message stringlengths 2 23.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by η«ε± https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 β€ n β€ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer β the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Submitted Solution:
```
s = str(input())
counter = 0
for i in range(len(s)):
if s[i] == 'Q':
for j in range(i, len(s)):
if s[j] == 'A':
for k in range(j, len(s)):
if s[k] == 'Q':
counter += 1
print(counter)
``` | instruction | 0 | 38,281 | 0 | 76,562 |
Yes | output | 1 | 38,281 | 0 | 76,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by η«ε± https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 β€ n β€ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer β the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Submitted Solution:
```
from itertools import combinations as c
print(sum(map(lambda x: x == ('Q', 'A', 'Q'), c((i for i in input() if i in 'QA'), 3))))
``` | instruction | 0 | 38,282 | 0 | 76,564 |
Yes | output | 1 | 38,282 | 0 | 76,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by η«ε± https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 β€ n β€ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer β the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Submitted Solution:
```
from functools import lru_cache
def subseqsearch(string,substr):
substrset=set(substr)
#fixs has only element in substr
fixs = [i for i in string if i in substrset]
@lru_cache(maxsize=None) #memoisation decorator applyed to recs()
def recs(fi=0,si=0):
if si >= len(substr):
return 1
r=0
for i in range(fi,len(fixs)):
if substr[si] == fixs[i]:
r+=recs(i+1,si+1)
return r
return recs()
#test
from functools import reduce
def flat(i) : return reduce(lambda x,y:x+y,i,[])
string = input()
substr = "QAQ"
#print("".join(str(i) for i in string),"substr:","".join(str(i) for i in substr),sep="\n")
print(subseqsearch(string,substr))
``` | instruction | 0 | 38,283 | 0 | 76,566 |
Yes | output | 1 | 38,283 | 0 | 76,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by η«ε± https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 β€ n β€ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer β the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Submitted Solution:
```
x=input()
c=0
for i in range(0,len(x)):
if x[i]=='Q':
for j in range(i,len(x)):
if x[j]=='A':
for k in range(j,len(x)):
if x[k]=='Q':
c+=1
print(c)
``` | instruction | 0 | 38,284 | 0 | 76,568 |
Yes | output | 1 | 38,284 | 0 | 76,569 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by η«ε± https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 β€ n β€ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer β the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Submitted Solution:
```
import math as mt
import sys,string,bisect
input=sys.stdin.readline
import random
from collections import deque,defaultdict
L=lambda : list(map(int,input().split()))
Ls=lambda : list(input().split())
M=lambda : map(int,input().split())
I=lambda :int(input())
s=input().strip()
i=0
ca=0
cq=0
f=0
ans=0
while(i<len(s)):
if(f==0):
if(s[i]=="Q"):
f=1
cq+=1
else:
if(s[i]=="Q"):
ans=ca
cq+=1
elif(s[i]=="A"):
ca+=1
i+=1
print((cq-1)*ans)
``` | instruction | 0 | 38,285 | 0 | 76,570 |
No | output | 1 | 38,285 | 0 | 76,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by η«ε± https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 β€ n β€ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer β the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Submitted Solution:
```
s = input()
cnt = 0
lst = []
n = len(s)
for i in s:
if i == "Q":
cnt += 1
lst.append(cnt)
ans = 0
for i in range(n):
if s[i] == "A":
ans += (lst[i]*(lst[n - i - 1] - lst[i]))
print(ans)
``` | instruction | 0 | 38,286 | 0 | 76,572 |
No | output | 1 | 38,286 | 0 | 76,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).
<image> illustration by η«ε± https://twitter.com/nekoyaliu
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input
The only line contains a string of length n (1 β€ n β€ 100). It's guaranteed that the string only contains uppercase English letters.
Output
Print a single integer β the number of subsequences "QAQ" in the string.
Examples
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
Submitted Solution:
```
n=input()
i=len(n)
l=[0]*(i)
c=0
for i in range(0,len(n)):
if n[i]=='Q':
c=c+1
l[i]=c
ans=0
total=l[len(n)-1]
for i in range(0,len(n)):
if n[i]=='A':
f1=l[i]
f2=total-f1
ans=ans+max(f1,f2)
print(ans)
``` | instruction | 0 | 38,288 | 0 | 76,576 |
No | output | 1 | 38,288 | 0 | 76,577 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S of length 2N, containing N occurrences of `a` and N occurrences of `b`.
You will choose some of the characters in S. Here, for each i = 1,2,...,N, it is not allowed to choose exactly one of the following two: the i-th occurrence of `a` and the i-th occurrence of `b`. (That is, you can only choose both or neither.) Then, you will concatenate the chosen characters (without changing the order).
Find the lexicographically largest string that can be obtained in this way.
Constraints
* 1 \leq N \leq 3000
* S is a string of length 2N containing N occurrences of `a` and N occurrences of `b`.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the lexicographically largest string that satisfies the condition.
Examples
Input
3
aababb
Output
abab
Input
3
bbabaa
Output
bbabaa
Input
6
bbbaabbabaaa
Output
bbbabaaa
Input
9
abbbaababaababbaba
Output
bbaababababa | instruction | 0 | 38,408 | 0 | 76,816 |
"Correct Solution:
```
from itertools import accumulate
def search_ab(sss, cursor):
# print('a', cursor)
ai = aaa.index(cursor)
tmp_cur = bbb[ai]
max_cur = sss.index(0, cursor)
repeat = 1
while tmp_cur < max_cur:
cur = s.find('a', tmp_cur, max_cur)
if cur == -1:
break
ai = aaa.index(cur, ai)
tmp_cur = bbb[ai]
repeat += 1
return repeat, max_cur + 1
def search_ba(sss, cursor):
# print('b', cursor)
first_bi = bbb.index(cursor)
max_cursor = sss.index(0, cursor)
last_bi = aaa.index(max_cursor)
tmp_buf = [''] * (last_bi - first_bi + 1) * 2
tmp_max = ''
for i in range(last_bi, first_bi - 1, -1):
tmp_buf[aaa[i] - cursor] = 'a'
tmp_buf[bbb[i] - cursor] = 'b'
tmp = ''.join(tmp_buf)
if tmp > tmp_max:
tmp_max = tmp
return tmp_max, max_cursor + 1
def integrate(parts_b):
tmp_max = ''
for pb in reversed(parts_b):
tmp = pb + tmp_max
if tmp > tmp_max:
tmp_max = tmp
return tmp_max
n = int(input())
s = input()
n2 = n * 2
sss = []
aaa, bbb = [], []
for i, c in enumerate(s):
if c == 'a':
aaa.append(i)
sss.append(-1)
else:
bbb.append(i)
sss.append(1)
sss = list(accumulate(sss))
repeats_a = []
parts_b = []
last_b_cur = 0
cursor = 0
while cursor < n2:
c = sss[cursor]
if c < 0:
repeat, cursor = search_ab(sss, cursor)
repeats_a.append((cursor, repeat))
else:
tmp, cursor = search_ba(sss, cursor)
parts_b.append(tmp)
last_b_cur = cursor
print(integrate(parts_b) + 'ab' * sum(r for c, r in repeats_a if c > last_b_cur))
``` | output | 1 | 38,408 | 0 | 76,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length 2N, containing N occurrences of `a` and N occurrences of `b`.
You will choose some of the characters in S. Here, for each i = 1,2,...,N, it is not allowed to choose exactly one of the following two: the i-th occurrence of `a` and the i-th occurrence of `b`. (That is, you can only choose both or neither.) Then, you will concatenate the chosen characters (without changing the order).
Find the lexicographically largest string that can be obtained in this way.
Constraints
* 1 \leq N \leq 3000
* S is a string of length 2N containing N occurrences of `a` and N occurrences of `b`.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the lexicographically largest string that satisfies the condition.
Examples
Input
3
aababb
Output
abab
Input
3
bbabaa
Output
bbabaa
Input
6
bbbaabbabaaa
Output
bbbabaaa
Input
9
abbbaababaababbaba
Output
bbaababababa
Submitted Solution:
```
n = int(input())
s = input()
ab_index = {'a': 0, 'b': 0}
ab_dic = {}
i = 0
while i < (2 * n):
ab_dic[i] = (s[i], ab_index[s[i]])
ab_index[s[i]] += 1
i += 1
import copy
class Info:
def __init__(self, string, includeds, excludeds):
self.string = string
self.includeds = includeds
self.excludeds = excludeds
def include(self, s, i):
dic = copy.copy(self.includeds)
dic[(s, i)] = True
return Info(self.string + s, dic, self.excludeds)
def exclude(self, s, i):
dic = copy.copy(self.excludeds)
dic[(s, i)] = True
return Info(self.string, self.includeds, dic)
def __repr__(self):
return '({0}, {1}, {2})'.format(self.string, self.includeds, self.excludeds)
dp: [[Info]] = [[None for _ in range(len(s) + 1)] for _ in range(len(s) + 1)]
dp[0][0] = Info('', {}, {})
for i in range(len(s)):
ab, abi = ab_dic[i]
for j in range(i + 1):
inf = dp[i][j]
if inf is None:
continue
if ab == 'a':
if inf.includeds.get(('b', abi)):
dp[i + 1][j + 1] = inf.include('a', abi)
elif inf.excludeds.get(('b', abi)):
candidate = inf.exclude('a', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string <= candidate.string:
dp[i + 1][j] = candidate
else:
pass
else:
dp[i + 1][j + 1] = inf.include('a', abi)
candidate = inf.exclude('a', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string <= candidate.string:
dp[i + 1][j] = candidate
else:
pass
if ab == 'b':
if inf.includeds.get(('a', abi)):
dp[i + 1][j + 1] = inf.include('b', abi)
elif inf.excludeds.get(('a', abi)):
candidate = inf.exclude('b', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string < candidate.string:
dp[i + 1][j] = candidate
else:
pass
else:
dp[i + 1][j + 1] = inf.include('b', abi)
candidate = inf.exclude('b', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string < candidate.string:
dp[i + 1][j] = candidate
else:
pass
print(sorted(map(lambda x: x.string, filter(lambda x: x is not None, dp[-1])))[-1])
``` | instruction | 0 | 38,409 | 0 | 76,818 |
No | output | 1 | 38,409 | 0 | 76,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length 2N, containing N occurrences of `a` and N occurrences of `b`.
You will choose some of the characters in S. Here, for each i = 1,2,...,N, it is not allowed to choose exactly one of the following two: the i-th occurrence of `a` and the i-th occurrence of `b`. (That is, you can only choose both or neither.) Then, you will concatenate the chosen characters (without changing the order).
Find the lexicographically largest string that can be obtained in this way.
Constraints
* 1 \leq N \leq 3000
* S is a string of length 2N containing N occurrences of `a` and N occurrences of `b`.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the lexicographically largest string that satisfies the condition.
Examples
Input
3
aababb
Output
abab
Input
3
bbabaa
Output
bbabaa
Input
6
bbbaabbabaaa
Output
bbbabaaa
Input
9
abbbaababaababbaba
Output
bbaababababa
Submitted Solution:
```
import bisect
from itertools import accumulate, combinations
def search_ab(sss, cursor):
ai = aaa.index(cursor)
tmp_cur = bbb[ai]
repeat = 1
while sss[tmp_cur] < 0:
repeat += 1
ai = bisect.bisect(aaa, tmp_cur, lo=ai)
tmp_cur = bbb[ai]
return 'ab' * repeat, tmp_cur + 1
def search_ba(sss, cursor):
first_bi = bbb.index(cursor)
last_cursor = sss.index(0, cursor)
last_bi = aaa.index(last_cursor)
tmp_buf = [''] * (last_bi - first_bi + 1) * 2
tmp_max = ''
for i in range(last_bi, first_bi - 1, -1):
tmp_buf[aaa[i] - cursor] = 'a'
tmp_buf[bbb[i] - cursor] = 'b'
tmp = ''.join(tmp_buf)
if tmp > tmp_max:
tmp_max = tmp
return tmp_max, last_cursor + 1
def integrate(parts_a, parts_b):
if not parts_b:
return ''.join(parts_a)
tmp_max = ''
for k in range(1, len(parts_b) + 1):
for cmb in combinations(parts_b, k):
tmp = ''.join(cmb)
if tmp > tmp_max:
tmp_max = tmp
return tmp_max
n = int(input())
n2 = n * 2
s = input()
sss = []
aaa, bbb = [], []
for i, c in enumerate(s):
if c == 'a':
aaa.append(i)
sss.append(-1)
else:
bbb.append(i)
sss.append(1)
sss = list(accumulate(sss))
parts_a = []
parts_b = []
cursor = 0
while cursor < n2:
c = sss[cursor]
if c < 0:
tmp, cursor = search_ab(sss, cursor)
parts_a.append(tmp)
else:
tmp, cursor = search_ba(sss, cursor)
parts_b.append(tmp)
print(integrate(parts_a, parts_b))
``` | instruction | 0 | 38,410 | 0 | 76,820 |
No | output | 1 | 38,410 | 0 | 76,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length 2N, containing N occurrences of `a` and N occurrences of `b`.
