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Provide tags and a correct Python 3 solution for this coding contest problem.
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string s consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string s[p1p2... pk] = sp1sp2... spk (1 ≤ p1 < p2 < ... < pk ≤ |s|) a subsequence of string s = s1s2... s|s|, where |s| is the length of string s. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String x = x1x2... x|x| is lexicographically larger than string y = y1y2... y|y| if either |x| > |y| and x1 = y1, x2 = y2, ..., x|y| = y|y|, or there exists such number r (r < |x|, r < |y|) that x1 = y1, x2 = y2, ..., xr = yr and xr + 1 > yr + 1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String s = s1s2... s|s| is a palindrome if it matches string rev(s) = s|s|s|s| - 1... s1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input
The only input line contains a non-empty string s consisting of lowercase English letters only. Its length does not exceed 10.
Output
Print the lexicographically largest palindromic subsequence of string s.
Examples
Input
radar
Output
rr
Input
bowwowwow
Output
wwwww
Input
codeforces
Output
s
Input
mississipp
Output
ssss
Note
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
|
instruction
| 0
| 15,346
| 0
| 30,692
|
Tags: binary search, bitmasks, brute force, greedy, implementation, strings
Correct Solution:
```
a = input()
print( max(a) * a.count(max(a)))
```
|
output
| 1
| 15,346
| 0
| 30,693
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string s consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string s[p1p2... pk] = sp1sp2... spk (1 ≤ p1 < p2 < ... < pk ≤ |s|) a subsequence of string s = s1s2... s|s|, where |s| is the length of string s. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String x = x1x2... x|x| is lexicographically larger than string y = y1y2... y|y| if either |x| > |y| and x1 = y1, x2 = y2, ..., x|y| = y|y|, or there exists such number r (r < |x|, r < |y|) that x1 = y1, x2 = y2, ..., xr = yr and xr + 1 > yr + 1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String s = s1s2... s|s| is a palindrome if it matches string rev(s) = s|s|s|s| - 1... s1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input
The only input line contains a non-empty string s consisting of lowercase English letters only. Its length does not exceed 10.
Output
Print the lexicographically largest palindromic subsequence of string s.
Examples
Input
radar
Output
rr
Input
bowwowwow
Output
wwwww
Input
codeforces
Output
s
Input
mississipp
Output
ssss
Note
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
|
instruction
| 0
| 15,347
| 0
| 30,694
|
Tags: binary search, bitmasks, brute force, greedy, implementation, strings
Correct Solution:
```
#!/usr/bin/python3.5
s=input()
l=list(s)
l.sort()
k=l[-1]
for i in l[::-1]:
if i!=k:
break
else:
print(k,end='')
```
|
output
| 1
| 15,347
| 0
| 30,695
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string s consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string s[p1p2... pk] = sp1sp2... spk (1 ≤ p1 < p2 < ... < pk ≤ |s|) a subsequence of string s = s1s2... s|s|, where |s| is the length of string s. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String x = x1x2... x|x| is lexicographically larger than string y = y1y2... y|y| if either |x| > |y| and x1 = y1, x2 = y2, ..., x|y| = y|y|, or there exists such number r (r < |x|, r < |y|) that x1 = y1, x2 = y2, ..., xr = yr and xr + 1 > yr + 1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String s = s1s2... s|s| is a palindrome if it matches string rev(s) = s|s|s|s| - 1... s1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input
The only input line contains a non-empty string s consisting of lowercase English letters only. Its length does not exceed 10.
Output
Print the lexicographically largest palindromic subsequence of string s.
Examples
Input
radar
Output
rr
Input
bowwowwow
Output
wwwww
Input
codeforces
Output
s
Input
mississipp
Output
ssss
Note
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
|
instruction
| 0
| 15,348
| 0
| 30,696
|
Tags: binary search, bitmasks, brute force, greedy, implementation, strings
Correct Solution:
```
y=input()
print( max(y)*y.count(max(y)) )
```
|
output
| 1
| 15,348
| 0
| 30,697
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string s consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string s[p1p2... pk] = sp1sp2... spk (1 ≤ p1 < p2 < ... < pk ≤ |s|) a subsequence of string s = s1s2... s|s|, where |s| is the length of string s. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String x = x1x2... x|x| is lexicographically larger than string y = y1y2... y|y| if either |x| > |y| and x1 = y1, x2 = y2, ..., x|y| = y|y|, or there exists such number r (r < |x|, r < |y|) that x1 = y1, x2 = y2, ..., xr = yr and xr + 1 > yr + 1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String s = s1s2... s|s| is a palindrome if it matches string rev(s) = s|s|s|s| - 1... s1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input
The only input line contains a non-empty string s consisting of lowercase English letters only. Its length does not exceed 10.
Output
Print the lexicographically largest palindromic subsequence of string s.
Examples
Input
radar
Output
rr
Input
bowwowwow
Output
wwwww
Input
codeforces
Output
s
Input
mississipp
Output
ssss
Note
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
Submitted Solution:
```
s=input()
ch=(max(s))
t=s.count(ch)
for i in range (t):
print(ch,end='')
```
|
instruction
| 0
| 15,349
| 0
| 30,698
|
Yes
|
output
| 1
| 15,349
| 0
| 30,699
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string s consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string s[p1p2... pk] = sp1sp2... spk (1 ≤ p1 < p2 < ... < pk ≤ |s|) a subsequence of string s = s1s2... s|s|, where |s| is the length of string s. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String x = x1x2... x|x| is lexicographically larger than string y = y1y2... y|y| if either |x| > |y| and x1 = y1, x2 = y2, ..., x|y| = y|y|, or there exists such number r (r < |x|, r < |y|) that x1 = y1, x2 = y2, ..., xr = yr and xr + 1 > yr + 1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String s = s1s2... s|s| is a palindrome if it matches string rev(s) = s|s|s|s| - 1... s1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input
The only input line contains a non-empty string s consisting of lowercase English letters only. Its length does not exceed 10.
Output
Print the lexicographically largest palindromic subsequence of string s.
Examples
Input
radar
Output
rr
Input
bowwowwow
Output
wwwww
Input
codeforces
Output
s
Input
mississipp
Output
ssss
Note
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
Submitted Solution:
```
def solve(s):
s = sorted(s, reverse=True)
res = s[0]
c = s[0]
for i in range(1,len(s)):
if s[i] == c:
res += s[i]
else:
break
return res
def main():
s = input()
print(solve(s))
main()
```
|
instruction
| 0
| 15,350
| 0
| 30,700
|
Yes
|
output
| 1
| 15,350
| 0
| 30,701
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string s consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string s[p1p2... pk] = sp1sp2... spk (1 ≤ p1 < p2 < ... < pk ≤ |s|) a subsequence of string s = s1s2... s|s|, where |s| is the length of string s. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String x = x1x2... x|x| is lexicographically larger than string y = y1y2... y|y| if either |x| > |y| and x1 = y1, x2 = y2, ..., x|y| = y|y|, or there exists such number r (r < |x|, r < |y|) that x1 = y1, x2 = y2, ..., xr = yr and xr + 1 > yr + 1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String s = s1s2... s|s| is a palindrome if it matches string rev(s) = s|s|s|s| - 1... s1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input
The only input line contains a non-empty string s consisting of lowercase English letters only. Its length does not exceed 10.
Output
Print the lexicographically largest palindromic subsequence of string s.
Examples
Input
radar
Output
rr
Input
bowwowwow
Output
wwwww
Input
codeforces
Output
s
Input
mississipp
Output
ssss
Note
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
Submitted Solution:
```
s = input()
m = max(s)
n = sum(map(lambda x: x==m, s))
print(m*n)
```
|
instruction
| 0
| 15,351
| 0
| 30,702
|
Yes
|
output
| 1
| 15,351
| 0
| 30,703
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string s consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string s[p1p2... pk] = sp1sp2... spk (1 ≤ p1 < p2 < ... < pk ≤ |s|) a subsequence of string s = s1s2... s|s|, where |s| is the length of string s. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String x = x1x2... x|x| is lexicographically larger than string y = y1y2... y|y| if either |x| > |y| and x1 = y1, x2 = y2, ..., x|y| = y|y|, or there exists such number r (r < |x|, r < |y|) that x1 = y1, x2 = y2, ..., xr = yr and xr + 1 > yr + 1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String s = s1s2... s|s| is a palindrome if it matches string rev(s) = s|s|s|s| - 1... s1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input
The only input line contains a non-empty string s consisting of lowercase English letters only. Its length does not exceed 10.
Output
Print the lexicographically largest palindromic subsequence of string s.
Examples
Input
radar
Output
rr
Input
bowwowwow
Output
wwwww
Input
codeforces
Output
s
Input
mississipp
Output
ssss
Note
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
Submitted Solution:
```
import sys
from functools import reduce
from collections import Counter
import time
import datetime
from math import sqrt,gcd
# def time_t():
# print("Current date and time: " , datetime.datetime.now())
# print("Current year: ", datetime.date.today().strftime("%Y"))
# print("Month of year: ", datetime.date.today().strftime("%B"))
# print("Week number of the year: ", datetime.date.today().strftime("%W"))
# print("Weekday of the week: ", datetime.date.today().strftime("%w"))
# print("Day of year: ", datetime.date.today().strftime("%j"))
# print("Day of the month : ", datetime.date.today().strftime("%d"))
# print("Day of week: ", datetime.date.today().strftime("%A"))
def ip(): return int(sys.stdin.readline())
def sip(): return sys.stdin.readline()
def mip(): return map(int,sys.stdin.readline().split())
def mips(): return map(str,sys.stdin.readline().split())
def lip(): return list(map(int,sys.stdin.readline().split()))
def matip(n,m):
lst=[]
for i in range(n):
arr = lip()
lst.insert(i,arr)
return lst
def factors(n): # find the factors of a number
return list(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def minJumps(arr, n): #to reach from 0 to n-1 in the array in minimum steps
jumps = [0 for i in range(n)]
if (n == 0) or (arr[0] == 0):
return float('inf')
jumps[0] = 0
for i in range(1, n):
jumps[i] = float('inf')
for j in range(i):
if (i <= j + arr[j]) and (jumps[j] != float('inf')):
jumps[i] = min(jumps[i], jumps[j] + 1)
break
return jumps[n-1]
def dic(arr): # converting list into dict of count
return Counter(arr)
def check_prime(n):
if n<2:
return False
for i in range(2,int(n**(0.5))+1,2):
if n%i==0:
return False
return True
# --------------------------------------------------------- #
# sys.stdin = open('input.txt','r')
# sys.stdout = open('output.txt','w')
# --------------------------------------------------------- #
s = sip()
lst = []
for item in s:
lst.append(item)
lst.sort()
val = lst.count(lst[-1])
print(lst[-1]*val)
```
|
instruction
| 0
| 15,352
| 0
| 30,704
|
Yes
|
output
| 1
| 15,352
| 0
| 30,705
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string s consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string s[p1p2... pk] = sp1sp2... spk (1 ≤ p1 < p2 < ... < pk ≤ |s|) a subsequence of string s = s1s2... s|s|, where |s| is the length of string s. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String x = x1x2... x|x| is lexicographically larger than string y = y1y2... y|y| if either |x| > |y| and x1 = y1, x2 = y2, ..., x|y| = y|y|, or there exists such number r (r < |x|, r < |y|) that x1 = y1, x2 = y2, ..., xr = yr and xr + 1 > yr + 1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String s = s1s2... s|s| is a palindrome if it matches string rev(s) = s|s|s|s| - 1... s1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input
The only input line contains a non-empty string s consisting of lowercase English letters only. Its length does not exceed 10.
Output
Print the lexicographically largest palindromic subsequence of string s.
Examples
Input
radar
Output
rr
Input
bowwowwow
Output
wwwww
Input
codeforces
Output
s
Input
mississipp
Output
ssss
Note
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
Submitted Solution:
```
s=input()
max=0
for i in s:
if ord(i)>max:
max=ord(i)
x=i
a=s.count(x)
print((chr(max))*2)
```
|
instruction
| 0
| 15,353
| 0
| 30,706
|
No
|
output
| 1
| 15,353
| 0
| 30,707
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string s consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string s[p1p2... pk] = sp1sp2... spk (1 ≤ p1 < p2 < ... < pk ≤ |s|) a subsequence of string s = s1s2... s|s|, where |s| is the length of string s. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String x = x1x2... x|x| is lexicographically larger than string y = y1y2... y|y| if either |x| > |y| and x1 = y1, x2 = y2, ..., x|y| = y|y|, or there exists such number r (r < |x|, r < |y|) that x1 = y1, x2 = y2, ..., xr = yr and xr + 1 > yr + 1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String s = s1s2... s|s| is a palindrome if it matches string rev(s) = s|s|s|s| - 1... s1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input
The only input line contains a non-empty string s consisting of lowercase English letters only. Its length does not exceed 10.
Output
Print the lexicographically largest palindromic subsequence of string s.
Examples
Input
radar
Output
rr
Input
bowwowwow
Output
wwwww
Input
codeforces
Output
s
Input
mississipp
Output
ssss
Note
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
Submitted Solution:
```
from itertools import chain, combinations
def fun(iterable):
lst = list(iterable)
return chain.from_iterable(combinations(lst, n) for n in range(len(lst) + 1))
s = input()
t = ''
for elem in fun(s):
x, y = ''.join(s), ''.join(s[::-1])
if x == y and x > t:
t = x
print(t)
```
|
instruction
| 0
| 15,354
| 0
| 30,708
|
No
|
output
| 1
| 15,354
| 0
| 30,709
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string s consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string s[p1p2... pk] = sp1sp2... spk (1 ≤ p1 < p2 < ... < pk ≤ |s|) a subsequence of string s = s1s2... s|s|, where |s| is the length of string s. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String x = x1x2... x|x| is lexicographically larger than string y = y1y2... y|y| if either |x| > |y| and x1 = y1, x2 = y2, ..., x|y| = y|y|, or there exists such number r (r < |x|, r < |y|) that x1 = y1, x2 = y2, ..., xr = yr and xr + 1 > yr + 1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String s = s1s2... s|s| is a palindrome if it matches string rev(s) = s|s|s|s| - 1... s1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input
The only input line contains a non-empty string s consisting of lowercase English letters only. Its length does not exceed 10.
Output
Print the lexicographically largest palindromic subsequence of string s.
Examples
Input
radar
Output
rr
Input
bowwowwow
Output
wwwww
Input
codeforces
Output
s
Input
mississipp
Output
ssss
Note
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
Submitted Solution:
```
from collections import Counter
c = Counter(input())
for letter, count in c.most_common(1):
print(letter*count)
```
|
instruction
| 0
| 15,355
| 0
| 30,710
|
No
|
output
| 1
| 15,355
| 0
| 30,711
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string s consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string s[p1p2... pk] = sp1sp2... spk (1 ≤ p1 < p2 < ... < pk ≤ |s|) a subsequence of string s = s1s2... s|s|, where |s| is the length of string s. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String x = x1x2... x|x| is lexicographically larger than string y = y1y2... y|y| if either |x| > |y| and x1 = y1, x2 = y2, ..., x|y| = y|y|, or there exists such number r (r < |x|, r < |y|) that x1 = y1, x2 = y2, ..., xr = yr and xr + 1 > yr + 1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String s = s1s2... s|s| is a palindrome if it matches string rev(s) = s|s|s|s| - 1... s1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input
The only input line contains a non-empty string s consisting of lowercase English letters only. Its length does not exceed 10.
Output
Print the lexicographically largest palindromic subsequence of string s.
Examples
Input
radar
Output
rr
Input
bowwowwow
Output
wwwww
Input
codeforces
Output
s
Input
mississipp
Output
ssss
Note
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr".
Submitted Solution:
```
s = input()
lista = []
for i in s:
lista.append(i)
for i in range(len(s)):
for j in range(i+1, len(s)):
lista.append(s[i]+s[j])
for k in range(j+1, len(s)):
lista.append(s[i]+s[j]+s[k])
for p in range(k+1, len(s)):
lista.append(s[i]+s[j]+s[k]+s[p])
for m in range(p+1, len(s)):
lista.append(s[i]+s[j]+s[k]+s[p]+s[m])
for n in range(m+1, len(s)):
lista.append(s[i]+s[j]+s[k]+s[p]+s[m]+s[n])
for o in range(n+1, len(s)):
lista.append(s[i]+s[j]+s[k]+s[p]+s[m]+s[n]+s[o])
for w in range(o+1, len(s)):
lista.append(s[i]+s[j]+s[k]+s[p]+s[m]+s[n]+s[o]+s[w])
lista.sort()
print(lista[-1])
```
|
instruction
| 0
| 15,356
| 0
| 30,712
|
No
|
output
| 1
| 15,356
| 0
| 30,713
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
|
instruction
| 0
| 15,391
| 0
| 30,782
|
Tags: dp, strings
Correct Solution:
```
#Not actually my code, solved mentally, only submitting to mark as solved so I don't do it again.
def gen(i, j):
if a[i][j] == -1:
a[i][j] = max(gen(i - 1, j - 1) + s2[i - 1] * (s2[i - 1] == s1[j - 1]), gen(i - 1, j), gen(i, j - 1), key = lambda x: [len(x), -x.count(viru)])
return a[i][j]
s1, s2, viru = [input() for x in range(3)]
a = [[''] * (len(s1) + 1)] + [[''] + [-1] * (len(s1)) for x in range(len(s2))]
ans = gen(len(s2), len(s1))
while viru in ans:
ans = min(ans.replace(viru, viru[:-1]), ans.replace(viru, viru[1:]), key = lambda x: x.count(viru))
print(ans + '0' * (ans == ''))
```
|
output
| 1
| 15,391
| 0
| 30,783
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
|
instruction
| 0
| 15,392
| 0
| 30,784
|
Tags: dp, strings
Correct Solution:
```
def dfs(i, j):
if a[i][j] != -1:
return a[i][j]
d1 = dfs(i - 1, j - 1) + (s1[i - 1] == s2[j - 1]) * s1[i - 1]
d2 = dfs(i - 1, j)
d3 = dfs(i, j - 1)
a[i][j] = max(d1, d2, d3, key=lambda x: [len(x), -x.count(virus)])
return a[i][j]
s1, s2, virus = input(), input(), input()
#s1, s2, virus = 'SLO', 'LL', 'asda'
n1, n2 = len(s1), len(s2)
a = [[''] * (n2 + 1)] + [[''] + [-1] * n2 for i in range(n1)]
#a[1][0] = 'hjkhj'
ans = dfs(n1, n2)
while virus in ans:
ans = min(ans.replace(virus, virus[:-1]), ans.replace(virus, virus[1:]), key=lambda x: x.count(virus))
print(0 if ans == '' else ans)
```
|
output
| 1
| 15,392
| 0
| 30,785
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
|
instruction
| 0
| 15,393
| 0
| 30,786
|
Tags: dp, strings
Correct Solution:
```
s1, s2, virus = [input().strip() for _ in range(3)]
n, m, v = len(s1), len(s2), len(virus)
# preprocessing (inneficient, but |virus| is small)
next_suffix = [v - 1 for i in range(v - 1)] + [v]
for i in range(v - 1):
j = v - i - 2
while j > 0:
if virus[i + 1:i + 1 + j] == virus[-j:]:
next_suffix[i] = v - j - 1
break
j -= 1
def memoize(function):
memo = dict()
def memoized(*args):
if args in memo:
return memo[args]
ans = function(*args)
memo[args] = ans
return ans
return memoized
@memoize
def lcs(i, j, k):
if k < 0:
return (float("-inf"), "")
if i < 0 or j < 0:
return (0, "")
if s1[i] == s2[j]:
if s1[i] != virus[k]:
newk = k
while newk < v and virus[newk] != s1[i]:
newk = next_suffix[newk]
r = lcs(i - 1, j - 1, newk - 1) # this is wrong, cant reset to v-1
return (r[0] + 1, r[1] + s1[i])
else:
r1 = lcs(i - 1, j - 1, k - 1)
r1 = (r1[0] + 1, r1[1] + s1[i])
r2 = lcs(i - 1, j - 1, k)
return max(r1, r2, key=lambda x: x[0])
return max(lcs(i, j - 1, k), lcs(i - 1, j, k), key=lambda x: x[0])
ans = lcs(n - 1, m - 1, v - 1)
if ans[0] > 0:
print(ans[1])
else:
print(0)
```
|
output
| 1
| 15,393
| 0
| 30,787
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
|
instruction
| 0
| 15,394
| 0
| 30,788
|
Tags: dp, strings
Correct Solution:
```
a, b, v = input(), input(), input()
h = [[-1] * len(b) for i in range(len(a))]
def f(ab):
return ab.count(v)
def g(i, j):
if i < 0 or j < 0:
return ""
if h[i][j] == -1:
s = g(i - 1, j - 1)
if (a[i] == b[j]):
s += a[i]
s1 = g(i - 1, j)
s2 = g(i, j - 1)
if (len(s1) - f(s1) > len(s) - f(s)):
s = s1
if (len(s2) - f(s2) > len(s) - f(s)):
s = s2
h[i][j] = s;
return h[i][j]
ss = g(len(a) - 1, len(b) - 1)
while v in ss :
ss1 = ss.replace(v, v[:-1])
ss2 = ss.replace(v, v[1:])
if (f(ss1) < f(ss2)):
ss = ss1
else:
ss = ss2
if len(ss) > 0:
print(ss)
else:
print(0)
```
|
output
| 1
| 15,394
| 0
| 30,789
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
|
instruction
| 0
| 15,395
| 0
| 30,790
|
Tags: dp, strings
Correct Solution:
```
def Solve(x,y,c):
if(c==len(virus)):
return 0
if(x==len(s1) or y==len(s2)):
return ""
if((x,y,c) in Mem):
return Mem[(x,y,c)]
ans=""
if(s1[x]==s2[y]):
q=0
if(s1[x]==virus[c]):
q=Solve(x+1,y+1,c+1)
if(q!=0):
q=s1[x]+q
else:
done=False
for match in T[c]:
if(s1[x]==virus[match]):
done=True
q=Solve(x+1,y+1,match+1)
if(q!=0):
q=s1[x]+q
break
if(not done):
q=Solve(x+1,y+1,0)
if(q!=0):
q=s1[x]+q
if(q!=0):
ans=q
q=Solve(x+1,y,c)
if(q!=0):
ans=max(ans,Solve(x+1,y,c),key=len)
q=Solve(x,y+1,c)
if(q!=0):
ans=max(ans,Solve(x,y+1,c),key=len)
Mem[(x,y,c)]=ans
return ans
Mem={}
s1=input()
s2=input()
virus=input()
T=[0]*len(virus)
T[0]=[0]
for i in range(1,len(virus)):
done=False
T[i]=[]
for j in range(1,i):
if(virus[j:i]==virus[:i-j]):
T[i].append(i-j)
T[i].append(0)
e=Solve(0,0,0)
if(e==""):
print(0)
else:
print(Solve(0,0,0))
```
|
output
| 1
| 15,395
| 0
| 30,791
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
|
instruction
| 0
| 15,396
| 0
| 30,792
|
Tags: dp, strings
Correct Solution:
```
s1, s2, virus = [input().strip() for _ in range(3)]
n, m, v = len(s1), len(s2), len(virus)
# preprocessing (inneficient, but |virus| is small)
next_suffix = [v - 1 for i in range(v - 1)] + [v]
for i in range(v - 1):
j = v - i - 2
while j > 0:
if virus[i + 1:i + 1 + j] == virus[-j:]:
next_suffix[i] = v - j - 1
break
j -= 1
def memoize(function):
memo = dict()
def memoized(*args):
if args in memo:
return memo[args]
ans = function(*args)
memo[args] = ans
return ans
return memoized
@memoize
def lcs(i, j, k):
if k < 0:
return (float("-inf"), "")
if i < 0 or j < 0:
return (0, "")
if s1[i] == s2[j]:
if s1[i] != virus[k]:
newk = k
while newk < v and virus[newk] != s1[i]:
newk = next_suffix[newk]
r = lcs(i - 1, j - 1, newk - 1)
return (r[0] + 1, r[1] + s1[i])
else:
r1 = lcs(i - 1, j - 1, k - 1)
r1 = (r1[0] + 1, r1[1] + s1[i])
r2 = lcs(i - 1, j - 1, k)
return max(r1, r2, key=lambda x: x[0])
return max(lcs(i, j - 1, k), lcs(i - 1, j, k), key=lambda x: x[0])
ans = lcs(n - 1, m - 1, v - 1)
if ans[0] > 0:
print(ans[1])
else:
print(0)
# 1540312184891
```
|
output
| 1
| 15,396
| 0
| 30,793
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
|
instruction
| 0
| 15,397
| 0
| 30,794
|
Tags: dp, strings
Correct Solution:
```
s1, s2, virus = [input().strip() for _ in range(3)]
n, m, v = len(s1), len(s2), len(virus)
next_suffix = [v for i in range(v)]
for i in range(v-2, -1, -1):
j = next_suffix[i + 1]
while j <= v-1 and virus[j] != virus[i + 1]:
j = next_suffix[j]
next_suffix[i] = j - 1
# print(virus, next_suffix)
# j = v - 1 - i - 1
# while j > 0:
# if virus[i + 1:i + 1 + j] == virus[-j:]:
# next_suffix[i] = v - j - 1
# break
# j -= 1
def memoize(function):
memo = dict()
def memoized(*args):
if args in memo:
return memo[args]
ans = function(*args)
memo[args] = ans
return ans
return memoized
@memoize
def lcs(i, j, k):
if k < 0:
return (float("-inf"), "")
if i < 0 or j < 0:
return (0, "")
if s1[i] == s2[j]:
if s1[i] != virus[k]:
newk = k
while newk < v and virus[newk] != s1[i]:
newk = next_suffix[newk]
r = lcs(i - 1, j - 1, newk - 1)
return (r[0] + 1, r[1] + s1[i])
else:
r1 = lcs(i - 1, j - 1, k - 1)
r1 = (r1[0] + 1, r1[1] + s1[i])
r2 = lcs(i - 1, j - 1, k)
return max(r1, r2, key=lambda x: x[0])
return max(lcs(i, j - 1, k), lcs(i - 1, j, k), key=lambda x: x[0])
ans = lcs(n - 1, m - 1, v - 1)
if ans[0] > 0:
print(ans[1])
else:
print(0)
```
|
output
| 1
| 15,397
| 0
| 30,795
|
Provide tags and a correct Python 3 solution for this coding contest problem.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
|
instruction
| 0
| 15,398
| 0
| 30,796
|
Tags: dp, strings
Correct Solution:
```
a, b, v = input(), input(), input()
t = [[-1] * len(b) for x in range(len(a))]
def g(i, j):
if i < 0 or j < 0: return ''
if t[i][j] == -1:
s = g(i - 1, j - 1)
if a[i] == b[j]: s += a[i]
t[i][j] = max(s, g(i - 1, j), g(i, j - 1), key=lambda q: len(q) - q.count(v))
return t[i][j]
s = g(len(a) - 1, len(b) - 1)
while v in s: s = min(s.replace(v, v[:-1]), s.replace(v, v[1:]), key=lambda q: q.count(v))
print(s if s else 0)
# Made By Mostafa_Khaled
```
|
output
| 1
| 15,398
| 0
| 30,797
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
Submitted Solution:
```
A = input()
B = input()
virus = input()
# DP[i][j][k] maximum length of common subsquence you can construct with A[i:], B[j:], having exactly k letter of the virus so far
# DP[0][0][0]
piTable = [0] * len(virus)
# ERROR 1 : this error though :(((
j = 0
for i in range(1, len(virus)):
while (virus[i] != virus[j] and j != 0):
j = piTable[j - 1]
if virus[i] == virus[j]:
piTable[i] = j + 1
j += 1
else:
piTable[i] = 0
memo = {}
parent = {}
def dp(i, j, k):
# print(i, j, k)
if k == len(virus):
return -10000
if i == len(A):
return 0
if j == len(B):
return 0
if (i, j, k) in memo:
return memo[(i, j, k)]
choices = [dp(i + 1, j, k), dp(i, j + 1, k)]
new_k = k
if A[i] == B[j]:
if A[i] == virus[k]:
choices.append(dp(i + 1, j + 1, k + 1) + 1)
new_k = k + 1
else:
# A[i] = j
# k=6, virus=abcjabcxxxxx virus[k]=x
# A[i] != virus[k]
while (A[i] != virus[new_k] and new_k != 0):
new_k = piTable[new_k - 1]
if A[i] == virus[new_k]:
# ERROR 2 : messed up with new_k and k, as well forgot updating new_k ...
new_k += 1
choices.append(dp(i + 1, j + 1, new_k) + 1)
else:
choices.append(dp(i + 1, j + 1, 0) + 1)
memo[(i, j, k)] = max(choices)
index = choices.index(memo[(i, j, k)])
if index == 0:
parent[(i, j, k)] = (i + 1, j, k)
elif index == 1:
parent[(i, j, k)] = (i, j + 1, k)
else:
parent[(i, j, k)] = (i + 1, j + 1, new_k)
# print(i,j,k)
return memo[(i, j, k)]
max_size_subsequence = dp(0, 0, 0)
if max_size_subsequence == 0:
print(0)
else:
# construct answer
ANSWER = ""
current_state = (0, 0, 0)
next_state = None
while current_state != None:
next_state = parent[current_state] if current_state in parent else None
if next_state is None:
break
# (i,j,k) -> (i+1,j+1,k')
if next_state[0] == current_state[0] + 1 and next_state[1] == current_state[1] + 1:
ANSWER += A[current_state[0]]
current_state = next_state
print(ANSWER)
```
|
instruction
| 0
| 15,399
| 0
| 30,798
|
Yes
|
output
| 1
| 15,399
| 0
| 30,799
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
Submitted Solution:
```
a, b, s = input(), input(), input()
i = 0
p = [-1] + [0] * len(s)
for j in range(1, len(s)):
while i + 1 and s[i] != s[j]: i = p[i]
i += 1
p[j + 1] = i
f = lambda a, b: a if len(a) > len(b) else b
g = lambda: [[''] * len(s) for i in range(len(b) + 1)]
u, v = g(), g()
for x in a:
for i, y in enumerate(b):
v[i] = [f(x, y) for x, y in zip(u[i], v[i - 1])]
if x == y:
for k, n in enumerate(u[i - 1]):
while k > -1 and x != s[k]: k = p[k]
k += 1
if k < len(s): v[i][k] = f(v[i][k], n + x)
u, v = v, g()
t = max(u[-2], key=lambda q: len(q))
print(t if t else 0)
```
|
instruction
| 0
| 15,400
| 0
| 30,800
|
Yes
|
output
| 1
| 15,400
| 0
| 30,801
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
Submitted Solution:
```
a, b, v = input(), input(), input()
t = [[-1] * len(b) for x in range(len(a))]
def g(i, j):
if i < 0 or j < 0: return ''
if t[i][j] == -1:
s = g(i - 1, j - 1)
if a[i] == b[j]: s += a[i]
t[i][j] = max(s, g(i - 1, j), g(i, j - 1), key=lambda q: len(q) - q.count(v))
return t[i][j]
s = g(len(a) - 1, len(b) - 1)
while v in s: s = min(s.replace(v, v[:-1]), s.replace(v, v[1:]), key=lambda q: q.count(v))
print(s if s else 0)
```
|
instruction
| 0
| 15,401
| 0
| 30,802
|
Yes
|
output
| 1
| 15,401
| 0
| 30,803
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
Submitted Solution:
```
s1, s2, virus = [input().strip() for _ in range(3)]
n, m, v = len(s1), len(s2), len(virus)
# preprocessing (inneficient, but |virus| is small)
next_suffix = [v - 1 for i in range(v - 1)] + [v]
for i in range(v - 1):
j = v - 1 - i - 1
while j > 0:
if virus[i + 1:i + 1 + j] == virus[-j:]:
next_suffix[i] = v - j - 1
break
j -= 1
def memoize(function):
memo = dict()
def memoized(*args):
if args in memo:
return memo[args]
ans = function(*args)
memo[args] = ans
return ans
return memoized
@memoize
def lcs(i, j, k):
if k < 0:
return (float("-inf"), "")
if i < 0 or j < 0:
return (0, "")
if s1[i] == s2[j]:
if s1[i] != virus[k]:
newk = k
while newk < v and virus[newk] != s1[i]:
newk = next_suffix[newk]
r = lcs(i - 1, j - 1, newk - 1)
return (r[0] + 1, r[1] + s1[i])
else:
r1 = lcs(i - 1, j - 1, k - 1)
r1 = (r1[0] + 1, r1[1] + s1[i])
r2 = lcs(i - 1, j - 1, k)
return max(r1, r2, key=lambda x: x[0])
return max(lcs(i, j - 1, k), lcs(i - 1, j, k), key=lambda x: x[0])
ans = lcs(n - 1, m - 1, v - 1)
if ans[0] > 0:
print(ans[1])
else:
print(0)
```
|
instruction
| 0
| 15,402
| 0
| 30,804
|
Yes
|
output
| 1
| 15,402
| 0
| 30,805
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
Submitted Solution:
```
def backtrk(arr, l1, l2, lcs, s1, s2):
while l1 > 0 and l2 > 0:
if s2[l1-1] == s1[l2-1]:
lcs.append(s2[l1-1])
l1 -= 1
l2 -= 1
elif arr[l1-1][l2] > arr[l1][l2-1]:
l1 -= 1
else:
l2 -= 1
s1 = list(input())
s2 = list(input())
virus = list(input())
arr, lcs= [], []
arr.append([0 for i in range(len(s1)+1)])
for i in range(1, len(s2)+1):
p = [0]*(len(s1)+1)
for j in range(1, len(s1)+1):
if(s2[i-1] == s1[j-1]):
p[j] = 1 + arr[i-1][j-1]
else:
p[j] = max(arr[i-1][j], p[j-1])
arr.append(p)
#print(arr[len(s2)][len(s1)])
if arr[len(s2)][len(s1)]:
backtrk(arr, len(s2), len(s1), lcs, s1, s2)
lcs.reverse()
if len(virus) <= len(lcs):
for i in range(len(lcs)):
count = 0
if lcs[i] == virus[0]:
for j in range(0, len(virus)):
if(i+j) < len(lcs) and (lcs[i+j] == virus[j]):
count += 1
else:
break
if(count == len(virus)):
lcs[i] = '0'
flag = True
for i in range(len(lcs)):
if lcs[i] != '0':
print(lcs[i], end='')
flag = False
if flag:
print("0")
```
|
instruction
| 0
| 15,403
| 0
| 30,806
|
No
|
output
| 1
| 15,403
| 0
| 30,807
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
Submitted Solution:
```
def LCS(x, y):
m = len(x)-1
n = len(y)-1
b = [0]*(m+1)
c = [0]*(m+1)
for i in range(m+1):
c[i] = [0]*(n+1)
b[i] = [0]*(n+1)
for i in range(1, m+1):
for j in range(1, n+1):
if x[i] == y[j]:
c[i][j] = c[i-1][j-1] + 1
b[i][j] = 1
elif c[i-1][j] >= c[i][j-1]:
c[i][j] = c[i-1][j]
b[i][j] = 2
else:
c[i][j] = c[i][j-1]
b[i][j] = 3
return c, b
def print_LCS(b, x, i, j):
global z
if i == 0 or j == 0:
return
if b[i][j] == 1:
print_LCS(b, x, i-1, j-1)
z = z + x[i]
elif b[i][j] == 2:
print_LCS(b, x, i-1, j)
else:
print_LCS(b, x, i, j-1)
s1 = input()
s2 = input()
v = input()
c, b = LCS('1'+s1, '1'+s2)
ans = ''
for i in range(1, len(s1)+1):
for j in range(1, len(s2)+1):
z=''; print_LCS(b, '1'+s1, i, j)
z = z.replace(v, '')
if len(z) > len(ans):
ans = z
if len(ans) > 0:
print(ans)
else:
print(0)
```
|
instruction
| 0
| 15,404
| 0
| 30,808
|
No
|
output
| 1
| 15,404
| 0
| 30,809
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
Submitted Solution:
```
# Why do we fall ? So we can learn to pick ourselves up.
def lcs(s1,s2):
m = len(s1)
n = len(s2)
gar = [[0]*(n+1) for i in range(0,m+1)]
for i in range(0,m+1):
for j in range(0,n+1):
if i == 0 or j == 0:
gar[i][j] = 0
elif s1[i-1] == s2[j-1]:
gar[i][j] = gar[i-1][j-1] + 1
else:
gar[i][j] = max(gar[i-1][j],gar[i][j-1])
ss = ""
i,j = m,n
while i > 0 and j > 0:
if s1[i-1] == s2[j-1]:
ss += s1[i-1]
i -= 1
j -= 1
elif gar[i-1][j] > gar[i][j-1]:
i -= 1
else:
j -= 1
return ss[::-1]
s1 = input()
s2 = input()
virus = input()
ss = lcs(s1,s2)
ind = ss.find(virus)
if ind != -1:
ss = ss[:ind] + ss[ind+len(virus):]
print([ss,0][len(ss) == 0])
"""
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
"""
```
|
instruction
| 0
| 15,405
| 0
| 30,810
|
No
|
output
| 1
| 15,405
| 0
| 30,811
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. For example, the sequence BDF is a subsequence of ABCDEF. A substring of a string is a continuous subsequence of the string. For example, BCD is a substring of ABCDEF.
You are given two strings s1, s2 and another string called virus. Your task is to find the longest common subsequence of s1 and s2, such that it doesn't contain virus as a substring.
Input
The input contains three strings in three separate lines: s1, s2 and virus (1 ≤ |s1|, |s2|, |virus| ≤ 100). Each string consists only of uppercase English letters.
Output
Output the longest common subsequence of s1 and s2 without virus as a substring. If there are multiple answers, any of them will be accepted.
If there is no valid common subsequence, output 0.
Examples
Input
AJKEQSLOBSROFGZ
OVGURWZLWVLUXTH
OZ
Output
ORZ
Input
AA
A
A
Output
0
Submitted Solution:
```
a, b, s = input(), input(), input()
i = 0
p = [-1] + [0] * len(s)
for j in range(1, len(s)):
while i + 1 and s[i] != s[j]: i = p[i]
i += 1
p[j + 1] = i
d = [[] for i in b] + [[]]
u, v = d[:], d[:]
for x in a:
for i, y in enumerate(b):
r = u[i] + v[i - 1]
if x == y:
for n, k in u[i - 1]:
while k > -1 and x != s[k]: k = p[k]
if k < len(s): r.append((n + x, k + 1))
r.append((x, s[0] == x))
if not r: continue
r.sort(key=lambda q: len(q[0]))
n, k = r.pop()
while r:
q = r.pop()
if len(q[0]) < len(n):
v[i].append((n, k))
n = q[0]
k = min(k, q[1])
v[i].append((n, k))
u, v = v, d[:]
print(max(u[-2], key=lambda q: len(q[0])) if u[-2] else 0)
```
|
instruction
| 0
| 15,406
| 0
| 30,812
|
No
|
output
| 1
| 15,406
| 0
| 30,813
|
Provide a correct Python 3 solution for this coding contest problem.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0
|
instruction
| 0
| 15,747
| 0
| 31,494
|
"Correct Solution:
```
def search_rolling_hash(S, T):
base = 1234
mod = 10**9+7
n, m = len(S), len(T)
rolling_hash = [0]*(n-m+1)
rolling_hash[0] = ord(S[0])
b_m = pow(base, m, mod)
T_hash = ord(T[0])
start_index = []
for i in range(1, m):
rolling_hash[0]*=base
rolling_hash[0]+=ord(S[i])
rolling_hash[0]%=mod
T_hash = (T_hash*base+ord(T[i]))%mod
if rolling_hash[0] == T_hash: start_index.append(0)
for i in range(1, ls):
rolling_hash[i] = (rolling_hash[i-1]*base-ord(S[i-1])*b_m+ord(S[i+m-1]))%mod
if rolling_hash[i] == T_hash: start_index.append(i)
return start_index
import sys
s = input()
t = input()
ls, lt = len(s), len(t)
S = s*(lt//ls+2)
index = search_rolling_hash(S, t)
flag = [False]*ls
for i in index:
flag[i] = True
stack = []
inf = float('inf')
cnt = [inf]*ls
for i in range(ls):
if flag[i] == False:
stack.append(i)
cnt[i] = 0
while stack:
x = stack.pop()
y = (x-lt)%ls
if cnt[y] == inf:
cnt[y] = cnt[x]+1
stack.append(y)
ans = max(cnt)
if ans == inf:
print(-1)
else:
print(ans)
```
|
output
| 1
| 15,747
| 0
| 31,495
|
Provide a correct Python 3 solution for this coding contest problem.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0
|
instruction
| 0
| 15,748
| 0
| 31,496
|
"Correct Solution:
```
S = input()
W = input()
M = len(S)
S *= 3 * (len(W)//len(S)+1)
N = len(W)
def primeFactor(N):
i, n, ret, d, sq = 2, N, {}, 2, 99
while i <= sq:
k = 0
while n % i == 0: n, k, ret[i] = n//i, k+1, k+1
if k > 0 or i == 97: sq = int(n**(1/2)+0.5)
if i < 4: i = i * 2 - 1
else: i, d = i+d, d^6
if n > 1: ret[n] = 1
return ret
def divisors(N):
pf = primeFactor(N)
ret = [1]
for p in pf:
ret_prev = ret
ret = []
for i in range(pf[p]+1):
for r in ret_prev:
ret.append(r * (p ** i))
return sorted(ret)
D = divisors(N)
p = 1
for d in D:
if W[:-d] == W[d:]:
W = W[:d]
p = N//d
N = d
break
W = W[:d]
T = [-1] * (len(W)+1)
ii = 2
jj = 0
T[0] = -1
T[1] = 0
while ii < len(W) + 1:
if W[ii - 1] == W[jj]:
T[ii] = jj + 1
ii += 1
jj += 1
elif jj > 0:
jj = T[jj]
else:
T[ii] = 0
ii += 1
m = 0
def KMP(i0):
global m, i
ret = -1
while m + i < len(S):
if W[i] == S[m + i]:
i += 1
if i == len(W):
t = m
m = m + i - T[i]
if i > 0:
i = T[i]
return t
else:
m = m + i - T[i]
if i > 0:
i = T[i]
return len(S)
i = 0
j = -10**9
c = 0
cmax = 0
f = 0
while m < len(S)-N:
k = KMP(0)
if k == len(S):
break
elif k == N + j:
c += 1
else:
c = 1
f += 1
cmax = max(cmax, c)
j = k
if f == 1:
print(-1)
else:
print(cmax//p)
```
|
output
| 1
| 15,748
| 0
| 31,497
|
Provide a correct Python 3 solution for this coding contest problem.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0
|
instruction
| 0
| 15,749
| 0
| 31,498
|
"Correct Solution:
```
# -*- coding: utf-8 -*-
import math
import sys
from collections import defaultdict
buff_readline = sys.stdin.buffer.readline
readline = sys.stdin.readline
INF = 2**60
class RollingHash():
"""
Original code is https://tjkendev.github.io/procon-library/python/string/rolling_hash.html
"""
class RH():
def __init__(self, s, base, mod):
self.base = base
self.mod = mod
self.rev = pow(base, mod-2, mod)
l = len(s)
self.h = h = [0]*(l+1)
tmp = 0
for i in range(l):
num = ord(s[i])
tmp = (tmp*base + num) % mod
h[i+1] = tmp
self.pw = pw = [1]*(len(s)+1)
v = 1
for i in range(l):
pw[i+1] = v = v * base % mod
def calc(self, l, r):
return (self.h[r] - self.h[l] * self.pw[r-l]) % self.mod
@staticmethod
def gen(a, b, num):
import random
result = set()
while 1:
while 1:
v = random.randint(a, b)//2*2+1
if v not in result:
break
for x in range(3, int(math.sqrt(v))+1, 2):
if v % x == 0:
break
else:
result.add(v)
if len(result) == num:
break
return result
def __init__(self, s, rand=False, num=5):
if rand:
bases = RollingHash.gen(2, 10**3, num)
else:
assert num <= 10
bases = [641, 103, 661, 293, 547, 311, 29, 457, 613, 599][:num]
MOD = 10**9+7
self.rhs = [self.RH(s, b, MOD) for b in bases]
def calc(self, l, r):
return tuple(rh.calc(l, r) for rh in self.rhs)
class UnionFind():
def __init__(self):
self.__table = {}
self.__size = defaultdict(lambda: 1)
self.__rank = defaultdict(lambda: 1)
def __root(self, x):
if x not in self.__table:
self.__table[x] = x
elif x != self.__table[x]:
self.__table[x] = self.__root(self.__table[x])
return self.__table[x]
def same(self, x, y):
return self.__root(x) == self.__root(y)
def union(self, x, y):
x = self.__root(x)
y = self.__root(y)
if x == y:
return False
if self.__rank[x] < self.__rank[y]:
self.__table[x] = y
self.__size[y] += self.__size[x]
else:
self.__table[y] = x
self.__size[x] += self.__size[y]
if self.__rank[x] == self.__rank[y]:
self.__rank[x] += 1
return True
def size(self, x):
return self.__size[self.__root(x)]
def num_of_group(self):
g = 0
for k, v in self.__table.items():
if k == v:
g += 1
return g
def slv(S, T):
lt = len(T)
ls = len(S)
SS = S * (-(-lt // ls) * 2 + 1)
srh = RollingHash(SS, num=2)
th = RollingHash(T, num=2).calc(0, len(T))
ok = []
for i in range(ls+lt):
if srh.calc(i, i+lt) == th:
ok.append(i)
ok = set(ok)
uf = UnionFind()
for i in ok:
if (i + lt) in ok:
if not uf.union(i, (i+lt) % ls):
return -1
ans = 0
for i in ok:
ans = max(ans, uf.size(i))
return ans
def main():
S = input()
T = input()
print(slv(S, T))
if __name__ == '__main__':
main()
```
|
output
| 1
| 15,749
| 0
| 31,499
|
Provide a correct Python 3 solution for this coding contest problem.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0
|
instruction
| 0
| 15,750
| 0
| 31,500
|
"Correct Solution:
```
import sys
sys.setrecursionlimit(10**7) #再帰関数の上限,10**5以上の場合python
import math
from copy import copy, deepcopy
from copy import deepcopy as dcp
from operator import itemgetter
from bisect import bisect_left, bisect, bisect_right#2分探索
#bisect_left(l,x), bisect(l,x)#aはソート済みである必要あり。aの中からx未満の要素数を返す。rightだと以下
from collections import deque, defaultdict
#deque(l), pop(), append(x), popleft(), appendleft(x)
#q.rotate(n)で → にn回ローテート
from collections import Counter#文字列を個数カウント辞書に、
#S=Counter(l),S.most_common(x),S.keys(),S.values(),S.items()
from itertools import accumulate,combinations,permutations,product#累積和
#list(accumulate(l))
from heapq import heapify,heappop,heappush
#heapify(q),heappush(q,a),heappop(q) #q=heapify(q)としないこと、返り値はNone
from functools import reduce,lru_cache#pypyでもうごく
#@lru_cache(maxsize = None)#maxsizeは保存するデータ数の最大値、2**nが最も高効率
from decimal import Decimal
import random
def input():
x=sys.stdin.readline()
return x[:-1] if x[-1]=="\n" else x
def printe(*x):print("## ",*x,file=sys.stderr)
def printl(li): _=print(*li, sep="\n") if li else None
def argsort(s, return_sorted=False):
inds=sorted(range(len(s)), key=lambda k: s[k])
if return_sorted: return inds, [s[i] for i in inds]
return inds
def alp2num(c,cap=False): return ord(c)-97 if not cap else ord(c)-65
def num2alp(i,cap=False): return chr(i+97) if not cap else chr(i+65)
def matmat(A,B):
K,N,M=len(B),len(A),len(B[0])
return [[sum([(A[i][k]*B[k][j]) for k in range(K)]) for j in range(M)] for i in range(N)]
def matvec(M,v):
N,size=len(v),len(M)
return [sum([M[i][j]*v[j] for j in range(N)]) for i in range(size)]
def T(M):
n,m=len(M),len(M[0])
return [[M[j][i] for j in range(n)] for i in range(m)]
def binr(x): return bin(x)[2:]
def bitcount(x): #xは64bit整数
x= x - ((x >> 1) & 0x5555555555555555)
x= (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)
x= (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f
x+= (x >> 8); x+= (x >> 16); x+= (x >> 32)
return x & 0x7f
def main():
mod = 1000000007
#w.sort(key=itemgetter(1),reverse=True) #二個目の要素で降順並び替え
#N = int(input())
#N, K = map(int, input().split())
#A = tuple(map(int, input().split())) #1行ベクトル
#L = tuple(int(input()) for i in range(N)) #改行ベクトル
#S = tuple(tuple(map(int, input().split())) for i in range(N)) #改行行列
class rh:#base,max_n,convertを指定すること
def __init__(self,base=26,max_n=5*10**5):#baseとmaxnを指定すること。baseは偶数(modが奇数)
self.mod=random.randrange(1<<54-1,1<<55-1,2)#奇数、衝突させられないよう乱数で生成
self.base=int(base*0.5+0.5)*2#偶数,英小文字なら26とか
self.max_n=max_n
self.pows1=[1]*(max_n+1)
cur1=1
for i in range(1,max_n+1):
cur1=cur1*base%self.mod
self.pows1[i]=cur1
def convert(self,c): return alp2num(c,False)#大文字の場合Trueにする
def get(self, s):#特定の文字のhash. O(|s|)
n=len(s)
mod=self.mod
si=[self.convert(ss) for ss in s]
h1=0
for i in range(n):h1=(h1+si[i]*self.pows1[n-1-i])%mod
return h1
def get_roll(self, s, k):#ローリングハッシュ,特定の文字の中から長さkのhashをすべて得る,O(|s|)
n=len(s)
si=[self.convert(ss) for ss in s]
mod=self.mod
h1=0
for i in range(k):h1=(h1+si[i%n]*self.pows1[k-1-i])%mod
hs=[0]*n
hs[0]=h1
mbase1=self.pows1[k]
base=self.base
for i in range(1,n):
front=si[i-1];back=si[(i+k-1)%n]
h1=(h1*base-front*mbase1+back)%mod
hs[i]=h1
return hs
s=input()
t=input()
n=len(s)
nt=len(t)
rhs=rh()
has=rhs.get(t)
hass=rhs.get_roll(s,nt)
l=[hass[i]==has for i in range(n)]
ds=[0]*n
for i in range(n):
ci=i
q=deque()
while l[ci] and ds[ci]==0:
q.append(ci)
ci=(ci+nt)%n
if ci==i:
print(-1)
return
tot=ds[ci]
while q:
tot+=1
ci=q.pop()
ds[ci]=tot
print(max(ds))
if __name__ == "__main__":
main()
```
|
output
| 1
| 15,750
| 0
| 31,501
|
Provide a correct Python 3 solution for this coding contest problem.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0
|
instruction
| 0
| 15,752
| 0
| 31,504
|
"Correct Solution:
```
def main():
import sys
input = sys.stdin.readline
S = input().rstrip('\n')
T = input().rstrip('\n')
ns = len(S)
nt = len(T)
mod1 = 1000000007
mod2 = 1000000009
mod3 = 998244353
b1 = 1007
b2 = 2009
b3 = 2999
H1 = 0
H2 = 0
H3 = 0
for i, t in enumerate(T):
t = ord(t)
H1 = ((H1 * b1)%mod1 + t)%mod1
H2 = ((H2 * b2)%mod2 + t)%mod2
H3 = ((H3 * b3)%mod3 + t)%mod3
h1 = 0
h2 = 0
h3 = 0
for i in range(nt):
s = ord(S[i%ns])
h1 = ((h1 * b1) % mod1 + s) % mod1
h2 = ((h2 * b2) % mod2 + s) % mod2
h3 = ((h3 * b3) % mod3 + s) % mod3
idx = []
if h1 == H1 and h2 == H2 and h3 == H3:
idx.append(0)
b1_nt = pow(b1, nt, mod1)
b2_nt = pow(b2, nt, mod2)
b3_nt = pow(b3, nt, mod3)
for i in range(1, ns):
s = ord(S[(i+nt-1)%ns])
s_prev = ord(S[i-1])
h1 = ((h1 * b1) % mod1 - (b1_nt * s_prev)%mod1 + s) % mod1
h2 = ((h2 * b2) % mod2 - (b2_nt * s_prev)%mod2 + s) % mod2
h3 = ((h3 * b3) % mod3 - (b3_nt * s_prev) % mod3 + s) % mod3
if h1 == H1 and h2 == H2 and h3 == H3:
idx.append(i%ns)
if len(idx) == 0:
print(0)
exit()
ans = 1
dic = {}
for i in range(len(idx)):
if (idx[i] - nt)%ns in dic:
tmp = dic[(idx[i]-nt)%ns] + 1
dic[idx[i]] = tmp
ans = max(ans, tmp)
else:
dic[idx[i]] = 1
for i in range(len(idx)):
if (idx[i] - nt)%ns in dic:
tmp = dic[(idx[i]-nt)%ns] + 1
dic[idx[i]] = tmp
ans = max(ans, tmp)
else:
dic[idx[i]] = 1
ans_ = ans
for i in range(len(idx)):
if (idx[i] - nt)%ns in dic:
tmp = dic[(idx[i]-nt)%ns] + 1
dic[idx[i]] = tmp
ans = max(ans, tmp)
else:
dic[idx[i]] = 1
if ans == ans_:
print(ans)
else:
print(-1)
if __name__ == '__main__':
main()
```
|
output
| 1
| 15,752
| 0
| 31,505
|
Provide a correct Python 3 solution for this coding contest problem.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0
|
instruction
| 0
| 15,753
| 0
| 31,506
|
"Correct Solution:
```
def kmp(s):
n = len(s)
kmp = [0] * (n+1)
kmp[0] = -1
j = -1
for i in range(n):
while j >= 0 and s[i] != s[j]:
j = kmp[j]
j += 1
kmp[i+1] = j
return kmp
from collections import deque
def topological_sort(graph: list, n_v: int) -> list:
# graph[node] = [(cost, to)]
indegree = [0] * n_v # 各頂点の入次数
for i in range(n_v):
for v in graph[i]:
indegree[v] += 1
cand = deque([i for i in range(n_v) if indegree[i] == 0])
res = []
while cand:
v1 = cand.popleft()
res.append(v1)
for v2 in graph[v1]:
indegree[v2] -= 1
if indegree[v2] == 0:
cand.append(v2)
return res
import sys
def main():
input = sys.stdin.readline
s = input().rstrip()
t = input().rstrip()
n0 = len(s)
m = len(t)
s = s * (m // n0 + 2)
res = kmp(t + '*' + s)
res = res[m+2:]
graph = [set() for _ in range(n0)]
for i in range(len(res)):
if res[i] >= m:
graph[(i-m+1)%n0].add( (i + 1) % n0 )
# print(graph)
ts = topological_sort(graph, n0)
if len(ts) != n0:
ans = -1
else:
# print(ts)
ans = 0
path = [0] * n0
for i in range(n0):
ans = max(ans, path[ts[i]])
for j in graph[ts[i]]:
path[j] = path[ts[i]] + 1
print(ans)
main()
```
|
output
| 1
| 15,753
| 0
| 31,507
|
Provide a correct Python 3 solution for this coding contest problem.
Given are two strings s and t consisting of lowercase English letters. Determine if the number of non-negative integers i satisfying the following condition is finite, and find the maximum value of such i if the number is finite.
* There exists a non-negative integer j such that the concatenation of i copies of t is a substring of the concatenation of j copies of s.
Constraints
* 1 \leq |s| \leq 5 \times 10^5
* 1 \leq |t| \leq 5 \times 10^5
* s and t consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
s
t
Output
If the number of non-negative integers i satisfying the following condition is finite, print the maximum value of such i; if the number is infinite, print `-1`.
Examples
Input
abcabab
ab
Output
3
Input
aa
aaaaaaa
Output
-1
Input
aba
baaab
Output
0
|
instruction
| 0
| 15,754
| 0
| 31,508
|
"Correct Solution:
```
def partial_match_table(word):
table = [0] * (len(word) + 1)
table[0] = -1
i, j = 0, 1
while j < len(word):
matched = word[i] == word[j]
if not matched and i > 0:
i = table[i]
else:
if matched:
i += 1
j += 1
table[j] = i
return table
def kmp_search(text, word):
table = partial_match_table(word)
i, p = 0, 0
results = []
while i < len(text) and p < len(word):
if text[i] == word[p]:
i += 1
p += 1
if p == len(word):
p = table[p]
results.append((i-len(word), i))
elif p == 0:
i += 1
else:
p = table[p]
return results
inf = 10**18
s = input().strip()
t = input().strip()
m=(len(t)+len(s)-1+len(s)-1)//len(s)
d = [-1] * (len(s)+1)
for a, b in kmp_search(m*s, t):
a, b = a%len(s)+1, b%len(s)+1
d[a] = b
ls = [-1]*(len(s)+1)
vs = set()
for i in range(1, len(s)+1):
if i in vs:
continue
c = 0
j = i
while True:
vs.add(i)
i = d[i]
if i == -1:
break
c += 1
if i == j:
c = inf
break
if ls[i] != -1:
c += ls[i]
break
if c == inf:
break
ls[j] = c
print(max(ls) if c != inf else -1)
```
|
output
| 1
| 15,754
| 0
| 31,509
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
|
instruction
| 0
| 15,983
| 0
| 31,966
|
Tags: dp
Correct Solution:
```
n=int(input())
s=input()
a=[int(o) for o in input().split()]
dp=[[0 for i in range(5)] for j in range(n+1)]
for i in range(n+1):
dp[i][4]=float('inf')
word=list("hard")
di={"h":0,"a":1,"r":2,"d":3}
for i in range(1,n+1):
dp[i][0]=dp[i-1][0]
dp[i][1]=dp[i-1][1]
dp[i][2]=dp[i-1][2]
dp[i][3]=dp[i-1][3]
if s[i-1] in word:
dp[i][di[s[i-1]]]=min(a[i-1]+dp[i][di[s[i-1]]],dp[i][di[s[i-1]]-1])
print(min(dp[n]))
```
|
output
| 1
| 15,983
| 0
| 31,967
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
|
instruction
| 0
| 15,984
| 0
| 31,968
|
Tags: dp
Correct Solution:
```
a = [0 for i in range(100010)]
dp = [[0 for i in range(4)] for j in range(100010)]
ans = 1e18
t = "hard"
n = int(input())
s = input()
s = ' ' + s
for i in range(n+1):
for j in range(4):
dp[i][j] = 1e18
dp[0][0] = 0
a = [0] + list(map(int, input().split()))
for x in a:
dp[0][0] += x
for i in range(1, n+1):
p = t.find(s[i])
for j in range(4):
if j==p:
if j<3:
dp[i][j+1] = min(dp[i][j+1], dp[i-1][j] - a[i])
dp[i][j] = min(dp[i][j], dp[i-1][j])
else:
dp[i][j] = min(dp[i][j], dp[i-1][j] - a[i])
for i in range(4):
ans = min(ans, dp[n][i])
print(ans)
```
|
output
| 1
| 15,984
| 0
| 31,969
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
|
instruction
| 0
| 15,985
| 0
| 31,970
|
Tags: dp
Correct Solution:
```
import sys
def eprint(*args, **kwargs):
print(*args, file=sys.stderr, **kwargs)
n = int(input())
s = input()
ambiguity = list(map(int, input().split()))
s_a = [y for y in list(zip([x for x in s], ambiguity)) if y[0] in "hard"]
if(len(s_a) < 4):
print(0)
quit()
l = len(s_a)+1
h = [0]*l
ha = [0]*l
har = [0]*l
hard = [0]*l
last_a = -1
last_r = -1
last_d = -1
for i, (c,a) in enumerate(s_a):
if c == 'h':
h[i] = a + h[i-1]
else:
h[i] = h[i-1]
if c == 'a':
ha[i] = min(h[i],a+ha[last_a])
last_a = i
else:
ha[i] = ha[i-1]
if c == 'r':
har[i] = min(ha[i], a+har[last_r])
last_r = i
else:
har[i] = har[i-1]
if c == 'd':
hard[i] = min(har[i], a+hard[last_d])
last_d = i
else:
hard[i] = hard[i-1]
eprint(h)
eprint(ha)
eprint(har)
eprint(hard)
print(hard[-2])
```
|
output
| 1
| 15,985
| 0
| 31,971
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
|
instruction
| 0
| 15,986
| 0
| 31,972
|
Tags: dp
Correct Solution:
```
n=int(input())
s=input()
a=list(map(int,input().split()))
h=[0]*n
ha=[0]*n
har=[0]*n
hard=[0]*n
if s[0]=="h":
h[0]=a[0]
for i in range(1,n):
if s[i]=="h":
h[i]=h[i-1]+a[i]
ha[i]=ha[i-1]
har[i]=har[i-1]
hard[i]=hard[i-1]
elif s[i]=="a":
h[i]=h[i-1]
ha[i]=min(ha[i-1]+a[i],h[i])
har[i]=har[i-1]
hard[i]=hard[i-1]
elif s[i]=="r":
h[i]=h[i-1]
ha[i]=ha[i-1]
har[i]=min(har[i-1]+a[i],ha[i],h[i])
hard[i]=hard[i-1]
elif s[i]=="d":
h[i]=h[i-1]
ha[i]=ha[i-1]
har[i]=har[i-1]
hard[i]=min(hard[i-1]+a[i],h[i],ha[i],har[i])
else:
h[i]=h[i-1]
ha[i]=ha[i-1]
har[i]=har[i-1]
hard[i]=hard[i-1]
print(hard[-1])
```
|
output
| 1
| 15,986
| 0
| 31,973
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
|
instruction
| 0
| 15,987
| 0
| 31,974
|
Tags: dp
Correct Solution:
```
n = int(input())
s = input()
arr = list(map(int, input().split()))
dp = {}
for i in range(n):
dp[i] = {}
for j in range(4):
if i == 0 and j == 0:
if s[i] == 'h': dp[i][j] = arr[i]
else: dp[i][j] = 0
elif i == 0: dp[i][j] = 0
elif j == 0:
dp[i][j] = dp[i - 1][j]
if s[i] == 'h': dp[i][j] += arr[i]
elif j == 1:
if s[i] != 'a': dp[i][j] = dp[i - 1][j]
else: dp[i][j] = min(arr[i] + dp[i - 1][j], dp[i - 1][j - 1])
elif j == 2:
if s[i] != 'r': dp[i][j] = dp[i - 1][j]
else: dp[i][j] = min(arr[i] + dp[i - 1][j], dp[i - 1][j - 1])
else:
if s[i] != 'd': dp[i][j] = dp[i - 1][j]
else: dp[i][j] = min(arr[i] + dp[i - 1][j], dp[i - 1][j - 1])
print(dp[n - 1][3])
```
|
output
| 1
| 15,987
| 0
| 31,975
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
|
instruction
| 0
| 15,988
| 0
| 31,976
|
Tags: dp
Correct Solution:
```
n = int(input())
s = input()
A = list(map(int, input().split()))
t = 'hard'
dph = [0] * (n + 1)
for i in range(n):
dph[i + 1] = dph[i] + A[i] * (s[i] == 'h')
dpha = [0] * (n + 1)
for i in range(n):
if s[i] != 'a':
dpha[i + 1] = dpha[i]
else:
dpha[i + 1] = min(dpha[i] + A[i], dph[i])
dphar = [0] * (n + 1)
for i in range(n):
if s[i] != 'r':
dphar[i + 1] = dphar[i]
else:
dphar[i + 1] = min(dphar[i] + A[i], dpha[i])
dphard = [0] * (n + 1)
for i in range(n):
if s[i] != 'd':
dphard[i + 1] = dphard[i]
else:
dphard[i + 1] = min(dphard[i] + A[i], dphar[i])
print(dphard[-1])
```
|
output
| 1
| 15,988
| 0
| 31,977
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
|
instruction
| 0
| 15,989
| 0
| 31,978
|
Tags: dp
Correct Solution:
```
import sys,math,itertools
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right
from heapq import heappop,heappush,heapify, nlargest
from copy import deepcopy
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_1(): return list(map(lambda x:int(x)-1, sys.stdin.readline().split()))
def inps(): return sys.stdin.readline()
def inpsl(x): tmp = sys.stdin.readline(); return list(tmp[:x])
def err(x): print(x); exit()
n = inp()
s = inpsl(n)
a = inpl()
dp = [[INF]*(n+1) for _ in range(5)]
dp[0][0] = 0
d = 'hard'
for i,x in enumerate(s):
for j in range(4):
tmp = a[i] if x == d[j] else 0
dp[j][i+1] = min(dp[j][i+1], dp[j][i]+tmp)
dp[j+1][i+1] = min(dp[j+1][i+1], dp[j][i])
res = INF
for j in range(4):
res = min(res, dp[j][-1])
print(res)
# if x == 'h':
# dp[0][i+1] = min(dp[0][i+1], dp[0][i]+a[i])
# dp[1][i+1] = min(dp[1][i+1], dp[0][i])
# if x == 'a':
# dp[1][i+1] = min(dp[1][i+1], dp[1][i+1]+a[i])
# dp[2][i+1] = min(dp[2][i+1], dp[1][i])
# if x == 'r':
# dp[2][i+1] = min(dp[2][i+1], dp[2][i+1]+a[i])
# dp[2][i+1] = min(dp[2][i+1], dp[1][i])
```
|
output
| 1
| 15,989
| 0
| 31,979
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
|
instruction
| 0
| 15,990
| 0
| 31,980
|
Tags: dp
Correct Solution:
```
import math
n = int(input())
s = input()
c = list(map(int, input().split()))
h = [0]*n
ta = 0
tr = 0
td = 0
for i in range(n):
if s[i]=='h':
h[i] = c[i]
if(i):
h[i] = h[i] + h[i-1]
if s[i]=='a':
ta = min(ta+c[i], h[i])
elif s[i]=='r':
tr = min(tr+c[i], ta)
elif s[i]=='d':
td = min(td+c[i], tr)
print(td)
```
|
output
| 1
| 15,990
| 0
| 31,981
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
Submitted Solution:
```
n = int(input())
s = input()
t = 'hard'
a = list(map(int, input().split()))
dp = [x[:] for x in [[0]*4]*(n+1)]
for i in range(1,n+1):
for j in range(4):
if s[i-1] == t[j]:
if j ==0:
dp[i][j] = dp[i-1][j]+a[i-1]
else:
dp[i][j] = min(dp[i-1][j]+a[i-1], dp[i-1][j-1])
else:
dp[i][j] = dp[i-1][j]
res = float('inf')
for i in range(4):
res = min(res, dp[n][i])
print(res)
```
|
instruction
| 0
| 15,991
| 0
| 31,982
|
Yes
|
output
| 1
| 15,991
| 0
| 31,983
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
s = input()
a = list(map(int, input().split()))
dp = [0] * 4
for i in range(n):
if s[i] == 'h':
dp[0] += a[i]
elif s[i] == 'a':
dp[1] = min(dp[0], a[i] + dp[1])
elif s[i] == 'r':
dp[2] = min(dp[1], a[i] + dp[2])
elif s[i] == 'd':
dp[3] = min(dp[2], a[i] + dp[3])
print(min(dp))
```
|
instruction
| 0
| 15,992
| 0
| 31,984
|
Yes
|
output
| 1
| 15,992
| 0
| 31,985
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
Submitted Solution:
```
input()
dp = [0] * 5
for i, c in zip(map('hard'.find, input()), map(int, input().split())) :
if i < 0 : continue
dp[i+1] = min(dp[i+1], dp[i])
dp[i] += c
# print('hard'[i],dp)
print(min(dp[:-1]))
```
|
instruction
| 0
| 15,993
| 0
| 31,986
|
Yes
|
output
| 1
| 15,993
| 0
| 31,987
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
Submitted Solution:
```
import sys
def solve():
'''
Time - O(n)
Space - O(n)
Ref - https://www.youtube.com/watch?v=xbNKWntnV20
'''
n = int(input())
s = input()
cost_str = input().split()
cost = [int(x) for x in cost_str]
#dp[i][j] = min cost so that s[..i + 1] doesnt have t[..j + 1] as subseq
dp = [[10**18 + 7] * (5) for _ in range(n + 1)]
t = "hard"
#dp[0][0] is invalid state since we dont have any char in t
for i in range(1, 5):
dp[0][i] = 0
for i in range(1, n + 1):
for j in range(1, 5):
#if ch dont match, then we an take same cost as prev i - 1
if s[i - 1] != t[j - 1]:
dp[i][j] = dp[i - 1][j]
else:
#We either dont take the s[i - 1] ch, in that mincost[i - 1] needs to be added to prev i - 1 state, when taking s[i - 1] ch, then we need to remove prev target j - 1 ch seq
dp[i][j] = min(dp[i - 1][j] + cost[i - 1], dp[i - 1][j - 1])
print(dp[n][4])
return 0
solve()
```
|
instruction
| 0
| 15,994
| 0
| 31,988
|
Yes
|
output
| 1
| 15,994
| 0
| 31,989
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
Submitted Solution:
```
n = int(input())
ambstr = input()
ambi = [int(x) for x in input().split()]
count = 0
h = 0
a = 0
r = 0
d = 0
hb = 0
ab = 0
rb = 0
db = 0
he = 0
ae = 0
re = 0
de = 0
for i in range(n):
if(count == 0 and ambstr[i] == 'h'):
count += 1
hb = i
elif(count == 1 and ambstr[i] == 'a'):
count += 1
ab = i
elif(count == 2 and ambstr[i] == 'r'):
count += 1
rb = i
elif(count == 3 and ambstr[i] == 'd'):
count += 1
db = i
count = 0
for i in range(n-1, -1, -1):
if(count == 0 and ambstr[i] == 'd'):
count += 1
de = i
elif(count == 1 and ambstr[i] == 'r'):
count += 1
re = i
elif(count == 2 and ambstr[i] == 'a'):
count += 1
ae = i
elif(count == 3 and ambstr[i] == 'h'):
count += 1
he = i
for i in range(hb,he+1):
if(ambstr[i] == 'h'):
h += ambi[i]
for i in range(ab,ae+1):
if(ambstr[i] == 'a'):
a += ambi[i]
for i in range(rb,re+1):
if(ambstr[i] == 'r'):
r += ambi[i]
for i in range(db,de+1):
if(ambstr[i] == 'd'):
d += ambi[i]
if(count < 4):
print(0)
elif(count == 4):
print(min(h, a ,r, d)%998244353)
```
|
instruction
| 0
| 15,995
| 0
| 31,990
|
No
|
output
| 1
| 15,995
| 0
| 31,991
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
Submitted Solution:
```
#########################################################################################################\
#########################################################################################################
###################################The_Apurv_Rathore#####################################################
#########################################################################################################
#########################################################################################################
import sys,os,io
from sys import stdin
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop
import math
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
def power(x, y, p) :
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def prefix_sum(arr):
r = [0] * (len(arr)+1)
for i, el in enumerate(arr):
r[i+1] = r[i] + el
return r
def divideCeil(n,x):
if (n%x==0):
return n//x
return n//x+1
def si():
return input()
def ii():
return int(input())
def li():
return list(map(int,input().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
#__________________________TEMPLATE__________________OVER_______________________________________________________
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
t = 1
# t = int(input())
for _ in range(t):
n = ii()
s = list(input())
l = li()
hard = "hard"
dp = [[10000000000000 for i in range(4)] for j in range(n+1)]
for i in range(n+1):
for j in range(4):
if (i==0):
dp[i][j]=0
continue
if (j==0):
if (s[i-1]==hard[j]):
dp[i][j] = l[i-1] + dp[i-1][j]
continue
if (s[i-1]!=hard[j]):
dp[i][j] = dp[i-1][j]
else:
dp[i][j] = min(dp[i-1][j-1] , dp[i-1][j] + l[i-1])
# print(dp)
print(min(dp[-1]))
```
|
instruction
| 0
| 15,996
| 0
| 31,992
|
No
|
output
| 1
| 15,996
| 0
| 31,993
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
Submitted Solution:
```
def f(a):
vis=set()
d={"h":0,"a":0,"r":0,"d":0}
ans=0
ans=0
st=set()
for cost,word in a:
# print(word)
if word in d:
d[word]+=cost
st.add(word)
if st==set(list("hard")):
mn=min(d.values())
for i in d:
if d[i]==mn:
d[i]=0
break
ans=max(ans,mn)
# d = {"h": 0, "a": 0, "r": 0, "d": 0}
return ans
n=int(input())
s=list(input())
cost=[*map(int,input().strip().split())]
a=[*zip(cost,s)]
print(f(a))
```
|
instruction
| 0
| 15,997
| 0
| 31,994
|
No
|
output
| 1
| 15,997
| 0
| 31,995
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.
Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0, and removing i-th character increases the ambiguity by a_i (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4 even though you delete it from the string had).
Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!
Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
The first line contains one integer n (1 ≤ n ≤ 10^5) — the length of the statement.
The second line contains one string s of length n, consisting of lowercase Latin letters — the statement written by Vasya.
The third line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 998244353).
Output
Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.
Examples
Input
6
hhardh
3 2 9 11 7 1
Output
5
Input
8
hhzarwde
3 2 6 9 4 8 7 1
Output
4
Input
6
hhaarr
1 2 3 4 5 6
Output
0
Note
In the first example, first two characters are removed so the result is ardh.
In the second example, 5-th character is removed so the result is hhzawde.
In the third example there's no need to remove anything.
Submitted Solution:
```
if __name__ == '__main__':
n = int(input())
s = list(input())
v = list(map(int, input().rstrip().split()))
i = 0
hard = list('hard$')
val = [0,0,0,0]
c = 0
while c < 4 and i < n:
if s[i] == hard[c + 1]:
c += 1
if s[i] == hard[c]:
val[c] += v[i]
i += 1
print (min(val))
```
|
instruction
| 0
| 15,998
| 0
| 31,996
|
No
|
output
| 1
| 15,998
| 0
| 31,997
|
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