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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one.
GukiZ has strings a, b, and c. He wants to obtain string k by swapping some letters in a, so that k should contain as many non-overlapping substrings equal either to b or c as possible. Substring of string x is a string formed by consecutive segment of characters from x. Two substrings of string x overlap if there is position i in string x occupied by both of them.
GukiZ was disappointed because none of his students managed to solve the problem. Can you help them and find one of possible strings k?
Input
The first line contains string a, the second line contains string b, and the third line contains string c (1 ≤ |a|, |b|, |c| ≤ 105, where |s| denotes the length of string s).
All three strings consist only of lowercase English letters.
It is possible that b and c coincide.
Output
Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.
Examples
Input
aaa
a
b
Output
aaa
Input
pozdravstaklenidodiri
niste
dobri
Output
nisteaadddiiklooprrvz
Input
abbbaaccca
ab
aca
Output
ababacabcc
Note
In the third sample, this optimal solutions has three non-overlaping substrings equal to either b or c on positions 1 – 2 (ab), 3 – 4 (ab), 5 – 7 (aca). In this sample, there exist many other optimal solutions, one of them would be acaababbcc. | instruction | 0 | 57,546 | 0 | 115,092 |
Tags: brute force, constructive algorithms, implementation, strings
Correct Solution:
```
import string
from collections import defaultdict
INF = 2147483647
string_a = input()
string_b = input()
string_c = input()
dict_a = defaultdict(int)
dict_b = defaultdict(int)
dict_c = defaultdict(int)
for ch in string_a:
dict_a[ch] += 1
for ch in string_b:
dict_b[ch] += 1
for ch in string_c:
dict_c[ch] += 1
bn, cn = 0, 0
for i in range(100001):
j = INF
for ch in string.ascii_lowercase:
a, b, c = dict_a[ch], dict_b[ch], dict_c[ch]
a -= b*i
if a < 0:
j = -1
if c != 0:
j = min(j, int(a/c) )
if j == -1:
break
if i+j > bn+cn:
(bn, cn) = (i, j)
res = ""
for i in range(bn):
res += string_b
for i in range(cn):
res += string_c
for ch in string.ascii_lowercase:
a, b, c = dict_a[ch], dict_b[ch], dict_c[ch]
a -= b*bn + c*cn
res += ch*a
print(res)
``` | output | 1 | 57,546 | 0 | 115,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one.
GukiZ has strings a, b, and c. He wants to obtain string k by swapping some letters in a, so that k should contain as many non-overlapping substrings equal either to b or c as possible. Substring of string x is a string formed by consecutive segment of characters from x. Two substrings of string x overlap if there is position i in string x occupied by both of them.
GukiZ was disappointed because none of his students managed to solve the problem. Can you help them and find one of possible strings k?
Input
The first line contains string a, the second line contains string b, and the third line contains string c (1 ≤ |a|, |b|, |c| ≤ 105, where |s| denotes the length of string s).
All three strings consist only of lowercase English letters.
It is possible that b and c coincide.
Output
Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.
Examples
Input
aaa
a
b
Output
aaa
Input
pozdravstaklenidodiri
niste
dobri
Output
nisteaadddiiklooprrvz
Input
abbbaaccca
ab
aca
Output
ababacabcc
Note
In the third sample, this optimal solutions has three non-overlaping substrings equal to either b or c on positions 1 – 2 (ab), 3 – 4 (ab), 5 – 7 (aca). In this sample, there exist many other optimal solutions, one of them would be acaababbcc. | instruction | 0 | 57,547 | 0 | 115,094 |
Tags: brute force, constructive algorithms, implementation, strings
Correct Solution:
```
import copy
a = input()
b = input()
c = input()
aa = {}
bb = {}
cc = {}
for i in 'abcdefghijklmnopqrstuvwxyz':
aa[i] = 0
bb[i] = 0
cc[i] = 0
for i in a:
aa[i] += 1
for i in b:
bb[i] += 1
for i in c:
cc[i] += 1
mx = 0
bmx = 0
cmx = 0
poss = 10**9
for i in 'abcdefghijklmnopqrstuvwxyz':
if bb[i] != 0:
poss = min(poss, aa[i]//bb[i])
for app in range(poss+1):
cct = 10**9
aa2 = copy.deepcopy(aa)
for i in 'abcdefghijklmnopqrstuvwxyz':
aa2[i] -= bb[i] * app
for i in 'abcdefghijklmnopqrstuvwxyz':
if cc[i] != 0:
cct = min(cct, aa2[i] // cc[i])
if app + cct > mx:
mx = app + cct
bmx = app
cmx = cct
for i in 'abcdefghijklmnopqrstuvwxyz':
aa[i] -= bb[i] * bmx
for i in 'abcdefghijklmnopqrstuvwxyz':
aa[i] -= cc[i] * cmx
fnstr = (bmx*b+cmx*c)
for i in 'abcdefghijklmnopqrstuvwxyz':
if aa[i] != 0:
fnstr += i * aa[i]
print(fnstr)
``` | output | 1 | 57,547 | 0 | 115,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one.
GukiZ has strings a, b, and c. He wants to obtain string k by swapping some letters in a, so that k should contain as many non-overlapping substrings equal either to b or c as possible. Substring of string x is a string formed by consecutive segment of characters from x. Two substrings of string x overlap if there is position i in string x occupied by both of them.
GukiZ was disappointed because none of his students managed to solve the problem. Can you help them and find one of possible strings k?