You will choose some of the characters in S. Here, for each i = 1,2,...,N, it is not allowed to choose exactly one of the following two: the i-th occurrence of `a` and the i-th occurrence of `b`. (That is, you can only choose both or neither.) Then, you will concatenate the chosen characters (without changing the order).
Find the lexicographically largest string that can be obtained in this way.
Constraints
* 1 \leq N \leq 3000
* S is a string of length 2N containing N occurrences of `a` and N occurrences of `b`.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the lexicographically largest string that satisfies the condition.
Examples
Input
3
aababb
Output
abab
Input
3
bbabaa
Output
bbabaa
Input
6
bbbaabbabaaa
Output
bbbabaaa
Input
9
abbbaababaababbaba
Output
bbaababababa
Submitted Solution:
```
from itertools import accumulate
def search_ab(sss, cursor):
# print('a', cursor)
ai = aaa.index(cursor)
tmp_cur = bbb[ai]
max_cur = sss.index(0, cursor)
repeat = 1
while tmp_cur < max_cur:
cur = s.find('a', tmp_cur, max_cur)
if cur == -1:
break
ai = aaa.index(cur, ai)
tmp_cur = bbb[ai]
repeat += 1
return repeat, max_cur + 1
def search_ba(sss, cursor):
# print('b', cursor)
first_bi = bbb.index(cursor)
max_cursor = sss.index(0, cursor)
last_bi = aaa.index(max_cursor)
tmp_buf = [''] * (last_bi - first_bi + 1) * 2
tmp_max = ''
for i in range(last_bi, first_bi - 1, -1):
tmp_buf[aaa[i] - cursor] = 'a'
tmp_buf[bbb[i] - cursor] = 'b'
tmp = ''.join(tmp_buf)
if tmp > tmp_max:
tmp_max = tmp
return tmp_max, max_cursor + 1
def integrate(parts_b):
tmp_max = ''
for pb in reversed(parts_b):
tmp = pb + tmp_max
if tmp > tmp_max:
tmp_max = tmp
return tmp_max
n = int(input())
s = input()
n2 = n * 2
sss = []
aaa, bbb = [], []
for i, c in enumerate(s):
if c == 'a':
aaa.append(i)
sss.append(-1)
else:
bbb.append(i)
sss.append(1)
sss = list(accumulate(sss))
repeat_a = 0
parts_b = []
cursor = 0
while cursor < n2:
c = sss[cursor]
if c < 0:
repeat, cursor = search_ab(sss, cursor)
repeat_a += repeat
else:
tmp, cursor = search_ba(sss, cursor)
parts_b.append(tmp)
if parts_b:
print(integrate(parts_b))
else:
print('ab' * repeat_a)
``` | instruction | 0 | 38,411 | 0 | 76,822 |
No | output | 1 | 38,411 | 0 | 76,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length 2N, containing N occurrences of `a` and N occurrences of `b`.
You will choose some of the characters in S. Here, for each i = 1,2,...,N, it is not allowed to choose exactly one of the following two: the i-th occurrence of `a` and the i-th occurrence of `b`. (That is, you can only choose both or neither.) Then, you will concatenate the chosen characters (without changing the order).
Find the lexicographically largest string that can be obtained in this way.
Constraints
* 1 \leq N \leq 3000
* S is a string of length 2N containing N occurrences of `a` and N occurrences of `b`.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the lexicographically largest string that satisfies the condition.
Examples
Input
3
aababb
Output
abab
Input
3
bbabaa
Output
bbabaa
Input
6
bbbaabbabaaa
Output
bbbabaaa
Input
9
abbbaababaababbaba
Output
bbaababababa
Submitted Solution:
```
n = int(input())
s = input()
ab_index = {'a': 0, 'b': 0}
ab_dic = {}
i = 0
while i < (2 * n):
ab_dic[i] = (s[i], ab_index[s[i]])
ab_index[s[i]] += 1
i += 1
import copy
class Info:
def __init__(self, string, includeds, excludeds):
self.string = string
self.includeds = includeds
self.excludeds = excludeds
def include(self, s, i):
dic = copy.copy(self.includeds)
dic[(s, i)] = True
return Info(self.string + s, dic, self.excludeds)
def exclude(self, s, i):
dic = copy.copy(self.excludeds)
dic[(s, i)] = True
return Info(self.string, self.includeds, dic)
def __repr__(self):
return '({0}, {1}, {2})'.format(self.string, self.includeds, self.excludeds)
dp = [[None for _ in range(len(s) + 1)] for _ in range(len(s) + 1)]
dp[0][0] = Info('', {}, {})
for i in range(len(s)):
ab, abi = ab_dic[i]
for j in range(i + 1):
inf = dp[i][j]
if inf is None:
continue
if inf.string == 'aa':
dp[i][j] = None
continue
if ab == 'a':
if inf.includeds.get(('b', abi)):
dp[i + 1][j + 1] = inf.include('a', abi)
elif inf.excludeds.get(('b', abi)):
candidate = inf.exclude('a', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string <= candidate.string:
dp[i + 1][j] = candidate
else:
pass
else:
dp[i + 1][j + 1] = inf.include('a', abi)
candidate = inf.exclude('a', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string <= candidate.string:
dp[i + 1][j] = candidate
else:
pass
if ab == 'b':
if inf.includeds.get(('a', abi)):
dp[i + 1][j + 1] = inf.include('b', abi)
elif inf.excludeds.get(('a', abi)):
candidate = inf.exclude('b', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string < candidate.string:
dp[i + 1][j] = candidate
else:
pass
else:
dp[i + 1][j + 1] = inf.include('b', abi)
candidate = inf.exclude('b', abi)
if dp[i + 1][j] is None:
dp[i + 1][j] = candidate
else:
if dp[i + 1][j].string < candidate.string:
dp[i + 1][j] = candidate
else:
pass
print(sorted(map(lambda x: x.string, filter(lambda x: x is not None, dp[-1])))[-1])
``` | instruction | 0 | 38,412 | 0 | 76,824 |
No | output | 1 | 38,412 | 0 | 76,825 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda". | instruction | 0 | 38,650 | 0 | 77,300 |
Tags: greedy, strings
Correct Solution:
```
n=int(input())
s=input()
f=1
for i in range(n-1):
if ord(s[i])>ord(s[i+1]) and f:
f=0
# print("DF",s[i])
pass
else:
print(s[i],end="")
if f==0:
print(s[i+1])
