arguments.
A: input array list
low: starting integer of range
high: ending integer of range
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10<sup>4</sup>
1 <= A[i] <= 10<sup>5</sup>
1 <= low <= high <= 10<sup>5</sup>
The Sum of N over all test case doesn't exceed 10^5For each test case return the modified array.Sample Input:
2
5
1 8 3 3 4
3 5
3
1 2 3
1 3
Sample Output:
1 3 3 4 8
1 2 3
<b>Explanation:</b>
Testcase 1: First, the array has elements less than or equal to 3. Then, elements between 3 and 5. And, finally elements greater than 5. So, one of the possible outputs is 1 3 3 4 8.
Testcase 2: First, the array has elements less than or equal to 1. Then, elements between 1 and 3. And, finally elements greater than 3. So, the output is 1 2 3., I have written this Solution Code:
public static ArrayList<Integer> threeWayPartition(ArrayList<Integer> A, int lowVal, int highVal)
{
int n = A.size();
ArrayList<Integer> arr = A;
int start = 0, end = n-1;
for (int i=0; i<=end;)
{
// swapping the element with those at start
// if array element is less than lowVal
if (arr.get(i) < lowVal){
int temp=arr.get(i);
arr.add(i,arr.get(start));
arr.remove(i+1);
arr.add(start,temp);
arr.remove(start+1);
i++;
start++;
}
// swapping the element with those at end
// if array element is greater than highVal
else if (arr.get(i) > highVal){
int temp=arr.get(i);
arr.add(i,arr.get(end));
arr.remove(i+1);
arr.add(end,temp);
arr.remove(end+1);
end--;
}
// else just move ahead
else
i++;
}
return arr;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] and a range [a, b]. The task is to partition the array around the range such that array is divided into three parts.
1) All elements smaller than a come first.
2) All elements in range a to b come next.
3) All elements greater than b appear in the end.
The individual elements of three sets can appear in any order. You are required to return the modified arranged array.
<b>Note:-</b>
In the case of custom input, you will get 1 if your code is correct else get a 0.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>threeWayPartition()</b> which contains following arguments.
A: input array list
low: starting integer of range
high: ending integer of range
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10<sup>4</sup>
1 <= A[i] <= 10<sup>5</sup>
1 <= low <= high <= 10<sup>5</sup>
The Sum of N over all test case doesn't exceed 10^5For each test case return the modified array.Sample Input:
2
5
1 8 3 3 4
3 5
3
1 2 3
1 3
Sample Output:
1 3 3 4 8
1 2 3
<b>Explanation:</b>
Testcase 1: First, the array has elements less than or equal to 3. Then, elements between 3 and 5. And, finally elements greater than 5. So, one of the possible outputs is 1 3 3 4 8.
Testcase 2: First, the array has elements less than or equal to 1. Then, elements between 1 and 3. And, finally elements greater than 3. So, the output is 1 2 3., I have written this Solution Code: def threewayPartition(arr,low,high):
i=low-1
pivot = arr[high]
for j in range(low,high):
if arr[j]<=pivot:
i+=1
arr[i],arr[j]=arr[j],arr[i]
i+=1
arr[i],arr[high]=arr[high],arr[i]
return i
def quickSort(arr,low,high):
if low<high:
key=threewayPartition(arr,low,high)
quickSort(arr,low,key-1)
quickSort(arr,key+1,high)
return arr
t= int(input())
while t>0:
s = int(input())
arr = list(map(int,input().split()))
a,b = list(map(int,input().split()))
print(1)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N*N dimensions. Take the input and display the matrix as an output.First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Print the given matrix.Input:
2
3 4
7 6
Output:
3 4
7 6
Input:
3
1 2 3
4 5 6
7 8 9
Output:
1 2 3
4 5 6
7 8 9, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main (String[] args)throws IOException {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int mat[][] = new int[N][N];
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++){
mat[i][j] = sc.nextInt();
}
}
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++){
System.out.print(mat[i][j]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |