message stringlengths 2 23.4k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 129 108k | cluster float64 6 6 | __index_level_0__ int64 258 216k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.
A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, `atcoder`, `zscoder` and `agc` are diverse words while `gotou` and `connect` aren't diverse words.
Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.
Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
Constraints
* 1 \leq |S| \leq 26
* S is a diverse word.
Input
Input is given from Standard Input in the following format:
S
Output
Print the next word that appears after S in the dictionary, or `-1` if it doesn't exist.
Examples
Input
atcoder
Output
atcoderb
Input
abc
Output
abcd
Input
zyxwvutsrqponmlkjihgfedcba
Output
-1
Input
abcdefghijklmnopqrstuvwzyx
Output
abcdefghijklmnopqrstuvx | instruction | 0 | 68,769 | 6 | 137,538 |
"Correct Solution:
```
S = input()
if len(S) != 26:
A = [0] * 26
offset = ord('a')
for s in S:
A[ord(s)-offset] = 1
print(S + chr(A.index(0)+offset))
elif S == 'zyxwvutsrqponmlkjihgfedcba':
print(-1)
else:
k = 25
while S[k-1] > S[k]:
k -= 1
X = sorted(S[k-1:])
print(S[:k-1] + X[X.index(S[k-1]) + 1])
``` | output | 1 | 68,769 | 6 | 137,539 |
Provide a correct Python 3 solution for this coding contest problem.
Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.
A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, `atcoder`, `zscoder` and `agc` are diverse words while `gotou` and `connect` aren't diverse words.
Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.
Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
Constraints
* 1 \leq |S| \leq 26
* S is a diverse word.
Input
Input is given from Standard Input in the following format:
S
Output
Print the next word that appears after S in the dictionary, or `-1` if it doesn't exist.
Examples
Input
atcoder
Output
atcoderb
Input
abc
Output
abcd
Input
zyxwvutsrqponmlkjihgfedcba
Output
-1
Input
abcdefghijklmnopqrstuvwzyx
Output
abcdefghijklmnopqrstuvx | instruction | 0 | 68,770 | 6 | 137,540 |
"Correct Solution:
```
s=input()
from bisect import bisect_left as bl,bisect_right as br
if len(s)==26:
if s=="zyxwvutsrqponmlkjihgfedcba":print(-1);exit()
f=s[-1]
p=[s[-1]]
for i in range(2,26):
if s[-i]>f:
f=s[-i];p.append(f)
else:print(s[:-i]+p[bl(p,s[-i])]);exit()
print(chr(ord(s[0])+1))
else:
f=sorted(list(set(list("aqzxswedcvfrtgbnhyujmkilop"))-set(list(s))))
print(s+f[0])
``` | output | 1 | 68,770 | 6 | 137,541 |
Provide a correct Python 3 solution for this coding contest problem.
Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.
A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, `atcoder`, `zscoder` and `agc` are diverse words while `gotou` and `connect` aren't diverse words.
Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.
Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
Constraints
* 1 \leq |S| \leq 26
* S is a diverse word.
Input
Input is given from Standard Input in the following format:
S
Output
Print the next word that appears after S in the dictionary, or `-1` if it doesn't exist.
Examples
Input
atcoder
Output
atcoderb
Input
abc
Output
abcd
Input
zyxwvutsrqponmlkjihgfedcba
Output
-1
Input
abcdefghijklmnopqrstuvwzyx
Output
abcdefghijklmnopqrstuvx | instruction | 0 | 68,771 | 6 | 137,542 |
"Correct Solution:
```
from string import ascii_lowercase
def solve(s):
if len(s) < 26:
for c in ascii_lowercase:
if c not in s:
return s + c
while s:
pi = ascii_lowercase.index(s[-1])
s = s[:-1]
for c in ascii_lowercase[pi + 1:]:
if c not in s:
return s + c
return -1
s = input()
print(solve(s))
``` | output | 1 | 68,771 | 6 | 137,543 |
Provide a correct Python 3 solution for this coding contest problem.
Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.
A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, `atcoder`, `zscoder` and `agc` are diverse words while `gotou` and `connect` aren't diverse words.
Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.
Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
Constraints
* 1 \leq |S| \leq 26
* S is a diverse word.
Input
Input is given from Standard Input in the following format:
S
Output
Print the next word that appears after S in the dictionary, or `-1` if it doesn't exist.
Examples
Input
atcoder
Output
atcoderb
Input
abc
Output
abcd
Input
zyxwvutsrqponmlkjihgfedcba
Output
-1
Input
abcdefghijklmnopqrstuvwzyx
Output
abcdefghijklmnopqrstuvx | instruction | 0 | 68,772 | 6 | 137,544 |
"Correct Solution:
```
S = str(input())
l = [chr(i) for i in range(97,97+26)]
flag = True
for s in l:
if s not in S:
ans = S + s
flag = False
break
if flag:
if S == 'zyxwvutsrqponmlkjihgfedcba':
ans = -1
else:
while flag:
num = ord(S[-1])
S = S[:-1]
for n in range(num+1, 123):
if chr(n) not in S:
ans = S + chr(n)
flag = False
break
print(ans)
``` | output | 1 | 68,772 | 6 | 137,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gotou just received a dictionary. However, he doesn't recognize the language used in the dictionary. He did some analysis on the dictionary and realizes that the dictionary contains all possible diverse words in lexicographical order.
A word is called diverse if and only if it is a nonempty string of English lowercase letters and all letters in the word are distinct. For example, `atcoder`, `zscoder` and `agc` are diverse words while `gotou` and `connect` aren't diverse words.
Given a diverse word S, determine the next word that appears after S in the dictionary, i.e. the lexicographically smallest diverse word that is lexicographically larger than S, or determine that it doesn't exist.
Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
Constraints
* 1 \leq |S| \leq 26
* S is a diverse word.
Input
Input is given from Standard Input in the following format:
S
Output
Print the next word that appears after S in the dictionary, or `-1` if it doesn't exist.
Examples
Input
atcoder
Output
atcoderb
Input
abc
Output
abcd
Input
zyxwvutsrqponmlkjihgfedcba
Output
-1
Input
abcdefghijklmnopqrstuvwzyx
Output
abcdefghijklmnopqrstuvx
Submitted Solution:
```
import sys
from collections import deque
from collections import Counter
s = deque(map(str, input().rstrip()))
if "".join(s) == "zyxwvutsrqponmlkjihgfedcba":
print(-1)
sys.exit()
s = [ord(x) for x in s]
if len(s) == 26:
s = s[::-1]
x = s[0]
for i, w in enumerate(s):
if x > w:
break
s[i] = s[i] + 1
s = sorted(s[:i], reverse=True) + s[i:]
s = s[::-1][:-i]
else:
can_use = set(range(97, 123)) - set(s)
s = s + [min(can_use)]
s = [chr(x) for x in s]
print("".join(s))
``` | instruction | 0 | 68,777 | 6 | 137,554 |
No | output | 1 | 68,777 | 6 | 137,555 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea | instruction | 0 | 68,927 | 6 | 137,854 |
Tags: hashing, strings
Correct Solution:
```
N = int(input())
repo = {} # empty dictionary
total_count = 0
for i in range(N):
val = input()
# we need to convert val into 01010101010101010101010101
# use bit-wise
num = 0
for c in val:
index = ord(c) - ord('a')
num ^= (1 << index) # this is to flip the binary at the index
# case of all paired
if num in repo:
total_count += repo[num]
# case of one non-paired
for j in range(26):
num2 = num ^ (1 << j)
if num2 in repo:
total_count += repo[num2]
# put num into repo
if num in repo:
repo[num] += 1
else:
repo[num] = 1
print(total_count)
``` | output | 1 | 68,927 | 6 | 137,855 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea | instruction | 0 | 68,928 | 6 | 137,856 |
Tags: hashing, strings
Correct Solution:
```
#!/usr/bin/env python3
"""
Created on Wed Feb 28 11:47:12 2018
@author: mikolajbinkowski
"""
import sys
N = int(input())
string_count = {}
for _ in range(N):
s = str(input())
char_count = {}
for c in s:
char_count[c] = char_count.get(c, 0) + 1
s0 = []
for a in 'abcdefghijklmnopqrstuvwxyz':
if char_count.get(a, 0) % 2 == 1:
s0.append(a)
s1 = ''.join(s0)
string_count[s1] = string_count.get(s1, 0) + 1
pairs = 0
for s, v in string_count.items():
pairs += v * (v-1) // 2
for i in range(len(s)):
pairs += v * string_count.get(s[:i] + s[i+1:], 0)
print(pairs)
``` | output | 1 | 68,928 | 6 | 137,857 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea | instruction | 0 | 68,929 | 6 | 137,858 |
Tags: hashing, strings
Correct Solution:
```
from collections import Counter
s=[]
for i in range(int(input())):
x=[0]*26
for j in input().strip():
x[ord(j)-97]^=1
s.append(''.join([str(y) for y in x ]))
z=Counter(s)
#print(s)
an=0
for j in s:
x=list(j)
for q in range(len(x)):
if x[q]=='1':
x[q]='0'
an+=z[''.join(x)]
# print( j,x,q)
x[q]='1'
for j in z:
an+=((z[j])*(z[j]-1))//2
print(an)
``` | output | 1 | 68,929 | 6 | 137,859 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea | instruction | 0 | 68,930 | 6 | 137,860 |
Tags: hashing, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
def main():
comp=[]
c=1
for i in range(27):
comp.append(c)
c*=2
n=int(input())
d={}
table=[]
for i in range(n):
k=[0]*27
s=input()
for j in s:
if ord(j)-ord('a')>=0:
k[ord(j)-ord('a')]+=1
#print(ord(j)-ord('a'))
key=0
for i in range(26):
if k[i]%2:
key+=comp[i]
table.append([k,key])
if key in d:
d[key]+=1
else:
d[key]=1
ans=0
for i in d.values():
ans+=(i)*(i-1)//2
#print(ans)
for i in table:
#print(i)
for j in range(len(i[0])):
if i[0][j]%2:
if i[1]-comp[j] in d.keys():
#print(i[1],i[1]-comp[j] )
ans+=(d[i[1]-comp[j]])
print(int(ans))
return
if __name__ == "__main__":
main()
``` | output | 1 | 68,930 | 6 | 137,861 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea | instruction | 0 | 68,931 | 6 | 137,862 |
Tags: hashing, strings
Correct Solution:
```
def zamien(S):
A = 26*[0]
for znak in S:
A[ord(znak)-ord('a')] += 1
return ''.join([str(i%2) for i in A])
def mainn():
A = {}
wynik = 0
for i in range(int(input())):
s = zamien(input())
wynik += A.get(s,0)
for i in range(26):
wynik += A.get(s[:i]+str((int(s[i])+1)%2)+s[i+1:],0)
A[s] = A.get(s,0) + 1
return wynik
print(mainn())
``` | output | 1 | 68,931 | 6 | 137,863 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea | instruction | 0 | 68,932 | 6 | 137,864 |
Tags: hashing, strings
Correct Solution:
```
n = int(input())
string_count = {}
for _ in range(n):
s = str(input())
item_count={}
for i,c in enumerate(s):
item_count[c]=item_count.get(c,0)+1
s0=[]
for i,x in enumerate('abcdefghijklmnopqrstuvwxyz'):
if item_count.get(x,0)%2==1:
s0.append(x)
s1 = ''.join(s0)
string_count[s1]=string_count.get(s1,0)+1
points=0
for j,a in enumerate(string_count):
x = string_count[a]
points+=x*(x-1)//2
for i in range(len(a)):
points+=x*string_count.get(a[:i]+a[i+1:],0)
print(points)
``` | output | 1 | 68,932 | 6 | 137,865 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea | instruction | 0 | 68,933 | 6 | 137,866 |
Tags: hashing, strings
Correct Solution:
```
from collections import Counter
n = int(input())
strings = []
for i in range(0,n):
counter = { x: 0 for x in range(ord('a'), ord('z')+1) }
napis = input()
for val in napis:
counter[ord(val)] = (counter[ord(val)] + 1) % 2
napis = ""
for key, val in counter.items():
if val != 0:
napis+=chr(key)
strings.append(napis)
c = Counter(strings)
strings = sorted(c.most_common(), key=lambda i: i[0])
count = 0
for key, val in strings:
if val > 1:
count += val*(val-1)/2
for charInt in range(ord('a'), ord('z')+1):
char = chr(charInt)
copy = {}
for key, val in strings:
if char in key:
copy[key.replace(char, "")] = copy.get(key.replace(char, ""), 0) + val
for key, val in strings:
if copy.get(key,0) != 0:
count+=val * copy[key]
print(int(count))
``` | output | 1 | 68,933 | 6 | 137,867 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea | instruction | 0 | 68,934 | 6 | 137,868 |
Tags: hashing, strings
Correct Solution:
```
import sys
import math
import collections
import heapq
import decimal
input=sys.stdin.readline
n=int(input())
d={}
ans=0
for i in range(n):
s=input()
k=len(s)-1
c=0
for j in range(k):
c^=(1<<ord(s[j])-97)
ans+=d.get(c,0)
for j in range(26):
ans+=d.get(c^(1<<j),0)
k=d.get(c,0)+1
d[c]=k
print(ans)
``` | output | 1 | 68,934 | 6 | 137,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea
Submitted Solution:
```
aw=[2**i for i in range(26)]
n=int(input())
d=dict()
ans=0
for ir in range(n):
st=input()
es=0
for j in st:
es=es^aw[ord(j)-97]
if es in d:
ans+=d[es]
for j in range(26):
es=es^aw[j]
if es in d:
ans+=d[es]
es=es^aw[j]
if es in d:
d[es]+=1
else:
d.update({es:1})
print(ans)
``` | instruction | 0 | 68,935 | 6 | 137,870 |
Yes | output | 1 | 68,935 | 6 | 137,871 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea
Submitted Solution:
```
def daj_maske(slowo):
maska = 0
for litera in slowo:
kod = ord(litera) - ord('a')
maska ^= (1 << kod)
return maska
n = int(input())
L = []
for _ in range(n):
slowo = input()
L.append(slowo)
wynik = 0
zlicz = {}
for slowo in L:
maska = daj_maske(slowo)
wynik += zlicz.get(maska, 0)
for i in range(26):
wynik += zlicz.get(maska ^ (1 << i), 0)
zlicz[maska] = zlicz.get(maska, 0) + 1
print(wynik)
``` | instruction | 0 | 68,936 | 6 | 137,872 |
Yes | output | 1 | 68,936 | 6 | 137,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea
Submitted Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 11/20/18
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward).
She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem,
so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet.
Your task is to find how many palindrome pairs are there in the array.
A palindrome pair is a pair of strings such that the following condition holds:
at least one permutation of the concatenation of the two strings is a palindrome.
In other words, if you have two strings, let's say "aab" and "abcac",
and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that
it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices.
The pair of strings with indices (𝑖,𝑗) is considered the same as the pair (𝑗,𝑖).
Input
The first line contains a positive integer 𝑁 (1≤𝑁≤100 000), representing the length of the input array.
Eacg of the next 𝑁 lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1000000.
Output
Output one number, representing how many palindrome pairs there are in the array.
# a, b can be transform to palindrome only when number of every characters in a+b is even, or only one is odd.
# only odd+even => odd
"""
import collections
N = int(input())
def hash(s):
wc = collections.Counter(s)
x = ''
for i in range(26):
w = chr(ord('a') + i)
c = wc[w]
if c % 2 == 0:
x += '0'
else:
x += '1'
return x
def neighbor(s):
for i in range(len(s)):
yield s[:i] + ('1' if s[i] == '0' else '0') + s[i + 1:]
def isneighbor(a, b):
return sum([0 if a[i] == b[i] else 1 for i in range(len(a))]) == 1
m = collections.defaultdict(list)
for i in range(N):
s = input()
m[hash(s)].append(i)
even = 0
odd = 0
for k, v in m.items():
lv = len(v)
even += lv * (lv - 1) // 2
for b in neighbor(k):
if b in m:
odd += lv * len(m[b])
print(even + odd // 2)
``` | instruction | 0 | 68,937 | 6 | 137,874 |
Yes | output | 1 | 68,937 | 6 | 137,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea
Submitted Solution:
```
n = int(input())
odd_occ = []
alphabet = 'abcdefghijklmnopqrstuvwxyz'
for k in range(n):
s = input()
occ = {}
for l in range(len(s)):
ltr = s[l]
if not occ.get(ltr):
occ[ltr] = 0
occ[ltr] += 1
odd_ltrs = ""
#print(occ)
for l in range(26):
ltr = alphabet[l]
if not occ.get(ltr):
continue
if occ[ltr] % 2 == 1:
odd_ltrs += ltr
odd_occ.append(odd_ltrs+'0')
#print(odd_occ)
count = {}
for i in odd_occ:
if not count.get(i):
count[i] = 0
count[i] += 1
res = 0
#print(count)
for i in count:
k = 0
for j in range(len(i)-1):
if not count.get(i[:j] + i[j+1:]):
continue
k += count[i[:j] + i[j+1:]]
res += count[i] * k
res += count[i] * (count[i] - 1) // 2
print(res)
``` | instruction | 0 | 68,938 | 6 | 137,876 |
Yes | output | 1 | 68,938 | 6 | 137,877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea
Submitted Solution:
```
def convert(S):
d = {}
for x in S:
if x not in d:
d[x] = 0
d[x]+=1
answer = []
ones = []
for x in d:
if d[x] % 2==1:
answer.append(x)
if d[x]==1:
ones.append(x)
answer = sorted(answer)
if len(answer) > 1:
final_answer = [tuple(answer[:-1]), tuple(answer[1:])]
else:
final_answer = [()]
n = len(answer)
for i in range(1, n-1):
a2 = answer[:i]+answer[(i+1):]
final_answer.append(tuple(a2))
return tuple(sorted(answer)), final_answer
def process(A):
d1 = {}
d2 = {}
for x in A:
C, D = convert(x)
if C not in d1:
d1[C] = 0
d1[C]+=1
for C2 in D:
if C2 not in d2:
d2[C2] = 0
d2[C2]+=1
answer = 0
for C in d1:
m = d1[C]
m2 = m*(m-1)//2
answer+=m2
if C in d2:
m2 = d2[C]
answer+=(m*m2)
return answer
n = int(input())
A = []
for i in range(n):
S = input()
A.append(S)
print(process(A))
``` | instruction | 0 | 68,939 | 6 | 137,878 |
No | output | 1 | 68,939 | 6 | 137,879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea
Submitted Solution:
```
def convert(S):
d = {}
for x in S:
if x not in d:
d[x] = 0
d[x]+=1
answer = []
ones = []
for x in d:
if d[x] % 2==1:
answer.append(x)
if d[x]==1:
ones.append(x)
answer = tuple(sorted(answer))
if len(ones) > 1 or len(ones)==0:
unique_ones = ''
else:
unique_ones = ones[0]
return (answer, unique_ones)
def process(A):
d = {}
ones = 0
for x in A:
C, D = convert(x)
if C not in d:
d[C] = 0
d[C]+=1
if D != '':
ones+=1
answer = 0
for C in d:
m = d[C]
m2 = m*(m-1)//2
answer+=m2
if () in d:
answer+=d[()]*ones
return answer
n = int(input())
A = []
for i in range(n):
S = input()
A.append(S)
print(process(A))
``` | instruction | 0 | 68,940 | 6 | 137,880 |
No | output | 1 | 68,940 | 6 | 137,881 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea
Submitted Solution:
```
def convert(S):
d = {}
for x in S:
if x not in d:
d[x] = 0
d[x]+=1
answer = []
ones = []
for x in d:
if d[x] % 2==1:
answer.append(x)
if d[x]==1:
ones.append(x)
answer = sorted(answer)
if len(answer) > 1:
final_answer = [tuple(answer[:-1]), tuple(answer[1:])]
elif len(answer)==1:
final_answer = [()]
else:
final_answer = []
n = len(answer)
for i in range(1, n-1):
a2 = answer[:i]+answer[(i+1):]
final_answer.append(tuple(a2))
return tuple(sorted(answer)), final_answer
def process(A):
d1 = {}
d2 = {}
for x in A:
C, D = convert(x)
if C not in d1:
d1[C] = 0
d1[C]+=1
for C2 in D:
if C2 not in d2:
d2[C2] = 0
d2[C2]+=1
answer = 0
for C in d1:
m = d1[C]
m2 = m*(m-1)//2
answer+=m2
if C in d2:
m2 = d2[C]
answer+=(m*m2)
print(d1, d2)
return answer
n = int(input())
A = []
for i in range(n):
S = input()
A.append(S)
print(process(A))
``` | instruction | 0 | 68,941 | 6 | 137,882 |
No | output | 1 | 68,941 | 6 | 137,883 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After learning a lot about space exploration, a little girl named Ana wants to change the subject.
Ana is a girl who loves palindromes (string that can be read the same backwards as forward). She has learned how to check for a given string whether it's a palindrome or not, but soon she grew tired of this problem, so she came up with a more interesting one and she needs your help to solve it:
You are given an array of strings which consist of only small letters of the alphabet. Your task is to find how many palindrome pairs are there in the array. A palindrome pair is a pair of strings such that the following condition holds: at least one permutation of the concatenation of the two strings is a palindrome. In other words, if you have two strings, let's say "aab" and "abcac", and you concatenate them into "aababcac", we have to check if there exists a permutation of this new string such that it is a palindrome (in this case there exists the permutation "aabccbaa").
Two pairs are considered different if the strings are located on different indices. The pair of strings with indices (i,j) is considered the same as the pair (j,i).
Input
The first line contains a positive integer N (1 ≤ N ≤ 100 000), representing the length of the input array.
Eacg of the next N lines contains a string (consisting of lowercase English letters from 'a' to 'z') — an element of the input array.
The total number of characters in the input array will be less than 1 000 000.
Output
Output one number, representing how many palindrome pairs there are in the array.
Examples
Input
3
aa
bb
cd
Output
1
Input
6
aab
abcac
dffe
ed
aa
aade
Output
6
Note
The first example:
1. aa + bb → abba.
The second example:
1. aab + abcac = aababcac → aabccbaa
2. aab + aa = aabaa
3. abcac + aa = abcacaa → aacbcaa
4. dffe + ed = dffeed → fdeedf
5. dffe + aade = dffeaade → adfaafde
6. ed + aade = edaade → aeddea
Submitted Solution:
```
import sys
input = sys.stdin.readline
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
def main():
comp=[]
c=1
for i in range(27):
comp.append(c)
c*=2
n=int(input())
d={}
table=[]
for i in range(n):
k=[0]*27
s=input()
for j in s:
if ord(j)-ord('a')>=0:
k[ord(j)-ord('a')]+=1
#print(ord(j)-ord('a'))
key=0
for i in range(26):
if k[i]%2:
key+=comp[i]
table.append([k,key])
if key in d:
d[key]+=1
else:
d[key]=1
ans=0
for i in d.values():
ans+=(i)*(i-1)//2
#print(ans)
for i in table:
for j in range(len(i[0])):
if i[0][j]:
if i[1]-comp[j] in d.keys():
ans+=(d[i[1]]*d[i[1]-comp[j]])/2
print(int(ans))
return
if __name__ == "__main__":
main()
``` | instruction | 0 | 68,942 | 6 | 137,884 |
No | output | 1 | 68,942 | 6 | 137,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct.
Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:
* If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great.
* If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice).
You are given a multiple-choice questions, can you predict child's choose?
Input
The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.
Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".
Output
Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).
Examples
Input
A.VFleaKing_is_the_author_of_this_problem
B.Picks_is_the_author_of_this_problem
C.Picking_is_the_author_of_this_problem
D.Ftiasch_is_cute
Output
D
Input
A.ab
B.abcde
C.ab
D.abc
Output
C
Input
A.c
B.cc
C.c
D.c
Output
B
Note
In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D.
In the second sample, no choice is great, so the child will choose the luckiest choice C.
In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B. | instruction | 0 | 69,329 | 6 | 138,658 |
Tags: implementation
Correct Solution:
```
a = []
b = []
for i in range(4):
x = input()
a.append(len(x[2:]))
b.append(x)
c = a.copy()
a.sort()
if 2 * a[0] <= a[1] and a[-1] < 2 * a[-2]:
print(b[c.index(a[0])][0])
elif a[-1] >= 2 * a[-2] and 2 * a[0] > a[1]:
print(b[c.index(a[-1])][0])
else:
print('C')
``` | output | 1 | 69,329 | 6 | 138,659 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct.
Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:
* If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great.
* If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice).
You are given a multiple-choice questions, can you predict child's choose?
Input
The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.
Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".
Output
Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).
Examples
Input
A.VFleaKing_is_the_author_of_this_problem
B.Picks_is_the_author_of_this_problem
C.Picking_is_the_author_of_this_problem
D.Ftiasch_is_cute
Output
D
Input
A.ab
B.abcde
C.ab
D.abc
Output
C
Input
A.c
B.cc
C.c
D.c
Output
B
Note
In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D.
In the second sample, no choice is great, so the child will choose the luckiest choice C.
In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B. | instruction | 0 | 69,330 | 6 | 138,660 |
Tags: implementation
Correct Solution:
```
# Codeforces Problemset
# 437A
a = len(input()) - 2
b = len(input()) - 2
c = len(input()) - 2
d = len(input()) - 2
great = []
if 2*a <= b and 2*a <= c and 2*a <= d:
great.append('A')
elif 2*b <= a and 2*b <= c and 2*b <= d:
great.append('B')
elif 2*c <= a and 2*c <= b and 2*c <= d:
great.append('C')
elif 2*d <= a and 2*d <= b and 2*d <= c:
great.append('D')
if a >= 2*b and a >= 2*c and a >= 2*d:
great.append('A')
elif b >= 2*a and b >= 2*c and b >= 2*d:
great.append('B')
elif c >= 2*a and c >= 2*b and c >= 2*d:
great.append('C')
elif d >= 2*a and d >= 2*b and d >= 2*c:
great.append('D')
if len(great) == 1:
print(great[0])
else:
print('C')
``` | output | 1 | 69,330 | 6 | 138,661 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct.
Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:
* If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great.
* If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice).
You are given a multiple-choice questions, can you predict child's choose?
Input
The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.
Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".
Output
Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).
Examples
Input
A.VFleaKing_is_the_author_of_this_problem
B.Picks_is_the_author_of_this_problem
C.Picking_is_the_author_of_this_problem
D.Ftiasch_is_cute
Output
D
Input
A.ab
B.abcde
C.ab
D.abc
Output
C
Input
A.c
B.cc
C.c
D.c
Output
B
Note
In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D.
In the second sample, no choice is great, so the child will choose the luckiest choice C.
In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B. | instruction | 0 | 69,331 | 6 | 138,662 |
Tags: implementation
Correct Solution:
```
p={ __:len(input())-2 for __ in 'ABCD'}
from math import floor
#print(p)
las=None
tot=0
for i in p:
if all([(2*p[i])<=p[j] for j in p if i!=j]) or\
all([p[i]>=(p[j]*2) for j in p if i!=j]):
las=i
tot+=1
if las!=None and tot==1:
print(las)
else:
print('C')
``` | output | 1 | 69,331 | 6 | 138,663 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct.
Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:
* If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great.
* If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice).
You are given a multiple-choice questions, can you predict child's choose?
Input
The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.
Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".
Output
Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).
Examples
Input
A.VFleaKing_is_the_author_of_this_problem
B.Picks_is_the_author_of_this_problem
C.Picking_is_the_author_of_this_problem
D.Ftiasch_is_cute
Output
D
Input
A.ab
B.abcde
C.ab
D.abc
Output
C
Input
A.c
B.cc
C.c
D.c
Output
B
Note
In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D.
In the second sample, no choice is great, so the child will choose the luckiest choice C.
In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B. | instruction | 0 | 69,332 | 6 | 138,664 |
Tags: implementation
Correct Solution:
```
a = [0] * 4
for i in range(4):
s = input().strip()
a[i] = len(s) - 2
ans = 'C'
f = 0
for i in range(4):
f1 = 1
f2 = 1
for j in range(4):
if i != j and a[i] < 2 * a[j]:
f1 = 0
if i != j and 2 * a[i] > a[j]:
f2 = 0
if f1 :
ans = chr(ord("A") + i)
f += 1
if f2:
ans = chr(ord("A") + i)
f += 1
if f == 1:
print(ans)
else:
print("C")
``` | output | 1 | 69,332 | 6 | 138,665 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct.
Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:
* If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great.
* If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice).
You are given a multiple-choice questions, can you predict child's choose?
Input
The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.
Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".
Output
Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).
Examples
Input
A.VFleaKing_is_the_author_of_this_problem
B.Picks_is_the_author_of_this_problem
C.Picking_is_the_author_of_this_problem
D.Ftiasch_is_cute
Output
D
Input
A.ab
B.abcde
C.ab
D.abc
Output
C
Input
A.c
B.cc
C.c
D.c
Output
B
Note
In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D.
In the second sample, no choice is great, so the child will choose the luckiest choice C.
In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B. | instruction | 0 | 69,333 | 6 | 138,666 |
Tags: implementation
Correct Solution:
```
answers = [len(input().split(".")[1]) for x in range(4)]
letters = ["A", "B", "C", "D"]
good = []
for x in range(4):
condition = True
for y in range(4):
if x != y and not answers[x] * 2 <= answers[y]:
condition = False
break
if not condition:
condition = True
for y in range(4):
if x != y and not answers[x] >= answers[y] * 2:
condition = False
break
if condition:
good.append(x)
if len(good) != 1:
print("C")
else:
print(letters[good[0]])
``` | output | 1 | 69,333 | 6 | 138,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct.
Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:
* If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great.
* If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice).
You are given a multiple-choice questions, can you predict child's choose?
Input
The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.
Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".
Output
Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).
Examples
Input
A.VFleaKing_is_the_author_of_this_problem
B.Picks_is_the_author_of_this_problem
C.Picking_is_the_author_of_this_problem
D.Ftiasch_is_cute
Output
D
Input
A.ab
B.abcde
C.ab
D.abc
Output
C
Input
A.c
B.cc
C.c
D.c
Output
B
Note
In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D.
In the second sample, no choice is great, so the child will choose the luckiest choice C.
In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B. | instruction | 0 | 69,334 | 6 | 138,668 |
Tags: implementation
Correct Solution:
```
a,b,c,d=input(),input(),input(),input()
a=len(a)-2
b=len(b)-2
c=len(c)-2
d=len(d)-2
e=False
if 2*a<=c and 2*a<=b and 2*a<=d or a>=2*c and a>=2*b and a>=2*d: e='A'
if 2*b<=a and 2*b<=c and 2*b<=d or b>=2*a and b>=2*c and b>=2*d:
if e==False:
e='B'
else:
e='C'
if 2*c<=a and 2*c<=b and 2*c<=d or c>=2*a and c>=2*b and c>=2*d:
if e==False:
e='C'
else:
e='C'
if 2*d<=a and 2*d<=b and 2*d<=c or d>=2*a and d>=2*b and d>=2*c:
if e==False:
e='D'
else:
e='C'
if e==False: e='C'
print(e)
``` | output | 1 | 69,334 | 6 | 138,669 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct.
Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:
* If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great.
* If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice).
You are given a multiple-choice questions, can you predict child's choose?
Input
The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.
Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".
Output
Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).
Examples
Input
A.VFleaKing_is_the_author_of_this_problem
B.Picks_is_the_author_of_this_problem
C.Picking_is_the_author_of_this_problem
D.Ftiasch_is_cute
Output
D
Input
A.ab
B.abcde
C.ab
D.abc
Output
C
Input
A.c
B.cc
C.c
D.c
Output
B
Note
In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D.
In the second sample, no choice is great, so the child will choose the luckiest choice C.
In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B. | instruction | 0 | 69,335 | 6 | 138,670 |
Tags: implementation
Correct Solution:
```
a={"A":len(input()[2:]),"B":len(input()[2:]),"C":len(input()[2:]),"D":len(input()[2:])}
mi=min(a.values())
ma=max(a.values())
l=list(a.values())
c=0
s=0
key_list = list(a.keys())
val_list = list(a.values())
for i in a.values():
if i==mi:
continue
elif mi<=i/2:
c+=1
for i in a.values():
if i== ma :
continue
if ma>=i*2:
s+=1
if c==3 and s!=3:
print(list(a.keys())[list(a.values()).index(mi)])
elif s==3 and c!=3:
print(list(a.keys())[list(a.values()).index(ma)])
else:
print("C")
``` | output | 1 | 69,335 | 6 | 138,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct.
Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:
* If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great.
* If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice).
You are given a multiple-choice questions, can you predict child's choose?
Input
The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.
Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".
Output
Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).
Examples
Input
A.VFleaKing_is_the_author_of_this_problem
B.Picks_is_the_author_of_this_problem
C.Picking_is_the_author_of_this_problem
D.Ftiasch_is_cute
Output
D
Input
A.ab
B.abcde
C.ab
D.abc
Output
C
Input
A.c
B.cc
C.c
D.c
Output
B
Note
In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D.
In the second sample, no choice is great, so the child will choose the luckiest choice C.
In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B. | instruction | 0 | 69,336 | 6 | 138,672 |
Tags: implementation
Correct Solution:
```
a = []
for i in range(4):
a.append(len(input()) - 2)
ans = list("ABCD")
tosc = ''
for i in range(4):
ok1 = 1;
ok2 = 1;
for j in range(4):
if i == j:
continue;
if a[i] < 2 * a[j]:
ok1 = 0
if a[i] * 2 > a[j]:
ok2 = 0
if any((ok1, ok2)):
if tosc:
tosc = 'C'
else:
tosc = ans[i]
print(tosc if tosc else 'C')
``` | output | 1 | 69,336 | 6 | 138,673 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout | instruction | 0 | 69,926 | 6 | 139,852 |
Tags: brute force, hashing, implementation, string suffix structures, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
L=list(input().split())
#n=2*10**5
#L=["aaa","bbb"]*(10**5)
ANS=[L[0]]
LEN2=len(L[0])
mod0=1<<16
mod1=(1<<16)-3
mod2=(1<<16)-2
p=75
TABLE0=[0]
TABLE1=[0]
TABLE2=[0]
def hash_ij0(i,j): # [i,j)のハッシュ値を求める
return (TABLE0[j]-TABLE0[i]*pow(p,j-i,mod0))%mod0
def hash_ij1(i,j): # [i,j)のハッシュ値を求める
return (TABLE1[j]-TABLE1[i]*pow(p,j-i,mod1))%mod1
def hash_ij2(i,j): # [i,j)のハッシュ値を求める
return (TABLE2[j]-TABLE2[i]*pow(p,j-i,mod2))%mod2
for s in L[0]:
TABLE0.append((p*TABLE0[-1]%mod0+ord(s)-48)%mod0)
TABLE1.append((p*TABLE1[-1]%mod1+ord(s)-48)%mod1)
TABLE2.append((p*TABLE2[-1]%mod2+ord(s)-48)%mod2)
for i in range(1,n):
NEXT=L[i]
LEN=len(NEXT)
ha0=0
ha1=0
ha2=0
plus=-1
#print(NEXT)
for j in range(min(LEN,LEN2)):
ha0=(p*ha0%mod0+ord(NEXT[j])-48)%mod0
ha1=(p*ha1%mod1+ord(NEXT[j])-48)%mod1
ha2=(p*ha2%mod2+ord(NEXT[j])-48)%mod2
#print(ha1,TABLE1)
#print(hash_ij1(LEN2-j-1,LEN2))
if ha0==hash_ij0(LEN2-j-1,LEN2) and ha1==hash_ij1(LEN2-j-1,LEN2) and ha2==hash_ij2(LEN2-j-1,LEN2):
plus=j
#print(plus)
if plus==-1:
ANS.append(NEXT)
LEN2+=len(NEXT)
for s in NEXT:
TABLE0.append((p*TABLE0[-1]%mod0+ord(s)-48)%mod0)
TABLE1.append((p*TABLE1[-1]%mod1+ord(s)-48)%mod1)
TABLE2.append((p*TABLE2[-1]%mod2+ord(s)-48)%mod2)
else:
NEXT=NEXT[plus+1:]
ANS.append(NEXT)
LEN2+=len(NEXT)
for s in NEXT:
TABLE0.append((p*TABLE0[-1]%mod0+ord(s)-48)%mod0)
TABLE1.append((p*TABLE1[-1]%mod1+ord(s)-48)%mod1)
TABLE2.append((p*TABLE2[-1]%mod2+ord(s)-48)%mod2)
sys.stdout.write("".join(ANS)+"\n")
``` | output | 1 | 69,926 | 6 | 139,853 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout | instruction | 0 | 69,927 | 6 | 139,854 |
Tags: brute force, hashing, implementation, string suffix structures, strings
Correct Solution:
```
from sys import stdin,stdout
from sys import setrecursionlimit as SRL; SRL(10**7)
rd = stdin.readline
rrd = lambda: map(int, rd().strip().split())
nxt = [0] * (1000005)
n = int(rd())
s = list(rd().split())
ans = []
for i in s[0]:
ans.append(i)
for i in range(1,n):
v = s[i]
t = 0
for i in range(2,len(v)):
while t and v[i-1] != v[t]:
t = nxt[t]
if v[i-1] == v[t]:
t+=1
nxt[i] = t
t = 0
for i in range(max(0,len(ans)-len(v)),len(ans)):
while t and ans[i] != v[t]:
t = nxt[t]
if ans[i] == v[t]:
t += 1
while t < len(v):
ans.append(v[t])
t += 1
stdout.write("".join(ans)+"\n")
``` | output | 1 | 69,927 | 6 | 139,855 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout | instruction | 0 | 69,928 | 6 | 139,856 |
Tags: brute force, hashing, implementation, string suffix structures, strings
Correct Solution:
```
def zalgo(s:str):
l=0;r=0;n=len(s);
z = [0]*n
for i in range(1,n):
z[i] = 0
if(i<=r): z[i]=min(r-i+1,z[i-l])
while(i+z[i]<n and s[z[i]]==s[i+z[i]]): z[i]+=1
if(i+z[i]-1>r):
l = i
r = i+z[i]-1
if(i+z[i]==n):
return z[i]
return 0
n = int(input())
s = input().split()
txt = []
for i in range(0,n):
x = len(txt)
y = len(s[i])
if x > y:
#print(x,y,(txt[x-y:y]))
mx = zalgo(s[i] + '#'+ ''.join((txt[x-y:x])))
else:
#print(x,y,txt[:])
mx = zalgo(s[i][:y]+'#'+''.join(txt))
# mx = zalgo(s[i] + '#' + ''.join((txt[-mn:])))
#print(mx)
# print(mx,mn,sep=' ')
txt.extend(s[i][mx:])
# print(txt)
print(''.join(txt))
``` | output | 1 | 69,928 | 6 | 139,857 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout | instruction | 0 | 69,929 | 6 | 139,858 |
Tags: brute force, hashing, implementation, string suffix structures, strings
Correct Solution:
```
from sys import stdin,stdout
from sys import setrecursionlimit as SRL; SRL(10**7)
rd = stdin.readline
rrd = lambda: map(int, rd().strip().split())
nxt = [0] * (1000005)
n = int(rd())
s = list(rd().split())
ans = []
for i in s[0]:
ans.append(i)
for i in range(1,n):
v = s[i]
t = 0
for i in range(2,len(v)):
while t and v[i-1] != v[t]:
t = nxt[t]
if v[i-1] == v[t]:
t+=1
nxt[i] = t
t = 0
for i in range(max(0,len(ans)-len(v)),len(ans)):
while t and ans[i] != v[t]:
t = nxt[t]
if ans[i] == v[t]:
t += 1
while t < len(v):
ans.append(v[t])
t += 1
print("".join(ans)+'\n')
``` | output | 1 | 69,929 | 6 | 139,859 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout | instruction | 0 | 69,930 | 6 | 139,860 |
Tags: brute force, hashing, implementation, string suffix structures, strings
Correct Solution:
```
def compress(words):
if not words:
return ''
def prefix(s):
table = [0] * len(s)
j = 0
for i in range(1, len(s)):
while 0 < j and s[i] != s[j]:
j = table[j - 1]
if s[i] == s[j]:
j += 1
table[i] = j
return table[-1]
result = [*words[0]]
for word in words[1:]:
n = min(len(result), len(word))
for _ in range(prefix([*word] + ['$'] + result[-n:])):
result.pop()
result.extend([*word])
return ''.join(result)
def solve():
_ = int(input())
words = input().split()
print(compress(words))
solve()
``` | output | 1 | 69,930 | 6 | 139,861 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout | instruction | 0 | 69,931 | 6 | 139,862 |
Tags: brute force, hashing, implementation, string suffix structures, strings
Correct Solution:
```
def prefix_func(s):
p = [0] * len(s)
for i in range(1, len(s)):
j = p[i - 1]
while j > 0 and s[i] != s[j]:
j = p[j - 1]
if s[i] == s[j]:
j += 1
p[i] = j
return p[-1]
n = int(input())
s = input().split()
ans = [*s[0]]
for i in s[1:]:
c = [*i] + ['$'] + ans[-min(len(i), len(ans)):]
j = prefix_func(c)
for _ in range(j):
ans.pop()
ans.extend([*i])
print(''.join(ans))
``` | output | 1 | 69,931 | 6 | 139,863 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout | instruction | 0 | 69,932 | 6 | 139,864 |
Tags: brute force, hashing, implementation, string suffix structures, strings
Correct Solution:
```
n = int(input())
words = input().split()[:n]
p_base = 1543
p_mod = 1300199
current = [c for c in words[0]]
for word in words[1:]:
cur_hash = 0
word_hash = 0
cur_base = 1
i_matches = []
same_i = 0
biggest_match = None
while same_i < len(current) and same_i < len(word):
cur_hash *= p_base
cur_hash %= p_mod
cur_hash += ord(current[len(current) - 1 - same_i])
cur_hash %= p_mod
word_hash += ord(word[same_i]) * cur_base
word_hash %= p_mod
cur_base *= p_base
cur_base %= p_mod
if cur_hash == word_hash:
i_matches.append(same_i)
#biggest_match = same_i
same_i += 1
for match in reversed(i_matches):
if ''.join(word[:match + 1]) == ''.join(current[-1 - match:]):
biggest_match = match
break
if biggest_match is None:
current.extend(list(word))
else:
current.extend(list(word[biggest_match + 1:]))
print(*current, sep='')
``` | output | 1 | 69,932 | 6 | 139,865 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout | instruction | 0 | 69,933 | 6 | 139,866 |
Tags: brute force, hashing, implementation, string suffix structures, strings
Correct Solution:
```
from sys import stdin, stdout
def compresswords(n, words):
a = []
for c in words[0]:
a.append(c)
for i in range(1, len(words)):
lps = getlps(words[i])
#print(lps)
idx = getsuffixmatchIdx(a, words[i], lps)
#print(idx)
#if idx == -1:
# idx = 0
for j in range(idx, len(words[i])):
a.append(words[i][j])
return ''.join(a)
def getlps(w):
lps = []
lps.append(-1)
for i in range(1, len(w)):
c = w[i]
idx = i-1
while idx >= 0 and w[lps[idx] + 1] != c:
idx = lps[idx]
if idx >= 0:
idx = lps[idx] + 1
lps.append(idx)
#for i in range(len(lps)):
# lps[i] += 1
return lps
def getsuffixmatchIdx(a, w, lps):
widx = 0
for i in range(max(0, len(a) - len(w)), len(a)):
c = a[i]
#print('w: ' + w[widx] + ' ' + str(widx))
while widx >= 0 and w[widx] != c:
widx -= 1
if widx > 0:
if lps[widx] >= 0:
widx = lps[widx] + 1
else:
widx = 0
#print('c: ' + str(c) + ' ' + str(widx) + ' | ' + str(i))
#print('-------------------------------')
#if widx >= 0:
## find match
#else:
## no match
widx += 1
return widx
if __name__ == '__main__':
n = int(stdin.readline())
words = list(stdin.readline().split())
if n != len(words):
print('length not match')
res = compresswords(n, words)
#stdout.write(res)
print(res)
#lps = getlps('ABABCABABX')
#print(lps)
#a = ['a','b','c','d','A','B']
#r = getsuffixmatchIdx(a, 'ABABCABABX', lps)
#print(r)
``` | output | 1 | 69,933 | 6 | 139,867 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout
Submitted Solution:
```
def zalgo(s:str):
l=0;r=0;n=len(s);
z = [0]*n
for i in range(1,n):
z[i] = 0
if(i<=r): z[i]=min(r-i+1,z[i-l])
while(i+z[i]<n and s[z[i]]==s[i+z[i]]): z[i]+=1
if(i+z[i]-1>r):
l = i
r = i+z[i]-1
if(i+z[i]==n):
return z[i]
return 0
n = int(input())
s = input().split()
txt = []
for i in range(0,n):
x = len(txt)
y = len(s[i])
mn = min(x,y)
#print(mn,txt[-mn:],sep=' ')
mx = zalgo(s[i] + '#' + ''.join((txt[-mn:])))
# print(mx,mn,sep=' ')
txt.extend(s[i][mx:])
print(''.join(txt))
``` | instruction | 0 | 69,934 | 6 | 139,868 |
Yes | output | 1 | 69,934 | 6 | 139,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout
Submitted Solution:
```
def merge(res, s):
pi = [0 for i in range(len(s)+1)]
pi[0] = -1
b = -1
for i in range(1, len(s)+1):
while b > -1 and s[i-1] != s[b]:
b = pi[b]
b+=1
pi[i] = b
b = 0
for i in range(max(0, len(res)-len(s)), len(res)):
while b > -1 and res[i] != s[b]:
b = pi[b]
b+=1
for i in range(b, len(s)):
res.append(s[i])
n = int(input())
s = input().split()
res = [*s[0]]
for i in range(1,n):
merge(res, s[i])
print(''.join(res))
``` | instruction | 0 | 69,935 | 6 | 139,870 |
Yes | output | 1 | 69,935 | 6 | 139,871 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout
Submitted Solution:
```
def zalgo(s:str):
l=0;r=0;n=len(s);
z = [0]*n
for i in range(1,n):
z[i] = 0
if(i<=r): z[i]=min(r-i+1,z[i-l])
while(i+z[i]<n and s[z[i]]==s[i+z[i]]): z[i]+=1
if(i+z[i]-1>r):
l = i
r = i+z[i]-1
if(i+z[i]==n):
return z[i]
return 0
n = int(input())
s = input().split()
txt = []
for i in range(0,n):
x = len(txt)
y = len(s[i])
if x > y:
mx = zalgo(s[i] + '#'+ ''.join((txt[x-y:x])))
else:
mx = zalgo(s[i][:y]+'#'+''.join(txt))
txt.extend(s[i][mx:])
print(''.join(txt))
``` | instruction | 0 | 69,936 | 6 | 139,872 |
Yes | output | 1 | 69,936 | 6 | 139,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout
Submitted Solution:
```
def zalgo(S):
L = len(S)
Z = [0]*L
l = 0
for i in range(1, L):
if i + Z[i-l] < l + Z[l]:
Z[i] = Z[i-l]
else:
cnt = max(0, l+Z[l]-i)
while i + cnt < L and S[i+cnt] == S[cnt]:
cnt += 1
Z[i] = cnt
l = i
Z[0] = L
return Z
N = int(input())
S = [list(map(ord, s)) for s in input().strip().split()]
Ans = [0]*(10**6)
for s in S:
n = len(s)
Z = zalgo(s+[-1]+Ans[-n:])[n+1:] + [0]
for i in range(n+1):
if n - i == Z[i]:
Ans += s[n-i:]
break
Ans = Ans[10**6:]
print(''.join(map(chr, Ans)))
``` | instruction | 0 | 69,937 | 6 | 139,874 |
Yes | output | 1 | 69,937 | 6 | 139,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout
Submitted Solution:
```
import fileinput as fs
def printSingle(words, size):
result = words[0]
resultSize = len(result)
for i in range(1,size):
wordSize = len(words[i])
startingIndex = max(0, resultSize - wordSize)
wordIndex = 0
while startingIndex < resultSize and wordIndex < wordSize:
if result[startingIndex] == words[i][wordIndex]:
startingIndex += 1
wordIndex += 1
else:
if result[startingIndex] == words[i][0]:
wordIndex = 1
else:
wordIndex = 0
startingIndex += 1
if wordIndex != wordSize:
result = result + words[i][wordIndex:]
resultSize = resultSize + wordSize - wordIndex
print(result)
if __name__ == '__main__':
first = True
for line in fs.input():
if first:
size = int(line)
first = not first
continue
else:
words = line.split(' ')
printSingle(words, size)
break
``` | instruction | 0 | 69,938 | 6 | 139,876 |
No | output | 1 | 69,938 | 6 | 139,877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout
Submitted Solution:
```
def mp():
return map(int, input().split())
mod = 10 ** 9 + 223
base = 79
p = [1] * (10 ** 5 + 1)
for i in range(1, 10 ** 5 + 1):
p[i] = (p[i - 1] * base) % mod
def hsh(c):
return ord(c) - 48
def find_i(s, t):
global p, mod, base
hc = [0] * (len(s) + 1)
ht = [0] * (len(t) + 1)
for i in range(len(s)):
hc[i] = (hc[i - 1] * base + hsh(s[i])) % mod
for i in range(len(t)):
ht[i] = (ht[i - 1] * base + hsh(t[i])) % mod
res = -1
for i in range(min(len(s), len(t))):
j = len(s) - i - 1
sv = (hc[len(s) - 1] + mod - hc[j - 1] * p[i + 1]) % mod
tv = ht[i]
#print(i, j, '->', sv, tv)
if sv == tv:
res = max(res, i)
return res
n = int(input())
ss = list(input().split())
print(ss[0], end = '')
for i in range(1, n):
idx = find_i(ss[i - 1], ss[i])
print(ss[i][idx + 1:], end = '')
``` | instruction | 0 | 69,939 | 6 | 139,878 |
No | output | 1 | 69,939 | 6 | 139,879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout
Submitted Solution:
```
_ = input()
s = input().split()
res=s[0][::-1]
for i in range(1, len(s)):
s[i]=s[i][::-1]
res=res+"#"
z = [0 for i in range(len(s[i]))]
l = 0
r = 0
for j in range(1, len(s[i])):
if j <= r:
z[j] = min(r - j + 1, z[j - l])
while (j + z[j] < len(s[i]) and res[z[j]] == s[i][j + z[j]]):
z[j] += 1
if (i + z[j] - 1 > r):
l = j
r = j + z[j] - 1
index=0
while index<len(z):
if z[index]+index == len(z):
break
index+=1
res=s[i][:index]+res
print (res[::-1][int(_)-1::])
``` | instruction | 0 | 69,940 | 6 | 139,880 |
No | output | 1 | 69,940 | 6 | 139,881 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".
Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.
Input
The first line contains an integer n (1 ≤ n ≤ 10^5), the number of the words in Amugae's sentence.
The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 10^6.
Output
In the only line output the compressed word after the merging process ends as described in the problem.
Examples
Input
5
I want to order pizza
Output
Iwantorderpizza
Input
5
sample please ease in out
Output
sampleaseinout
Submitted Solution:
```
lens = int(input())
given = input().split()
answer = given[0]
for i in range(1, len(given)):
curr = given[i]
j = 0
if curr[j] in answer:
indx = answer.rindex(curr[j])
need = answer[indx:]
if curr.startswith(need):
balance = curr[len(need):]
else:
balance = curr
for c in balance:
answer += c
print(answer)
``` | instruction | 0 | 69,941 | 6 | 139,882 |
No | output | 1 | 69,941 | 6 | 139,883 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nam is playing with a string on his computer. The string consists of n lowercase English letters. It is meaningless, so Nam decided to make the string more beautiful, that is to make it be a palindrome by using 4 arrow keys: left, right, up, down.
There is a cursor pointing at some symbol of the string. Suppose that cursor is at position i (1 ≤ i ≤ n, the string uses 1-based indexing) now. Left and right arrow keys are used to move cursor around the string. The string is cyclic, that means that when Nam presses left arrow key, the cursor will move to position i - 1 if i > 1 or to the end of the string (i. e. position n) otherwise. The same holds when he presses the right arrow key (if i = n, the cursor appears at the beginning of the string).
When Nam presses up arrow key, the letter which the text cursor is pointing to will change to the next letter in English alphabet (assuming that alphabet is also cyclic, i. e. after 'z' follows 'a'). The same holds when he presses the down arrow key.
Initially, the text cursor is at position p.
Because Nam has a lot homework to do, he wants to complete this as fast as possible. Can you help him by calculating the minimum number of arrow keys presses to make the string to be a palindrome?
Input
The first line contains two space-separated integers n (1 ≤ n ≤ 105) and p (1 ≤ p ≤ n), the length of Nam's string and the initial position of the text cursor.
The next line contains n lowercase characters of Nam's string.
Output
Print the minimum number of presses needed to change string into a palindrome.
Examples
Input
8 3
aeabcaez
Output
6
Note
A string is a palindrome if it reads the same forward or reversed.
In the sample test, initial Nam's string is: <image> (cursor position is shown bold).
In optimal solution, Nam may do 6 following steps:
<image>
The result, <image>, is now a palindrome. | instruction | 0 | 70,195 | 6 | 140,390 |
Tags: brute force, greedy, implementation
Correct Solution:
```
n, p = map(int, input().split())
p -= 1
st = input().strip()
l = 0
h = n // 2 - 1
if p > h:
p = n - 1 - p
inside = False
ans = 0
i = 0
H = -1
while i <= h:
if st[i] != st[n - 1 - i]:
if not inside:
l = i
inside = True
H = i
ans += min(abs(ord(st[i]) - ord(st[n - 1 - i])), abs(ord('z') - ord('a') + 1 - abs(ord(st[i]) - ord(st[n - 1 - i]))))
i += 1
h = H
if h == -1:
print("0")
exit(0)
ans += h - l
ans += min(abs(l - p), abs(p - h))
print(ans)
``` | output | 1 | 70,195 | 6 | 140,391 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nam is playing with a string on his computer. The string consists of n lowercase English letters. It is meaningless, so Nam decided to make the string more beautiful, that is to make it be a palindrome by using 4 arrow keys: left, right, up, down.
There is a cursor pointing at some symbol of the string. Suppose that cursor is at position i (1 ≤ i ≤ n, the string uses 1-based indexing) now. Left and right arrow keys are used to move cursor around the string. The string is cyclic, that means that when Nam presses left arrow key, the cursor will move to position i - 1 if i > 1 or to the end of the string (i. e. position n) otherwise. The same holds when he presses the right arrow key (if i = n, the cursor appears at the beginning of the string).
When Nam presses up arrow key, the letter which the text cursor is pointing to will change to the next letter in English alphabet (assuming that alphabet is also cyclic, i. e. after 'z' follows 'a'). The same holds when he presses the down arrow key.
Initially, the text cursor is at position p.
Because Nam has a lot homework to do, he wants to complete this as fast as possible. Can you help him by calculating the minimum number of arrow keys presses to make the string to be a palindrome?
Input
The first line contains two space-separated integers n (1 ≤ n ≤ 105) and p (1 ≤ p ≤ n), the length of Nam's string and the initial position of the text cursor.
The next line contains n lowercase characters of Nam's string.
Output
Print the minimum number of presses needed to change string into a palindrome.
Examples
Input
8 3
aeabcaez
Output
6
Note
A string is a palindrome if it reads the same forward or reversed.
In the sample test, initial Nam's string is: <image> (cursor position is shown bold).
In optimal solution, Nam may do 6 following steps:
<image>
The result, <image>, is now a palindrome. | instruction | 0 | 70,197 | 6 | 140,394 |
Tags: brute force, greedy, implementation
Correct Solution:
```
n, p = map(int, input().split())
p -= 1
if p >= n // 2 : p = n - 1 - p
s = input()
d = [abs(ord(s[i]) - ord(s[n - i - 1])) for i in range (n // 2)]
d = [min(x, 26 - x) for x in d]
first = last = -1
for i in range (len(d)) :
if d[i] > 0 :
if first == -1 : first = i
last = i
print(0 if first == -1 else min(abs(p - first), abs(p-last)) + (last - first) + sum(d))
``` | output | 1 | 70,197 | 6 | 140,395 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nam is playing with a string on his computer. The string consists of n lowercase English letters. It is meaningless, so Nam decided to make the string more beautiful, that is to make it be a palindrome by using 4 arrow keys: left, right, up, down.
There is a cursor pointing at some symbol of the string. Suppose that cursor is at position i (1 ≤ i ≤ n, the string uses 1-based indexing) now. Left and right arrow keys are used to move cursor around the string. The string is cyclic, that means that when Nam presses left arrow key, the cursor will move to position i - 1 if i > 1 or to the end of the string (i. e. position n) otherwise. The same holds when he presses the right arrow key (if i = n, the cursor appears at the beginning of the string).
When Nam presses up arrow key, the letter which the text cursor is pointing to will change to the next letter in English alphabet (assuming that alphabet is also cyclic, i. e. after 'z' follows 'a'). The same holds when he presses the down arrow key.
Initially, the text cursor is at position p.
Because Nam has a lot homework to do, he wants to complete this as fast as possible. Can you help him by calculating the minimum number of arrow keys presses to make the string to be a palindrome?
Input
The first line contains two space-separated integers n (1 ≤ n ≤ 105) and p (1 ≤ p ≤ n), the length of Nam's string and the initial position of the text cursor.
The next line contains n lowercase characters of Nam's string.
Output
Print the minimum number of presses needed to change string into a palindrome.
Examples
Input
8 3
aeabcaez
Output
6
Note
A string is a palindrome if it reads the same forward or reversed.
In the sample test, initial Nam's string is: <image> (cursor position is shown bold).
In optimal solution, Nam may do 6 following steps:
<image>
The result, <image>, is now a palindrome. | instruction | 0 | 70,198 | 6 | 140,396 |
Tags: brute force, greedy, implementation
Correct Solution:
```
n,p=map(int,input().split())
s=input()
ans=[]
for i in range(len(s)):
ans.append(s[i])
sm1=0
la1=0
flag=0
for i in range(len(ans)//2):
if(ans[i]!=ans[-1*(i+1)] and flag==0):
sm1=i+1
flag=1
if(ans[i]!=ans[-1*(i+1)]):
la1=i+1
if(sm1==0):
print(0)
exit()
if(len(ans)==1):
print(0)
exit()
if(p>len(ans)//2):
p=len(ans)+1-p
if(p<sm1):
count=0
for i in range(p-1,la1):
diff=max(ord(ans[i]),ord(ans[-1*(i+1)]))-min(ord(ans[i]),ord(ans[-1*(i+1)]))
count+=min(diff,26-diff)
print(count+la1-p)
elif(sm1<=p<=la1):
count=la1-sm1
for i in range(sm1-1,la1):
diff=max(ord(ans[i]),ord(ans[-1*(i+1)]))-min(ord(ans[i]),ord(ans[-1*(i+1)]))
count+=min(diff,26-diff)
print(count+min(p-sm1,la1-p))
else:
count=0
for i in range(p-1,sm1-2,-1):
diff=max(ord(ans[i]),ord(ans[-1*(i+1)]))-min(ord(ans[i]),ord(ans[-1*(i+1)]))
count+=min(diff,26-diff)
print(count+p-sm1)
``` | output | 1 | 70,198 | 6 | 140,397 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nam is playing with a string on his computer. The string consists of n lowercase English letters. It is meaningless, so Nam decided to make the string more beautiful, that is to make it be a palindrome by using 4 arrow keys: left, right, up, down.
There is a cursor pointing at some symbol of the string. Suppose that cursor is at position i (1 ≤ i ≤ n, the string uses 1-based indexing) now. Left and right arrow keys are used to move cursor around the string. The string is cyclic, that means that when Nam presses left arrow key, the cursor will move to position i - 1 if i > 1 or to the end of the string (i. e. position n) otherwise. The same holds when he presses the right arrow key (if i = n, the cursor appears at the beginning of the string).
When Nam presses up arrow key, the letter which the text cursor is pointing to will change to the next letter in English alphabet (assuming that alphabet is also cyclic, i. e. after 'z' follows 'a'). The same holds when he presses the down arrow key.
Initially, the text cursor is at position p.
Because Nam has a lot homework to do, he wants to complete this as fast as possible. Can you help him by calculating the minimum number of arrow keys presses to make the string to be a palindrome?
Input
The first line contains two space-separated integers n (1 ≤ n ≤ 105) and p (1 ≤ p ≤ n), the length of Nam's string and the initial position of the text cursor.
The next line contains n lowercase characters of Nam's string.
Output
Print the minimum number of presses needed to change string into a palindrome.
Examples
Input
8 3
aeabcaez
Output
6
Note
A string is a palindrome if it reads the same forward or reversed.
In the sample test, initial Nam's string is: <image> (cursor position is shown bold).
In optimal solution, Nam may do 6 following steps:
<image>
The result, <image>, is now a palindrome. | instruction | 0 | 70,199 | 6 | 140,398 |
Tags: brute force, greedy, implementation
Correct Solution:
```
n,p=map(int,input().split())
s=input()
ans=0
p-=1
if(n%2==1 and p==n//2 and n!=1):
p-=1
ans+=1
cnt=0
for i in range(n):
if(n-1-i<=i):
break
if(s[i]!=s[n-1-i]):
cnt+=1
if(p>=n//2 and n!=1):
h=""
for i in range(n-1,-1,-1):
h+=s[i]
s=h
p=n-1-p
r=0
l=0
rr=0
ll=0
for i in range(p):
if(s[i]!=s[n-1-i]):
ans+=min(26-abs(ord(s[i])-ord(s[n-1-i])),abs(ord(s[i])-ord(s[n-1-i])))
r+=1
rr=max(rr,p-i)
for i in range(p+1,n//2):
if(s[i]!=s[n-1-i]):
ans+=min(26-abs(ord(s[i])-ord(s[n-1-i])),abs(ord(s[i])-ord(s[n-1-i])))
l+=1
ll=max(ll,i-p)
ans+=min(26-abs(ord(s[p])-ord(s[n-1-p])),abs(ord(s[p])-ord(s[n-1-p])))
ans+=ll+rr+min(ll,rr)
print(ans)
``` | output | 1 | 70,199 | 6 | 140,399 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nam is playing with a string on his computer. The string consists of n lowercase English letters. It is meaningless, so Nam decided to make the string more beautiful, that is to make it be a palindrome by using 4 arrow keys: left, right, up, down.
There is a cursor pointing at some symbol of the string. Suppose that cursor is at position i (1 ≤ i ≤ n, the string uses 1-based indexing) now. Left and right arrow keys are used to move cursor around the string. The string is cyclic, that means that when Nam presses left arrow key, the cursor will move to position i - 1 if i > 1 or to the end of the string (i. e. position n) otherwise. The same holds when he presses the right arrow key (if i = n, the cursor appears at the beginning of the string).
When Nam presses up arrow key, the letter which the text cursor is pointing to will change to the next letter in English alphabet (assuming that alphabet is also cyclic, i. e. after 'z' follows 'a'). The same holds when he presses the down arrow key.
Initially, the text cursor is at position p.
Because Nam has a lot homework to do, he wants to complete this as fast as possible. Can you help him by calculating the minimum number of arrow keys presses to make the string to be a palindrome?
Input
The first line contains two space-separated integers n (1 ≤ n ≤ 105) and p (1 ≤ p ≤ n), the length of Nam's string and the initial position of the text cursor.
The next line contains n lowercase characters of Nam's string.
Output
Print the minimum number of presses needed to change string into a palindrome.
Examples
Input
8 3
aeabcaez
Output
6
Note
A string is a palindrome if it reads the same forward or reversed.
In the sample test, initial Nam's string is: <image> (cursor position is shown bold).
In optimal solution, Nam may do 6 following steps:
<image>
The result, <image>, is now a palindrome. | instruction | 0 | 70,200 | 6 | 140,400 |
Tags: brute force, greedy, implementation
Correct Solution:
```
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
# from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
# import numpy as np
starttime = time.time()
# import numpy as np
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
# sys.setrecursionlimit(int(pow(10,6)))
sys.stdin = open("input.txt", "r")
# sys.stdout = open("../output.txt", "w")
except:
pass
n,p=map(int,input().split())
l=list(input())
p-=1
k=n//2
d=[0]*(k+1)
for i in range(k):
d[i]=min(abs(ord(l[i])-ord(l[n-i-1])),26-abs(ord(l[i])-ord(l[n-i-1])))
if p>=k:
p=n-1-p
if sum(d)==0:
print(0)
exit()
i=0
j=k
while d[i]==0:
i += 1
while d[j]==0:
j -= 1
if i>p:
i=p
if j<p:
j=p
print(sum(d)+min(p-2*i+j,2*j-i-p))
endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
``` | output | 1 | 70,200 | 6 | 140,401 |
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