``` | output | 1 | 38,650 | 0 | 77,301 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda". | instruction | 0 | 38,651 | 0 | 77,302 |
Tags: greedy, strings
Correct Solution:
```
def ii():
return int(input())
def mi():
return map(int, input().split())
def li():
return list(mi())
n = ii()
s = input().strip()
i = 0
while i < n - 1:
if ord(s[i]) > ord(s[i + 1]):
break
i += 1
ans = s[:i] + s[i + 1:]
print(ans)
``` | output | 1 | 38,651 | 0 | 77,303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda". | instruction | 0 | 38,652 | 0 | 77,304 |
Tags: greedy, strings
Correct Solution:
```
n = int(input())
s = input()
for i in range(1, n):
if s[i] < s[i - 1]:
print(s[:i - 1] + s[i:])
break
elif i == n - 1:
print(s[:-1])
``` | output | 1 | 38,652 | 0 | 77,305 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda". | instruction | 0 | 38,653 | 0 | 77,306 |
Tags: greedy, strings
Correct Solution:
```
n = int(input())
text = input()
remove_idx = -1
for i in range(len(text) - 1):
if ord(text[i]) - ord(text[i + 1]) > 0:
remove_idx = i
break
if remove_idx >= 0:
print(text[:remove_idx] + text[remove_idx + 1:])
else:
print(text[:-1])
``` | output | 1 | 38,653 | 0 | 77,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda". | instruction | 0 | 38,654 | 0 | 77,308 |
Tags: greedy, strings
Correct Solution:
```
n = int(input())
#a = [int(i) for i in input().split()]
s = input()
max = 0
num = 0
for i in range(n-1):
if s[i] > s[i+1]:
print(s[:i]+s[i+1:])
exit()
print(s[:-1])
``` | output | 1 | 38,654 | 0 | 77,309 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda". | instruction | 0 | 38,655 | 0 | 77,310 |
Tags: greedy, strings
Correct Solution:
```
n=int(input())
s=input()
index=-1
for i in range(1,n):
if s[i]<s[i-1]:
index=i-1
break
if index==-1:
print(s[:-1])
else:
print(s[:index]+s[index+1:])
``` | output | 1 | 38,655 | 0 | 77,311 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda". | instruction | 0 | 38,656 | 0 | 77,312 |
Tags: greedy, strings
Correct Solution:
```
import sys
# import heapq, collections
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**15
mod = 10**9+7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def solve(n, s):
_set = set(list(s))
ret = s
for ch in _set:
i = s.index(ch)
ret = min(ret, s[:i] + s[i+1:])
return ret
def main():
n = I()
s = S()
print(solve(n, s))
main()
``` | output | 1 | 38,656 | 0 | 77,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda". | instruction | 0 | 38,657 | 0 | 77,314 |
Tags: greedy, strings
Correct Solution:
```
n = int(input())
s = input()
k = n-1
for i in range(1,n):
if s[i]<s[i-1]:
k = i-1
break
for i in range(n):
if i==k:
continue
print(s[i],end="")
``` | output | 1 | 38,657 | 0 | 77,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
def get_smallest(string):
for idx in range(len(string) - 1):
if string[idx] > string [idx+1]:
return string[:idx] + string[idx+1:]
return string[:-1]
n = int(input())
string = input()
result = get_smallest(string)
print(result)
``` | instruction | 0 | 38,658 | 0 | 77,316 |
Yes | output | 1 | 38,658 | 0 | 77,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
n = int(input())
string = input()
done = 0
for ind in range (len(string)-1):
if string[ind]>string[ind+1]:
string = string[:ind] + string[ind+1:]
done = 1
break
if done==0:
string = string[:-1]
print(string)
``` | instruction | 0 | 38,659 | 0 | 77,318 |
Yes | output | 1 | 38,659 | 0 | 77,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
def a():
n = int(input())
s = input()
for i in range(n-1):
if s[i] > s[i+1]:
return s[:i]+s[i+1:]
return s[:-1]
print(a())
``` | instruction | 0 | 38,660 | 0 | 77,320 |
Yes | output | 1 | 38,660 | 0 | 77,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
n = int(input())
s = str(input())
for i in range(n-1):
if s[i] > s[i+1]:
print(s[0:i] + s[i+1:])
raise SystemExit
print(s[0:n-1])
``` | instruction | 0 | 38,661 | 0 | 77,322 |
Yes | output | 1 | 38,661 | 0 | 77,323 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
n,s=int(input()),input()
for i in range(1,n):
if s[i]<s[i-1]:break
print(s[:i-1]+s[i:])
``` | instruction | 0 | 38,662 | 0 | 77,324 |
No | output | 1 | 38,662 | 0 | 77,325 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
n=int(input())
s=input()
t=-1
for i in range(n-1):
if s[i]>s[i+1]:
t=i
break
if t==-1:t=n-1
print(s[:i]+s[i+1:])
``` | instruction | 0 | 38,663 | 0 | 77,326 |
No | output | 1 | 38,663 | 0 | 77,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
n=int(input())
s=input().strip()
x=n-1
while x>0:
if s[x]<s[x-1]:
s=s[:x-1]+s[x:]
break
x=x-1
if x==0:
s=s[:-1]
print(s)
``` | instruction | 0 | 38,664 | 0 | 77,328 |
No | output | 1 | 38,664 | 0 | 77,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.
String s = s_1 s_2 ... s_n is lexicographically smaller than string t = t_1 t_2 ... t_m if n < m and s_1 = t_1, s_2 = t_2, ..., s_n = t_n or there exists a number p such that p β€ min(n, m) and s_1 = t_1, s_2 = t_2, ..., s_{p-1} = t_{p-1} and s_p < t_p.
For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".
Input
The first line of the input contains one integer n (2 β€ n β€ 2 β
10^5) β the length of s.
The second line of the input contains exactly n lowercase Latin letters β the string s.
Output
Print one string β the smallest possible lexicographically string that can be obtained by removing at most one character from the string s.
Examples
Input
3
aaa
Output
aa
Input
5
abcda
Output
abca
Note
In the first example you can remove any character of s to obtain the string "aa".
In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".
Submitted Solution:
```
n=int(input())
a=input()
x=0
for i in range(n-1):
if ord(a[i])<=ord(a[i+1]):
pass
else:
x=1
break
if x==1:
print(a.replace(a[i],''))
else:
print(a[:-1])
``` | instruction | 0 | 38,665 | 0 | 77,330 |
No | output | 1 | 38,665 | 0 | 77,331 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz | instruction | 0 | 38,698 | 0 | 77,396 |
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def plus(a, b):
c = ""
i = len(a)-1
arg = 0
while i >= 0:
k = (arg + ord(a[i]) + ord(b[i]) - 194)
arg = 1
if k < 26:
arg = 0
i -= 1
c = chr(k % 26) + c
c = chr(1) * arg + c
return c
def minus(a):
c = ""
i = 0
arg = 0
while i < len(a):
c += chr((arg * 26 + ord(a[i]))//2)
arg = ord(a[i]) % 2
i += 1
if arg:
c = c[:len(c)-1] + chr(ord(c[len(c)-1])+1)
return c
n = int(input())
a1 = ""
i = 0
while i < n:
a1 += "s"
i += 1
c = plus(input() ,input())
c = minus(c)
c1 = ""
for q in c:
c1 += chr(ord(q) + 97)
print(c1[len(c1)-n:])
``` | output | 1 | 38,698 | 0 | 77,397 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz | instruction | 0 | 38,699 | 0 | 77,398 |
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def plus(a, b):
c = ""
i = len(a)-1
arg = 0
while i >= 0:
k = (arg + ord(a[i]) + ord(b[i]) - 194)
arg = 1
if k < 26:
arg = 0
i -= 1
c = chr(k % 26) + c
c = chr(1) * arg + c
return c
def minus(a):
c = ""
arg = 0
for q in a:
c += chr((arg * 26 + ord(q))//2)
arg = ord(q) % 2
if arg:
c = c[:len(c)-1] + chr(ord(c[len(c)-1])+1)
return c
n = int(input())
a1 = ""
i = 0
while i < n:
a1 += "s"
i += 1
c = minus(plus(input() ,input()))
c1 = ""
for q in c:
c1 += chr(ord(q) + 97)
print(c1[len(c1)-n:])
``` | output | 1 | 38,699 | 0 | 77,399 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz | instruction | 0 | 38,700 | 0 | 77,400 |
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def arr(ls):
l = [0]*len(ls)
for i,v in enumerate(ls):
l[i] = ord(v)-97
return l
def div2(ls, n):
carry = 0
for i in range(n):
rem = ls[i]%2 == 1
ls[i] = ls[i]//2 + carry
if rem: carry = 13
else: carry = 0
return ls
def subtr(ls1, ls2,n):
carry = 0
for i in range(n-1, -1, -1):
ls1[i]-=carry
carry = int(ls1[i] < ls2[i])
ls1[i] = (ls1[i]-ls2[i])%26
return ls1
def add(ls1, ls2, n):
for i in range(n-1, -1, -1):
ls1[i] += ls2[i]
if ls1[i] > 25:
ls1[i-1]+=1
ls1[i]-=26
return ls1
import io, os
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# input = io.StringIO(os.read(0, os.fstat(0).st_size).decode()).readline
kk=lambda:map(int,input().split())
k2=lambda:map(lambda x:int(x)-1, input().split())
ll=lambda:list(kk())
n,w1, w2 =int(input()), list(input().strip()), list(input().strip())
ls1, ls2 = arr(w1), arr(w2)
subtr(ls2, ls1, n)
div2(ls2, n)
add(ls1, ls2, n)
print("".join(map(lambda x: chr(97+x), [x for x in ls1])))
``` | output | 1 | 38,700 | 0 | 77,401 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz | instruction | 0 | 38,701 | 0 | 77,402 |
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def divide_2(a, m):
r = 0
q = []
for x in a:
cur = r * m + x
q.append(cur // 2)
r = cur % 2
return q
def add(s, t, m):
r = 0
a = []
for x, y in zip(s[::-1], t[::-1]):
cur = r+x+y
a.append(cur % m )
r = cur // m
if r != 0:
a.append(r)
return a[::-1]
def to_num(s):
a = []
for x in s:
a.append(ord(x)-ord('a'))
return a
def to_char(s, k):
a = []
for x in s[-k:]:
a.append(chr(x+ord('a')))
return ''.join(a)
k = int(input())
x = to_num(input())
y = to_num(input())
print(to_char(divide_2(add(x, y, 26), 26), k))
``` | output | 1 | 38,701 | 0 | 77,403 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz | instruction | 0 | 38,702 | 0 | 77,404 |
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def arr(ls):
l = [0]*len(ls)
for i,v in enumerate(ls):
l[i] = v-97
return l
def div2(ls, n):
carry = 0
for i in range(n):
rem = ls[i]%2 == 1
ls[i] = ls[i]//2 + carry
if rem: carry = 13
else: carry = 0
return ls
def subtr(ls1, ls2,n):
carry = 0
for i in range(n-1, -1, -1):
ls1[i]-=carry
carry = int(ls1[i] < ls2[i])
ls1[i] = (ls1[i]-ls2[i])%26
return ls1
def add(ls1, ls2, n):
for i in range(n-1, -1, -1):
ls1[i] += ls2[i]
if ls1[i] > 25:
ls1[i-1]+=1
ls1[i]-=26
return ls1
import io, os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# input = io.StringIO(os.read(0, os.fstat(0).st_size).decode()).readline
kk=lambda:map(int,input().split())
k2=lambda:map(lambda x:int(x)-1, input().split())
ll=lambda:list(kk())
n,w1, w2 =int(input()), list(input().strip()), list(input().strip())
ls1, ls2 = arr(w1), arr(w2)
subtr(ls2, ls1, n)
div2(ls2, n)
add(ls1, ls2, n)
print("".join(map(lambda x: chr(97+x), [x for x in ls1])))
``` | output | 1 | 38,702 | 0 | 77,405 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz | instruction | 0 | 38,703 | 0 | 77,406 |
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def plus(a, b):
c = ""
i = len(a)-1
arg = 0
while i >= 0:
k = (arg + ord(a[i]) + ord(b[i]) - 194)
arg = k >= 26
i -= 1
c = chr(k % 26) + c
c = chr(1) * arg + c
return c
def minus(a):
c = ""
arg = 0
for q in a:
c += chr(ord(q)//2 + arg*13)
arg = ord(q) % 2
if arg:
c = c[:len(c)-1] + chr(ord(c[len(c)-1])+1)
return c
n = int(input())
a1 = ""
a1 = "s"*n
c = minus(plus(input() ,input()))
c1 = ""
for q in c:
c1 += chr(ord(q) + 97)
print(c1[len(c1)-n:])
``` | output | 1 | 38,703 | 0 | 77,407 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz | instruction | 0 | 38,704 | 0 | 77,408 |
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def plus(a, b):
c = ""
i = len(a)-1
arg = 0
while i >= 0:
k = (arg + ord(a[i]) + ord(b[i]) - 194)
arg = 1
if k < 26:
arg = 0
i -= 1
c = chr(k % 26) + c
c = chr(1) * arg + c
return c
def minus(a):
c = ""
arg = 0
for q in a:
c += chr(ord(q)//2 + arg*13)
arg = ord(q) % 2
if arg:
c = c[:len(c)-1] + chr(ord(c[len(c)-1])+1)
return c
n = int(input())
a1 = ""
i = 0
while i < n:
a1 += "s"
i += 1
c = minus(plus(input() ,input()))
c1 = ""
for q in c:
c1 += chr(ord(q) + 97)
print(c1[len(c1)-n:])
``` | output | 1 | 38,704 | 0 | 77,409 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz | instruction | 0 | 38,705 | 0 | 77,410 |
Tags: bitmasks, math, number theory, strings
Correct Solution:
```
def plus(a, b):
c = ""
i = len(a)-1
arg = 0
while i >= 0:
k = (arg + ord(a[i]) + ord(b[i]) - 194)
arg = 1
if k < 26:
arg = 0
i -= 1
c = chr(k % 26) + c
c = chr(1) * arg + c
return c
def minus(a):
c = ""
i = 0
arg = 0
while i < len(a):
c += chr((arg * 26 + ord(a[i]))//2)
arg = ord(a[i]) % 2
i += 1
if arg:
c = c[:len(c)-1] + chr(ord(c[len(c)-1])+1)
return c
n = int(input())
aa = input()
bb = input()
a1 = ""
b1 = ""
i = 0
while i < n:
a1 += "s"
i += 1
c = plus(aa,bb)
c = minus(c)
c1 = ""
for q in c:
c1 += chr(ord(q) + 97)
print(c1[len(c1)-n:])
``` | output | 1 | 38,705 | 0 | 77,411 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
from math import ceil
k = int(input())
s1 = input()
s2 = input()
so1 = []
so2 = []
for i in range(k):
so1.append(ord(s1[i]) - ord("a"))
so2.append(ord(s2[i]) - ord("a"))
for i in range(k-1, -1, -1):
so1[i] += so2[i]
if i != 0:
so1[i - 1] += so1[i] // 26
so1[i] %= 26
med = ""
for i in range(k):
v = so1[i] // 2
if i < k - 1:
so1[i + 1] += (so1[i] % 2) * 26
med += chr(ord("a") + v)
print(med)
``` | instruction | 0 | 38,706 | 0 | 77,412 |
Yes | output | 1 | 38,706 | 0 | 77,413 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
#threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
#sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**30, func=lambda a, b: min(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b:a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] > k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
#-----------------------------------------trie---------------------------------
class Node:
def __init__(self, data):
self.data = data
self.count=0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count+=1
if not val:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += 1
self.temp.data = pre_xor
def query(self, xor):
self.temp = self.root
for i in range(31, -1, -1):
val = xor & (1 << i)
if not val:
if self.temp.left and self.temp.left.count>0:
self.temp = self.temp.left
elif self.temp.right:
self.temp = self.temp.right
else:
if self.temp.right and self.temp.right.count>0:
self.temp = self.temp.right
elif self.temp.left:
self.temp = self.temp.left
self.temp.count-=1
return xor ^ self.temp.data
#-------------------------bin trie-------------------------------------------
n=int(input())
s=input()
t=input()
d1=defaultdict(chr)
d=defaultdict(int)
for i in range(97,123):
d[chr(i)]=i-97
d1[i-97]=chr(i)
carry=0
ans=deque([])
for i in range(n-1,-1,-1):
su=carry+d[s[i]]+d[t[i]]
ans.appendleft(d1[su%26])
carry=su//26
if carry!=0:
ans.appendleft(d1[carry])
rem=0
fi=[]
#print(ans)
for i in range(len(ans)):
qu=(rem*26+d[ans[i]])//2
rem=d[ans[i]]%2
if i==0 and qu==0 and d[s[i]]!=0:
continue
fi.append(d1[qu])
print(*fi,sep="")
``` | instruction | 0 | 38,707 | 0 | 77,414 |
Yes | output | 1 | 38,707 | 0 | 77,415 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
def plus(a, b):
c = ""
i = len(a)-1
arg = 0
while i >= 0:
k = (arg + ord(a[i]) + ord(b[i]) - 194)
arg = 1
if k < 26:
arg = 0
i -= 1
c = chr(k % 26) + c
c = chr(1) * arg + c
return c
def minus(a):
c = ""
i = 0
arg = 0
while i < len(a):
c += chr((arg * 26 + ord(a[i]))//2)
arg = ord(a[i]) % 2
i += 1
if arg:
c = c[:len(c)-1] + chr(ord(c[len(c)-1])+1)
return c
n = int(input())
aa = input()
bb = input()
a1 = ""
b1 = ""
i = 0
while i < n:
a1 += aa[i]
i += 1
c = plus(aa,bb)
c = minus(c)
c1 = ""
for q in c:
c1 += chr(ord(q) + ord('a'))
print(c1[len(c1)-n:])
``` | instruction | 0 | 38,708 | 0 | 77,416 |
Yes | output | 1 | 38,708 | 0 | 77,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
k = int(input())
s = input()
t = input()
val = [0 for i in range(0, k)]
nval = [0 for i in range(0, k)]
for i in range(0, k):
val[i] = ord(s[i]) - ord('a') + ord(t[i]) - ord('a')
ans = ['a' for i in range(0, k)]
carry = 0
for i in range(0, k):
ncarry = 0
t = val[i] // 2 + carry
if val[i] % 2 == 1:
ncarry += 13
nval[i] = t
carry = ncarry
for i in range(k - 1, -1, -1):
if nval[i] >= 26:
nval[i - 1] += nval[i] // 26
nval[i] %= 26
ans[i] = chr(ord('a') + nval[i])
print("".join(x for x in ans))
``` | instruction | 0 | 38,709 | 0 | 77,418 |
Yes | output | 1 | 38,709 | 0 | 77,419 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
k=int(input())
s1=str(input())
s2=str(input())
ans=''
rem=0
s='abcdefghijklmnopqrstuvwxyz'
for i in range(len(s1)):
val1=ord(s1[i])-96
val2=ord(s2[i])-96
val=(val1+val2)+(26*rem)
#print(val)
if(val>53):
indexx=len(ans)-1
flag=0
while(flag==0):
index1=s.index(ans[indexx])
if(index1==25):
ans=ans[:indexx]+'a'+ans[indexx+1:]
indexx-=1
else:
ans=ans[:indexx]+s[index1+1]+ans[indexx+1:]
flag=1
break
rem+=(val//26)
val=val%26
rem+=val%2
val=val//2
ans+=s[val-1]
print(ans)
``` | instruction | 0 | 38,710 | 0 | 77,420 |
No | output | 1 | 38,710 | 0 | 77,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
k=int(input())
a=input("")
b=input("")
xx=1
aa=0
bb=0
for i in range(k-1,-1,-1):
aa+=xx*(ord(a[i])-ord('a'))
xx*=26
xx=1
for i in range(k-1,-1,-1):
bb+=xx*(ord(b[i])-ord('a'))
xx*=26
cc=(int)(aa+bb)
cc=(int)(cc/2)
xx=1
print(aa)
print(bb)
for i in range(k-1):
xx*=26
for i in range(k):
yy=(int)(cc/xx)
print((chr)(ord('a')+yy),end='')
cc%=xx
xx=(int)(xx/26)
``` | instruction | 0 | 38,711 | 0 | 77,422 |
No | output | 1 | 38,711 | 0 | 77,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
k = int(input())
a = input()
b = input()
o = ord('a')
f = 0
s = 0
if (k == 159000):
exit(0)
for i in a:
f = f * 26 + (ord(i) - o)
for i in b:
s = s * 26 + (ord(i) - o)
t = (f + s) // 2
str = []
while k > 0:
str.append(chr(t % 26 + o))
t //= 26
k -= 1
print(*str[::-1], sep="")
``` | instruction | 0 | 38,712 | 0 | 77,424 |
No | output | 1 | 38,712 | 0 | 77,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t, both consisting of exactly k lowercase Latin letters, s is lexicographically less than t.
Let's consider list of all strings consisting of exactly k lowercase Latin letters, lexicographically not less than s and not greater than t (including s and t) in lexicographical order. For example, for k=2, s="az" and t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Input
The first line of the input contains one integer k (1 β€ k β€ 2 β
10^5) β the length of strings.
The second line of the input contains one string s consisting of exactly k lowercase Latin letters.
The third line of the input contains one string t consisting of exactly k lowercase Latin letters.
It is guaranteed that s is lexicographically less than t.
It is guaranteed that there is an odd number of strings lexicographically not less than s and not greater than t.
Output
Print one string consisting exactly of k lowercase Latin letters β the median (the middle element) of list of strings of length k lexicographically not less than s and not greater than t.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz
Submitted Solution:
```
n = int(input())
a = [ord(c) - 96 for c in input()]
b = [ord(c) - 96 for c in input()]
ans = []
for i in range(n):
sum = a[i] + b[i]
if i+1<n:
a[i+1]+=(sum%2)*26
sum//=2
ans.append(sum)
for i in range(n-1,-1,-1):
ans[i]%=26
if ans[i]==0:
ans[i]=26
if i-1 >= 0:
ans[i-1] += (ans[i]//26)
ans = [chr(x+96) for x in ans]
print(''.join(ans))
``` | instruction | 0 | 38,713 | 0 | 77,426 |
No | output | 1 | 38,713 | 0 | 77,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a uppercase Latin letters 'A' and b letters 'B'.
The period of the string is the smallest such positive integer k that s_i = s_{i~mod~k} (0-indexed) for each i. Note that this implies that k won't always divide a+b = |s|.
For example, the period of string "ABAABAA" is 3, the period of "AAAA" is 1, and the period of "AABBB" is 5.
Find the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Input
The first line contains two integers a and b (1 β€ a, b β€ 10^9) β the number of letters 'A' and 'B', respectively.
Output
Print the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Examples
Input
2 4
Output
4
Input
5 3
Output
5
Note
All the possible periods for the first example:
* 3 "BBABBA"
* 4 "BBAABB"
* 5 "BBBAAB"
* 6 "AABBBB"
All the possible periods for the second example:
* 3 "BAABAABA"
* 5 "BAABABAA"
* 6 "BABAAABA"
* 7 "BAABAAAB"
* 8 "AAAAABBB"
Note that these are not the only possible strings for the given periods. | instruction | 0 | 38,730 | 0 | 77,460 |
Tags: binary search, implementation, math
Correct Solution:
```
a,b= map(int,input().split())
ans,l,n=0,1,a+b
while l<=n:
g= n//l
if a<g or b<g:
l= (n//g) +1
continue
r= n//g
al = (a+g)//(g+1)
ah = a//g
bl=(b+g)//(g+1)
bh = b//g
if (al <= ah and bl <= bh):
ans += max(0,min(r,ah+bh)- max(l,al +bl)+1)
l=r+1
print(ans)
``` | output | 1 | 38,730 | 0 | 77,461 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a uppercase Latin letters 'A' and b letters 'B'.
The period of the string is the smallest such positive integer k that s_i = s_{i~mod~k} (0-indexed) for each i. Note that this implies that k won't always divide a+b = |s|.
For example, the period of string "ABAABAA" is 3, the period of "AAAA" is 1, and the period of "AABBB" is 5.
Find the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Input
The first line contains two integers a and b (1 β€ a, b β€ 10^9) β the number of letters 'A' and 'B', respectively.
Output
Print the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Examples
Input
2 4
Output
4
Input
5 3
Output
5
Note
All the possible periods for the first example:
* 3 "BBABBA"
* 4 "BBAABB"
* 5 "BBBAAB"
* 6 "AABBBB"
All the possible periods for the second example:
* 3 "BAABAABA"
* 5 "BAABABAA"
* 6 "BABAAABA"
* 7 "BAABAAAB"
* 8 "AAAAABBB"
Note that these are not the only possible strings for the given periods. | instruction | 0 | 38,731 | 0 | 77,462 |
Tags: binary search, implementation, math
Correct Solution:
```
import math
a,b= map(int,input().split())
n=a+b
ans,l=0,1
while l<=n:
g= n//l
if a<g or b<g:
l= (n//g) +1
continue
r= n//g
a_low = (a+g)//(g+1)
a_high = a//g
b_low=(b+g)//(g+1)
b_high = b//g
if (a_low <= a_high and b_low <= b_high):
ans += max(0,min(r,a_high+b_high)- max(l,a_low +b_low)+1)
l=r+1
print(ans)
``` | output | 1 | 38,731 | 0 | 77,463 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a uppercase Latin letters 'A' and b letters 'B'.
The period of the string is the smallest such positive integer k that s_i = s_{i~mod~k} (0-indexed) for each i. Note that this implies that k won't always divide a+b = |s|.
For example, the period of string "ABAABAA" is 3, the period of "AAAA" is 1, and the period of "AABBB" is 5.
Find the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Input
The first line contains two integers a and b (1 β€ a, b β€ 10^9) β the number of letters 'A' and 'B', respectively.
Output
Print the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Examples
Input
2 4
Output
4
Input
5 3
Output
5
Note
All the possible periods for the first example:
* 3 "BBABBA"
* 4 "BBAABB"
* 5 "BBBAAB"
* 6 "AABBBB"
All the possible periods for the second example:
* 3 "BAABAABA"
* 5 "BAABABAA"
* 6 "BABAAABA"
* 7 "BAABAAAB"
* 8 "AAAAABBB"
Note that these are not the only possible strings for the given periods. | instruction | 0 | 38,732 | 0 | 77,464 |
Tags: binary search, implementation, math
Correct Solution:
```
import math
a, b = [int(x) for x in input().split()]
cnt = 0
ab = a + b
k = 2
while k <= ab:
t = ab // k
aPeriod = a // t
bPeriod = b // t
aLeft = (a + t) // (t + 1)
bLeft = (b + t) // (t + 1)
arem = a - t * aPeriod
brem = b - t * bPeriod
if aPeriod >= arem and bPeriod >= brem:
validK = aPeriod + bPeriod - aLeft - bLeft + 1
if (t + 1) * (aLeft + bLeft) == ab:
validK -= 1
if validK > 0:
cnt += validK
k = ab // t + 1
print(cnt)
``` | output | 1 | 38,732 | 0 | 77,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a uppercase Latin letters 'A' and b letters 'B'.
The period of the string is the smallest such positive integer k that s_i = s_{i~mod~k} (0-indexed) for each i. Note that this implies that k won't always divide a+b = |s|.
For example, the period of string "ABAABAA" is 3, the period of "AAAA" is 1, and the period of "AABBB" is 5.
Find the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Input
The first line contains two integers a and b (1 β€ a, b β€ 10^9) β the number of letters 'A' and 'B', respectively.
Output
Print the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Examples
Input
2 4
Output
4
Input
5 3
Output
5
Note
All the possible periods for the first example:
* 3 "BBABBA"
* 4 "BBAABB"
* 5 "BBBAAB"
* 6 "AABBBB"
All the possible periods for the second example:
* 3 "BAABAABA"
* 5 "BAABABAA"
* 6 "BABAAABA"
* 7 "BAABAAAB"
* 8 "AAAAABBB"
Note that these are not the only possible strings for the given periods.
Submitted Solution:
```
a = [int(i) for i in input().split()]
print(max(a))
``` | instruction | 0 | 38,733 | 0 | 77,466 |
No | output | 1 | 38,733 | 0 | 77,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a uppercase Latin letters 'A' and b letters 'B'.
The period of the string is the smallest such positive integer k that s_i = s_{i~mod~k} (0-indexed) for each i. Note that this implies that k won't always divide a+b = |s|.
For example, the period of string "ABAABAA" is 3, the period of "AAAA" is 1, and the period of "AABBB" is 5.
Find the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Input
The first line contains two integers a and b (1 β€ a, b β€ 10^9) β the number of letters 'A' and 'B', respectively.
Output
Print the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Examples
Input
2 4
Output
4
Input
5 3
Output
5
Note
All the possible periods for the first example:
* 3 "BBABBA"
* 4 "BBAABB"
* 5 "BBBAAB"
* 6 "AABBBB"
All the possible periods for the second example:
* 3 "BAABAABA"
* 5 "BAABABAA"
* 6 "BABAAABA"
* 7 "BAABAAAB"
* 8 "AAAAABBB"
Note that these are not the only possible strings for the given periods.
Submitted Solution:
```
import fileinput
def calc(a, b):
ans = 0
l = a + b
i = 1
while i < l:
P = l // i
r = l // P
if P <= a and P <= b:
an = (a + P) // (P + 1)
ax = a // P
bn = (b + P) // (P + 1)
bx = b // P
if an <= ax and bn <= bx:
ans += min(ax + bx, r) - max(an + bn, i) + 1
i = r + 1
return ans
if __name__ == '__main__':
it = fileinput.input()
a, b = [int(x) for x in next(it).split()]
print(calc(a, b))
``` | instruction | 0 | 38,734 | 0 | 77,468 |
No | output | 1 | 38,734 | 0 | 77,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a uppercase Latin letters 'A' and b letters 'B'.
The period of the string is the smallest such positive integer k that s_i = s_{i~mod~k} (0-indexed) for each i. Note that this implies that k won't always divide a+b = |s|.
For example, the period of string "ABAABAA" is 3, the period of "AAAA" is 1, and the period of "AABBB" is 5.
Find the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Input
The first line contains two integers a and b (1 β€ a, b β€ 10^9) β the number of letters 'A' and 'B', respectively.
Output
Print the number of different periods over all possible strings with a letters 'A' and b letters 'B'.
Examples
Input
2 4
Output
4
Input
5 3
Output
5
Note
All the possible periods for the first example:
* 3 "BBABBA"
* 4 "BBAABB"
* 5 "BBBAAB"
* 6 "AABBBB"
All the possible periods for the second example:
* 3 "BAABAABA"
* 5 "BAABABAA"
* 6 "BABAAABA"
* 7 "BAABAAAB"
* 8 "AAAAABBB"
Note that these are not the only possible strings for the given periods.
Submitted Solution:
```
a,b = map(int, input().split())
total = a+b
result = 0
i = 1
while i <= total:
groups = total//i
rem = total%i
if a%groups <= rem:
if rem > groups:
result += rem//groups
i += rem//groups
else:
result += 1
i += 1
else:
i += 1
print(result)
``` | instruction | 0 | 38,735 | 0 | 77,470 |
No | output | 1 | 38,735 | 0 | 77,471 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.