Input
The first line contains string a, the second line contains string b, and the third line contains string c (1 ≤ |a|, |b|, |c| ≤ 105, where |s| denotes the length of string s).
All three strings consist only of lowercase English letters.
It is possible that b and c coincide.
Output
Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.
Examples
Input
aaa
a
b
Output
aaa
Input
pozdravstaklenidodiri
niste
dobri
Output
nisteaadddiiklooprrvz
Input
abbbaaccca
ab
aca
Output
ababacabcc
Note
In the third sample, this optimal solutions has three non-overlaping substrings equal to either b or c on positions 1 – 2 (ab), 3 – 4 (ab), 5 – 7 (aca). In this sample, there exist many other optimal solutions, one of them would be acaababbcc.
Submitted Solution:
```
def read_data():
a = input()
b = input()
c = input()
return a, b, c
def count_alphabets(S):
counts = [0] * 26
ord_a = ord('a')
for s in S:
counts[ord(s) - ord_a] += 1
return counts
def solve(a, b, c):
c_a = count_alphabets(a)
c_a_copy = c_a[:]
c_b = count_alphabets(b)
c_c = count_alphabets(c)
max_x = min(ca // cb for ca, cb in zip(c_a, c_b) if cb)
record = 0
best_pair = (0, 0)
for x in range(max_x + 1):
y = min(ca // cc for ca, cc in zip(c_a, c_c) if cc)
if x + y > record:
record = x + y
best_pair = (x, y)
c_a = [ca - cb for ca, cb in zip(c_a, c_b)]
x, y = best_pair
residue = [ca - x * cb - y * cc for ca, cb, cc in zip(c_a_copy, c_b, c_c)]
residue = ''.join([chr(ord('a') + i) * r for i, r in enumerate(residue)])
return b * x + c * y + residue
if __name__ == '__main__':
a, b, c = read_data()
print(solve(a, b, c))
``` | instruction | 0 | 57,548 | 0 | 115,096 |
Yes | output | 1 | 57,548 | 0 | 115,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one.
GukiZ has strings a, b, and c. He wants to obtain string k by swapping some letters in a, so that k should contain as many non-overlapping substrings equal either to b or c as possible. Substring of string x is a string formed by consecutive segment of characters from x. Two substrings of string x overlap if there is position i in string x occupied by both of them.
GukiZ was disappointed because none of his students managed to solve the problem. Can you help them and find one of possible strings k?
Input
The first line contains string a, the second line contains string b, and the third line contains string c (1 ≤ |a|, |b|, |c| ≤ 105, where |s| denotes the length of string s).
All three strings consist only of lowercase English letters.
It is possible that b and c coincide.
Output
Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.
Examples
Input
aaa
a
b
Output
aaa
Input
pozdravstaklenidodiri
niste
dobri
Output
nisteaadddiiklooprrvz
Input
abbbaaccca
ab
aca
Output
ababacabcc
Note
In the third sample, this optimal solutions has three non-overlaping substrings equal to either b or c on positions 1 – 2 (ab), 3 – 4 (ab), 5 – 7 (aca). In this sample, there exist many other optimal solutions, one of them would be acaababbcc.
Submitted Solution:
```
a = input().rstrip()
b = input().rstrip()
c = input().rstrip()
d = {}
d1 = {}
d2 = {}
db = {}
dc = {}
sb = set()
sc = set()
s = set()
for i in range(len(a)):
if a[i] not in s:
s.add(a[i])
d[a[i]] = 1
d1[a[i]] = 1
d2[a[i]] = 1
else:
d[a[i]] += 1
d1[a[i]] += 1
d2[a[i]] += 1
for i in range(len(b)):
if b[i] not in sb:
sb.add(b[i])
db[b[i]] = 1
else:
db[b[i]] += 1
for i in range(len(c)):
if c[i] not in sc:
sc.add(c[i])
dc[c[i]] = 1
else:
dc[c[i]] += 1
maxib = 10**6
maxic = 10**6
for i in range(len(b)):
if b[i] in s:
maxib = min(maxib, d[b[i]]//db[b[i]])
else:
maxib = 0
break
for i in range(len(c)):
if c[i] in s:
maxic = min(maxic, d[c[i]]//dc[c[i]])
else:
maxic = 0
break
ans = -1
lol = 0
pop = 0
for i in range(maxib + 1):
minb = i
for key, value in db.items():
if key in s:
d[key] -= i*value
minc = 10**6
for key, value in dc.items():
if key in s:
if d[key]//dc[key] < minc:
minc = d[key]//dc[key]
else:
minc = 0
if minc + minb > ans:
ans = minc + minb
lol = minc
pop = minb
for key, value in db.items():
if key in s:
d[key] += i*value
for i in range(lol):
print(c, end = '')
for i in range(pop):
print(b, end = '')
for i in c:
if i in s:
d[i] -= lol
for i in b:
if i in s:
d[i] -= pop
for i,j in d.items():
print(i*j, end = '')
``` | instruction | 0 | 57,549 | 0 | 115,098 |
Yes | output | 1 | 57,549 | 0 | 115,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one.
GukiZ has strings a, b, and c. He wants to obtain string k by swapping some letters in a, so that k should contain as many non-overlapping substrings equal either to b or c as possible. Substring of string x is a string formed by consecutive segment of characters from x. Two substrings of string x overlap if there is position i in string x occupied by both of them.
GukiZ was disappointed because none of his students managed to solve the problem. Can you help them and find one of possible strings k?
Input
The first line contains string a, the second line contains string b, and the third line contains string c (1 ≤ |a|, |b|, |c| ≤ 105, where |s| denotes the length of string s).
All three strings consist only of lowercase English letters.
It is possible that b and c coincide.
Output
Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.
Examples
Input
aaa
a
b
Output
aaa
Input
pozdravstaklenidodiri
niste
dobri
Output
nisteaadddiiklooprrvz
Input
abbbaaccca
ab
aca
Output
ababacabcc
Note
In the third sample, this optimal solutions has three non-overlaping substrings equal to either b or c on positions 1 – 2 (ab), 3 – 4 (ab), 5 – 7 (aca). In this sample, there exist many other optimal solutions, one of them would be acaababbcc.
Submitted Solution:
```
def ci(c):
return ord(c) - 97
def ic(i):
return chr(i + 97)
a, b, c = input(), input(), input()
ina = 26*[0]
for i in a:
ina[ci(i)] += 1
inb = 26*[0]
for i in b:
inb[ci(i)] += 1
inc = 26*[0]
for i in c:
inc[ci(i)] += 1
bl = 26*[10**7]
for i in range(26):
if inb[i] != 0:
bl[i] = ina[i]//inb[i]
blmin = min(bl)
best = 0
best_val = (0, 0)
for nb in range(blmin + 1):
inac = ina[:]
for i in range(26):
inac[i] -= inb[i]*nb
assert inac[i] > -1
cl = 26*[10**7]
for i in range(26):
if inc[i] != 0:
cl[i] = inac[i]//inc[i]
clmin = min(cl)
if nb + clmin > best:
best = nb + clmin
best_val = (nb, clmin)
for i in b:
ina[ci(i)] -= best_val[0]
for i in c:
ina[ci(i)]-= best_val[1]
s = ''
for i in range(26):
s += ic(i)*ina[i]
print(best_val[0]*b + best_val[1]*c + s)
``` | instruction | 0 | 57,550 | 0 | 115,100 |
Yes | output | 1 | 57,550 | 0 | 115,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one.
GukiZ has strings a, b, and c. He wants to obtain string k by swapping some letters in a, so that k should contain as many non-overlapping substrings equal either to b or c as possible. Substring of string x is a string formed by consecutive segment of characters from x. Two substrings of string x overlap if there is position i in string x occupied by both of them.
GukiZ was disappointed because none of his students managed to solve the problem. Can you help them and find one of possible strings k?
Input
The first line contains string a, the second line contains string b, and the third line contains string c (1 ≤ |a|, |b|, |c| ≤ 105, where |s| denotes the length of string s).
All three strings consist only of lowercase English letters.
It is possible that b and c coincide.
Output
Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.
Examples
Input
aaa
a
b
Output
aaa
Input
pozdravstaklenidodiri
niste
dobri
Output
nisteaadddiiklooprrvz
Input
abbbaaccca
ab
aca
Output
ababacabcc
Note
In the third sample, this optimal solutions has three non-overlaping substrings equal to either b or c on positions 1 – 2 (ab), 3 – 4 (ab), 5 – 7 (aca). In this sample, there exist many other optimal solutions, one of them would be acaababbcc.
Submitted Solution:
```
from collections import Counter
from copy import deepcopy
def getMax(c1, c2):
ret = 1 << 32
for k in c2.keys():
ret = min(ret, c1[k]//c2[k])
return ret
a, b, c = input(), input(), input()
counterA, counterB, counterC = Counter(a), Counter(b), Counter(c)
cntB, cntC = 0, getMax(counterA, counterC)
counterAA = deepcopy(counterA)
cntBB = 0
while counterAA & counterB == counterB:
cntBB += 1
counterAA -= counterB
cntCC = getMax(counterAA, counterC)
if cntB + cntC < cntBB + cntCC:
cntB, cntC = cntBB, cntCC
ans = b * cntB + c * cntC
ans += ''.join((counterA - Counter(ans)).elements())
print(ans)
``` | instruction | 0 | 57,551 | 0 | 115,102 |
Yes | output | 1 | 57,551 | 0 | 115,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one.
GukiZ has strings a, b, and c. He wants to obtain string k by swapping some letters in a, so that k should contain as many non-overlapping substrings equal either to b or c as possible. Substring of string x is a string formed by consecutive segment of characters from x. Two substrings of string x overlap if there is position i in string x occupied by both of them.
GukiZ was disappointed because none of his students managed to solve the problem. Can you help them and find one of possible strings k?
Input
The first line contains string a, the second line contains string b, and the third line contains string c (1 ≤ |a|, |b|, |c| ≤ 105, where |s| denotes the length of string s).
All three strings consist only of lowercase English letters.
It is possible that b and c coincide.
Output
Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.
Examples
Input
aaa
a
b
Output
aaa
Input
pozdravstaklenidodiri
niste
dobri
Output
nisteaadddiiklooprrvz
Input
abbbaaccca
ab
aca
Output
ababacabcc
Note
In the third sample, this optimal solutions has three non-overlaping substrings equal to either b or c on positions 1 – 2 (ab), 3 – 4 (ab), 5 – 7 (aca). In this sample, there exist many other optimal solutions, one of them would be acaababbcc.
Submitted Solution:
```
from collections import Counter
a, b, c = Counter(input()), input(), input()
cb = Counter(b)
cc = Counter(c)
out = ''
if len(b) > len(c):
b, c = c, b
cb, cc = cc, cb
while (a & cb) == cb:
a -= cb
out += b
while (a & cc) == cc:
a -= cc
out += c
print(out + ''.join(a.elements()))
``` | instruction | 0 | 57,552 | 0 | 115,104 |
No | output | 1 | 57,552 | 0 | 115,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one.
GukiZ has strings a, b, and c. He wants to obtain string k by swapping some letters in a, so that k should contain as many non-overlapping substrings equal either to b or c as possible. Substring of string x is a string formed by consecutive segment of characters from x. Two substrings of string x overlap if there is position i in string x occupied by both of them.
GukiZ was disappointed because none of his students managed to solve the problem. Can you help them and find one of possible strings k?
Input
The first line contains string a, the second line contains string b, and the third line contains string c (1 ≤ |a|, |b|, |c| ≤ 105, where |s| denotes the length of string s).
All three strings consist only of lowercase English letters.
It is possible that b and c coincide.
Output
Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.
Examples
Input
aaa
a
b
Output
aaa
Input
pozdravstaklenidodiri
niste
dobri
Output
nisteaadddiiklooprrvz
Input
abbbaaccca
ab
aca
Output
ababacabcc
Note
In the third sample, this optimal solutions has three non-overlaping substrings equal to either b or c on positions 1 – 2 (ab), 3 – 4 (ab), 5 – 7 (aca). In this sample, there exist many other optimal solutions, one of them would be acaababbcc.
Submitted Solution:
```
from sys import stdin, stdout
import copy
"""
n = stdin.readline()
arr = [int(x) for x in stdin.readline().split()]
stdout.write(str(summation))
"""
a = input()
b = input()
c = input()
dicta = {}
dictb = {}
dictc = {}
for i in a :
if i not in dicta :
dicta[i] = 1
else :
dicta[i] = dicta[i]+1
dicta2 = copy.deepcopy(dicta)
for i in b :
if i not in dictb :
dictb[i] = 1
else :
dictb[i] = dictb[i]+1
for i in c :
if i not in dictc :
dictc[i] = 1
else :
dictc[i] = dictc[i]+1
"""###"""
minib = 1000001
for i in dictb :
if i in dicta and dicta[i]//dictb[i] > 0:
if dicta[i]//dictb[i] <= minib :
minib = dicta[i]//dictb[i]
else :
minib = 0
break
for i in dicta :
if i in dictb :
dicta[i] = dicta[i] - minib*dictb[i]
minic1 = 1000001
for i in dictc :
if i in dicta and dicta[i]//dictc[i]>0:
if dicta[i]//dictc[i] <= minic1 :
minic1 = dicta[i]//dictc[i]
else :
minic1 = 0
break
ans1 = minib*b + minic1*c
"""###"""
minic = 1000001
for i in dictc :
if i in dicta2 and dicta2[i]//dictc[i]>0:
if dicta2[i]//dictc[i] <= minic :
minic = dicta2[i]//dictc[i]
else :
minic = 0
break
for i in dicta2 :
if i in dictc :
dicta2[i] = dicta2[i] - minic*dictc[i]
minib2 = 1000001
for i in dictb :
if i in dicta2 and dicta2[i]//dictb[i]>0:
if dicta2[i]//dictb[i] <= minib2 :
minib2 = dicta2[i]//dictb[i]
else :
minib2 = 0
break
ans2 = minib2*b + minic*c
"""###"""
if minib+minic1>=minib2+minic :
for i in dicta :
if i in dictc :
dicta[i] = dicta[i] - minic1*dictc[i]
ans1 = ans1+i*dicta[i]
else :
ans1 = ans1+i*dicta[i]
stdout.write(ans1)
else :
for i in dicta2 :
if i in dictb :
dicta2[i] = dicta2[i] - minib2*dictb[i]
ans2 = ans2+i*dicta2[i]
else :
ans2 = ans2+i*dicta2[i]
stdout.write(ans2)
``` | instruction | 0 | 57,553 | 0 | 115,106 |
No | output | 1 | 57,553 | 0 | 115,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one.
GukiZ has strings a, b, and c. He wants to obtain string k by swapping some letters in a, so that k should contain as many non-overlapping substrings equal either to b or c as possible. Substring of string x is a string formed by consecutive segment of characters from x. Two substrings of string x overlap if there is position i in string x occupied by both of them.
GukiZ was disappointed because none of his students managed to solve the problem. Can you help them and find one of possible strings k?
Input
The first line contains string a, the second line contains string b, and the third line contains string c (1 ≤ |a|, |b|, |c| ≤ 105, where |s| denotes the length of string s).
All three strings consist only of lowercase English letters.
It is possible that b and c coincide.
Output
Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.
Examples
Input
aaa
a
b
Output
aaa
Input
pozdravstaklenidodiri
niste
dobri
Output
nisteaadddiiklooprrvz
Input
abbbaaccca
ab
aca
Output
ababacabcc
Note
In the third sample, this optimal solutions has three non-overlaping substrings equal to either b or c on positions 1 – 2 (ab), 3 – 4 (ab), 5 – 7 (aca). In this sample, there exist many other optimal solutions, one of them would be acaababbcc.
Submitted Solution:
```
import copy
a = input()
b = input()
c = input()
aa = {}
bb = {}
cc = {}
for i in 'abcdefghijklmnopqrstuvwxyz':
aa[i] = 0
bb[i] = 0
cc[i] = 0
for i in a:
aa[i] += 1
for i in b:
bb[i] += 1
for i in c:
cc[i] += 1
mx = 0
bmx = 0
cmx = 0
poss = 10**9
for i in 'abcdefghijklmnopqrstuvwxyz':
if bb[i] != 0:
poss = min(poss, aa[i]//bb[i])
for app in range(poss+1):
cct = 10**9
aa2 = copy.deepcopy(aa)
for i in b:
aa2[i] -= bb[i] * app
for i in 'abcdefghijklmnopqrstuvwxyz':
if cc[i] != 0:
cct = min(cct, aa2[i] // cc[i])
if app + cct > mx:
mx = app + cct
bmx = app
cmx = cct
for i in bb:
aa[i] -= bb[i] * bmx
for i in cc:
aa[i] -= cc[i] * cmx
fnstr = (bmx*b+cmx*c)
for i in 'abcdefghijklmnopqrstuvwxyz':
if aa[i] != 0:
fnstr += i * aa[i]
print(fnstr)
``` | instruction | 0 | 57,554 | 0 | 115,108 |
No | output | 1 | 57,554 | 0 | 115,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one.
GukiZ has strings a, b, and c. He wants to obtain string k by swapping some letters in a, so that k should contain as many non-overlapping substrings equal either to b or c as possible. Substring of string x is a string formed by consecutive segment of characters from x. Two substrings of string x overlap if there is position i in string x occupied by both of them.
GukiZ was disappointed because none of his students managed to solve the problem. Can you help them and find one of possible strings k?
Input
The first line contains string a, the second line contains string b, and the third line contains string c (1 ≤ |a|, |b|, |c| ≤ 105, where |s| denotes the length of string s).
All three strings consist only of lowercase English letters.
It is possible that b and c coincide.
Output
Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.
Examples
Input
aaa
a
b
Output
aaa
Input
pozdravstaklenidodiri
niste
dobri
Output
nisteaadddiiklooprrvz
Input
abbbaaccca
ab
aca
Output
ababacabcc
Note
In the third sample, this optimal solutions has three non-overlaping substrings equal to either b or c on positions 1 – 2 (ab), 3 – 4 (ab), 5 – 7 (aca). In this sample, there exist many other optimal solutions, one of them would be acaababbcc.
Submitted Solution:
```
from collections import Counter
import io
def div(a, b):
result = 10000000
for k, v in b.items():
result = min(result, a[k] // v)
return result
a = input()
b = input()
c = input()
ac = Counter(a)
bc = Counter(b)
cc = Counter(c)
best = 0
best_ij = (None, None)
for i in range(div(ac, bc) + 1):
j = div(ac, cc)
if i + j > best:
best_ij = (i, j)
best = i + j
for k in bc:
ac[k] -= bc[k]
ac = Counter(a)
for k in bc:
ac[k] -= best_ij[0] * bc[k]
for k in cc:
ac[k] -= best_ij[1] * cc[k]
s = io.StringIO()
s.write(best_ij[0] * b)
s.write(best_ij[1] * c)
for k, v in ac.items():
s.write(k * v)
print(best_ij)
print(s.getvalue())
``` | instruction | 0 | 57,555 | 0 | 115,110 |
No | output | 1 | 57,555 | 0 | 115,111 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times. | instruction | 0 | 57,748 | 0 | 115,496 |
Tags: implementation, strings
Correct Solution:
```
matrix = [[0 for j in range(26)] for i in range(26)]
n = int(input())
s = input()
max_i, max_j, max_v = 0, 0, 0
for i in range(1, n):
x, y = s[i - 1], s[i]
i_alph = ord(x) - ord('A')
j_alph = ord(y) - ord('A')
matrix[i_alph][j_alph] += 1
if max_v < matrix[i_alph][j_alph]:
max_v, max_i, max_j = matrix[i_alph][j_alph], i_alph, j_alph
first = chr(ord('A') + max_i)
second = chr(ord('A') + max_j)
print(first + second)
``` | output | 1 | 57,748 | 0 | 115,497 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times. | instruction | 0 | 57,749 | 0 | 115,498 |
Tags: implementation, strings
Correct Solution:
```
"""n=int(input())
f=input().split(' ')
v=list()
max=0
for i in range(0,n):
if int(f[i])>max:
max=int(f[i])
v.append(int(f[i]))
ans=0
#dp={i:0 for i in range(0,max+1)}
dp=dict()
for el in v:
if el-1 not in dp:
dp[el-1]=0
dp[el]=dp[el-1]+1
if dp[el]>ans:
ans=dp[el]
pred=el
sol=list()
i=n-1
for el in v[::-1]:
if el==pred:
sol.append(i+1)
pred-=1
i-=1
print(ans)
for i in sol[::-1]:
print(i,end=' ')
"""
"""n=int(input())
f=input().split(' ')
f=[int(x) for x in f]
def deg3(x):
y=0
while x%3==0:
y+=1
x/=3
return y
def comp(x,y):
if x[0]!=y[0]:
return x[0]>y[0]
else:
return x[1]<y[1]
v=[(-deg3(x),x) for x in f]
v=sorted(v)
sol=[x[1] for x in v]
for i in sol:
print(i,end=' ') """
"""import random
f = input().split(' ')
n = int(f[0])
k = int(f[1])
c = input().replace('\n','').split(' ')
f = [int(c[i]) for i in range(0,n)]
def quicksel(L, k):
piv = random.choice(L)
ls = [x for x in L if x < piv]
eq = [x for x in L if x == piv]
gr = [x for x in L if x > piv]
if k <= len(ls):
return quicksel(ls, k)
elif k <= len(ls) + len(eq):
return eq[0]
else:
return quicksel(gr, k - len(ls) - len(eq))
if k!=0:
x = quicksel(f, k)
else:
x = quicksel(f, 1)-1
cnt = 0
for i in f:
if i <= x:
cnt += 1
if (cnt != k) or x < 1 or x > 10 ** 9:
print(-1)
else:
print(x)"""
n=int(input())
sir=str(input())
def hashcode(s):
return 31*(ord(s[0])-65)+ord(s[1])-65
def decode(cod):
return (chr(cod//31+65),chr(cod%31+65))
v=dict()
maxsub=1
index=(sir[0],sir[1])
v[hashcode(index)]=1
for i in range(1,len(sir)-1):
t=(sir[i],sir[i+1])
cod=hashcode(t)
if cod not in v:
v[cod]=1
else:
v[cod]+=1
if v[cod]>maxsub:
maxsub=v[cod]
index=decode(cod)
sol=''.join(index)
print(sol)
``` | output | 1 | 57,749 | 0 | 115,499 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times. | instruction | 0 | 57,750 | 0 | 115,500 |
Tags: implementation, strings
Correct Solution:
```
n=int(input())
s=input()
dic={}
for i in range(0,n-1):
a=s[i:i+2]
if a in dic:
dic[a]+=1
else:
dic[a]=1
l=len(dic)
h=sorted((value,key) for (key,value) in dic.items())
print(h[l-1][1])
``` | output | 1 | 57,750 | 0 | 115,501 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times. | instruction | 0 | 57,751 | 0 | 115,502 |
Tags: implementation, strings
Correct Solution:
```
from statistics import mode
def most_common(name):
return mode(name)
n = int(input(""))
word = input("")
word1 = []
for i in range(len(word)-1):
word1.append(word[i:i+2])
a = most_common(word1)
print(a)
``` | output | 1 | 57,751 | 0 | 115,503 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times. | instruction | 0 | 57,752 | 0 | 115,504 |
Tags: implementation, strings
Correct Solution:
```
n=int(input())
s=input()
m=[]
for i in range(n-1):
m.append(s[i:i+2])
print(max(m,key=m.count))
``` | output | 1 | 57,752 | 0 | 115,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times. | instruction | 0 | 57,753 | 0 | 115,506 |
Tags: implementation, strings
Correct Solution:
```
n = int(input())
s = input()
d = []
ans = ""
max_ = 0
for i in range(n-1):
if not s[i] + s[i+1] in d:
d.append(s[i] + s[i+1])
f = 0
for j in range(i,n-1):
if s[j] + s[j+1] == s[i] + s[i+1]:
f += 1
if f > max_:
max_ = f
ans = s[i] + s[i+1]
print(ans)
``` | output | 1 | 57,753 | 0 | 115,507 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times. | instruction | 0 | 57,754 | 0 | 115,508 |
Tags: implementation, strings
Correct Solution:
```
input()
xs = input()
prev = ""
d = {}
for x in xs:
a = prev + x
if a in d:
d[a] += 1
else:
d[a] = 1
prev = x
max_a = ""
max = -float('inf')
for k, v in d.items():
if v >= max:
max = v
max_a = k
print(max_a)
``` | output | 1 | 57,754 | 0 | 115,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times. | instruction | 0 | 57,755 | 0 | 115,510 |
Tags: implementation, strings
Correct Solution:
```
n = int(input())
s = input()
D = dict()
for i in range(len(s) - 1):
if s[i] + s[i + 1] in D:
D[s[i] + s[i + 1]] += 1
else:
D[s[i] + s[i + 1]] = 1
Max = -1
ans = ""
for key in D:
if D[key] > Max:
Max = D[key]
ans = key
print(ans)
``` | output | 1 | 57,755 | 0 | 115,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times.
Submitted Solution:
```
def get_key(val , freq):
for key, value in freq.items():
if val == value:
return key
n = int(input())
s = input()
freq = {}
y = []
for i in range(n-1):
x = s[i:i+2]
if x in freq:
freq[x]+=1
else:
freq[x] = 1
val = freq.values()
m = max(val)
print(get_key(m,freq))
``` | instruction | 0 | 57,756 | 0 | 115,512 |
Yes | output | 1 | 57,756 | 0 | 115,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times.
Submitted Solution:
```
n = int(input())
l = input()
from collections import Counter
# print(l)
counter = []
for i in range(len(l)-1):
counter.append(l[i]+l[i+1])
print(Counter(counter).most_common()[0][0])
``` | instruction | 0 | 57,757 | 0 | 115,514 |
Yes | output | 1 | 57,757 | 0 | 115,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times.
Submitted Solution:
```
n = int(input())
a = input()
d = {}
for i in range(1,n):
if a[i-1]+a[i] in d:
d[a[i-1]+a[i]] += 1
else:
d[a[i-1]+a[i]] = 1
print(max(d, key = d.get))
``` | instruction | 0 | 57,758 | 0 | 115,516 |
Yes | output | 1 | 57,758 | 0 | 115,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times.
Submitted Solution:
```
import sys
def f(n, array):
d2gram = {}
for i, c in enumerate(array):
if i<=0:
continue
ngram = array[i-1] + c
if ngram in d2gram:
d2gram[ngram] += 1
else:
d2gram[ngram] = 1
return max(d2gram, key=d2gram.get)
if __name__ == '__main__':
for line in sys.stdin:
n = int(line)
array = sys.stdin.readline()
ans = f(n, array)
print(ans)
``` | instruction | 0 | 57,759 | 0 | 115,518 |
Yes | output | 1 | 57,759 | 0 | 115,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times.
Submitted Solution:
```
input()
s = input()
data = {}
maxi = -1
while len(s) != 0:
if s[0:2] not in data.keys():
data[s[0:2]] = 0
else:
data[s[0:2]] += 1
s = s[2:]
#print(data)
for i in data:
if data[i] > maxi:
maxi = data[i]
ans = i
print(ans)
``` | instruction | 0 | 57,760 | 0 | 115,520 |
No | output | 1 | 57,760 | 0 | 115,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times.
Submitted Solution:
```
a=int(input())
b=input()
l=[]
for x in range (0, a-1):
sub=b[x:x+2]
l.append(sub)
print(max(l))
``` | instruction | 0 | 57,761 | 0 | 115,522 |
No | output | 1 | 57,761 | 0 | 115,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times.
Submitted Solution:
```
n = int(input())
s = input()
pairs = []
for i in range(0, len(s)-1):
pairs.append("{}{}".format(s[i], s[i+1]))
nums = {}
for pair in pairs:
nums[pair] = 0
for pair in pairs:
nums[pair] += 1
ex = sorted(nums.values())[-1]
for item in nums.keys():
if nums[item] == ex:
print(item)
``` | instruction | 0 | 57,762 | 0 | 115,524 |
No | output | 1 | 57,762 | 0 | 115,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.
You are given a string s consisting of n capital Latin letters. Your task is to find any two-gram contained in the given string as a substring (i.e. two consecutive characters of the string) maximal number of times. For example, for string s = "BBAABBBA" the answer is two-gram "BB", which contained in s three times. In other words, find any most frequent two-gram.
Note that occurrences of the two-gram can overlap with each other.
Input
The first line of the input contains integer number n (2 ≤ n ≤ 100) — the length of string s. The second line of the input contains the string s consisting of n capital Latin letters.
Output
Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string s as a substring (i.e. two consecutive characters of the string) maximal number of times.
Examples
Input
7
ABACABA
Output
AB
Input
5
ZZZAA
Output
ZZ
Note
In the first example "BA" is also valid answer.
In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times.
Submitted Solution:
```
from functools import reduce
# import sys
# sys.stdin = open('input.txt','r')
# sys.stdout = open('output.txt','w')
######
# Matrix Multiplication Function
def matmult(a,b):
zip_b = list(zip(*b))
return [[sum(ele_a*ele_b for ele_a, ele_b in zip(row_a, col_b))
for col_b in zip_b] for row_a in a]
######
# isPrime
def Prime(n):
if n & 1 == 0:
return False
d= 3
while d * d <= n:
if n % d == 0:
return False
d= d + 2
return True
# This function returns True if a number is Prime else returns False
######
# factors of a number
def factors(n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
# This function returns set of all the factors
######
def isEven(n):
if (n%2==0):
return True
return False
######
def sieve_of_eratosthenes(x , y):
l = [2] # list of prime numbers
for n in range(3,y+1,2): # iterations over odd numbers
isprime = True
for e in l:
if n % e == 0:
isprime = False
break
if(isprime):
l.append(n)
return [e for e in l if e >= x]
######
# for _ in range(int(input())):
n = int(input())
s = input()
# n,k = map(int,input().split())
# lst = list(map(int,input().split()))
d = {}
ans,val = '',0
for i in range(n-2):
if s[i:i+2] in d.keys():
d[s[i:i+2]]+=1
if (d[s[i:i+2]]>val):
val = d[s[i:i+2]]
ans = s[i:i+2]
else:
d[s[i:i+2]]=1
print(ans,val)
``` | instruction | 0 | 57,763 | 0 | 115,526 |
No | output | 1 | 57,763 | 0 | 115,527 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4 | instruction | 0 | 57,768 | 0 | 115,536 |
"Correct Solution:
```
S = input()
max0 = 0
max1 = 0
amax0 = []
amax1 = []
mi0 = 0
mi1 = 0
h0 = 0
h1 = 0
for i, c in enumerate(S):
if c=="1":
h1 += 1
if h1-mi1 > max1:
max1 = h1-mi1
amax1 = [i]
elif h1-mi1 == max1:
amax1.append(i)
h0 -= 1
if h0 < mi0:
mi0 = h0
elif c=="0":
h1 -= 1
if h1 < mi1:
mi1 = h1
h0 += 1
if h0-mi0 > max0:
max0 = h0-mi0
amax0 = [i]
elif h0-mi0 == max0:
amax0.append(i)
else:
h1 -= 1
if h1 < mi1:
mi1 = h1
h0 -= 1
if h0 < mi0:
mi0 = h0
# if max0 < max1:
# ma = max1
# idx_max = amax1
# else:
# ma = max0
# idx_max = amax0
if max0 < max1:
if len(set([i%2 for i in amax1]))==1:
print(max1)
else:
print(max1+1)
elif max0 > max1:
if len(set([i%2 for i in amax0]))==1:
print(max0)
else:
print(max0+1)
else:
if len(set([i%2 for i in amax1]+[(max0+i)%2 for i in amax0]))==1:
print(max0)
else:
print(max0+1)
``` | output | 1 | 57,768 | 0 | 115,537 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4 | instruction | 0 | 57,769 | 0 | 115,538 |
"Correct Solution:
```
S = input()
N = len(S)
A = [0]*(N+1)
for i,s in enumerate(S,1):
if s=="1":
A[i] = A[i-1]+1
else:
A[i] = A[i-1]-1
m = max(A)
B = [A[-1]]*(N+1)
C = [m-A[-1]]*(N+1)
for i in range(N):
i = N-i-1
B[i] = max(A[i],B[i+1])
C[i] = m-B[i]
d = 0
e = 0
D = A[:]
E = A[:]
for i,s in enumerate(S,1):
if s=='?' and C[i] >= d+2:
d += 2
if s=='?' and C[i] >= e+1:
e += 2
D[i] += d
E[i] += e
ans = min(max(D)-min(D),max(E)-min(E))
print(ans)
``` | output | 1 | 57,769 | 0 | 115,539 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4 | instruction | 0 | 57,770 | 0 | 115,540 |
"Correct Solution:
```
def main():
S = input()
size = [[[[0, 0]], []], [[], []]]
length = 0
for s in S:
size2 = [[[], []], [[], []], [[], []]]
if s != "0":
for i in range(2):
for j in range(2):
if size[i][j]:
for x, y in size[i][j]:
if y == length+i:
size2[i+1][j ^ 1].append([y+1, y+1])
if x != y:
size2[i][j ^ 1].append([x+1, y-1])
else:
size2[i][j ^ 1].append([x+1, y+1])
if s != "1":
for i in range(2):
for j in range(2):
if size[i][j]:
for x, y in size[i][j]:
if x == 0:
size2[i+1][j].append([0, 0])
if x != y:
size2[i][j ^ 1].append([x+1, y-1])
else:
size2[i][j ^ 1].append([x-1, y-1])
if len(size2[0][0]+size2[0][1]) == 0:
size2 = size2[1:]
length += 1
else:
size2.pop()
for i in range(2):
for j in range(2):
size2[i][j] = sorted(size2[i][j])
size = [[[], []], [[], []]]
for i in range(2):
for j in range(2):
if len(size2[i][j]) == 0:
continue
elif len(size2[i][j]) == 1:
size[i][j] = size2[i][j]
continue
size[i][j] = [size2[i][j][0]]
for x, y in size2[i][j][1:]:
p, q = size[i][j][-1]
if y <= q:
continue
if x <= q+2:
size[i][j][-1][1] = y
else:
size[i][j].append([x, y])
print(length)
main()
``` | output | 1 | 57,770 | 0 | 115,541 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4 | instruction | 0 | 57,771 | 0 | 115,542 |
"Correct Solution:
```
def main():
S = input()
size = [[[[0, 0]], []], [[], []]]
length = 0
for s in S:
size2 = [[[], []], [[], []], [[], []]]
if s != "0":
for i in range(2):
for j in range(2):
if size[i][j]:
for x, y in size[i][j]:
if y == length+i:
size2[i+1][j ^ 1].append([y+1, y+1])
if x != y:
size2[i][j ^ 1].append([x+1, y-1])
else:
size2[i][j ^ 1].append([x+1, y+1])
if s != "1":
for i in range(2):
for j in range(2):
if size[i][j]:
for x, y in size[i][j]:
if x == 0:
size2[i+1][j].append([0, 0])
if x != y:
size2[i][j ^ 1].append([x+1, y-1])
else:
size2[i][j ^ 1].append([x-1, y-1])
if len(size2[0][0]+size2[0][1]) == 0:
size2 = size2[1:]
length += 1
else:
size2.pop()
for i in range(2):
for j in range(2):
size2[i][j] = sorted(size2[i][j])
size = [[[], []], [[], []]]
for i in range(2):
for j in range(2):
if len(size2[i][j]) == 0:
continue
size[i][j] = [size2[i][j][0]]
for x, y in size2[i][j][1:]:
size[i][j][-1][1] = max(y, size[i][j][-1][1])
print(length)
main()
``` | output | 1 | 57,771 | 0 | 115,543 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4 | instruction | 0 | 57,772 | 0 | 115,544 |
"Correct Solution:
```
S = input()
c = 0
cum = [c]
for s in S:
if s == "1":
c += 1
else:
c -= 1
cum.append(c)
max_cum = [None] * (len(S) + 1)
max_cum[-1] = cum[-1]
for i in reversed(range(len(S))):
max_cum[i] = max(cum[i], max_cum[i + 1])
z = max_cum[0]
def f(m):
c = 0
fz = c
add = 0
for i, s in enumerate(S):
if s == "1":
c += 1
elif s == "0":
c -= 1
fz = min(fz, c)
elif add + max_cum[i + 1] + 2 <= m:
c += 1
add += 2
else:
c -= 1
fz = min(fz, c)
return fz
fz = f(z)
fz1 = f(z + 1)
ans = min(z - fz, z + 1 - fz1)
print(ans)
``` | output | 1 | 57,772 | 0 | 115,545 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4 | instruction | 0 | 57,773 | 0 | 115,546 |
"Correct Solution:
```
def bisect(ng, ok, judge):
while abs(ng-ok) > 1:
m = (ng+ok)//2
if judge(m):
ok = m
else:
ng = m
return ok
def solve(S):
d = {'0':0,'1':1,'?':2}
S = tuple(d[c] for c in S)
lookup = ((-1,-1),(1,1), (-1,1))
def judge(target_lo, target_hi):
lo,hi = 0,0
pl = target_lo%2 == 0
ph = target_hi%2 == 0
for s in S:
a,b = lookup[s]
lo = max(lo+a, target_lo+pl)
hi = min(hi+b, target_hi-ph)
if hi < lo:
return False
pl = not pl
ph = not ph
return True
best = 10**6*2
n0,n1,n2 = S.count(0), S.count(1), S.count(2)
lo = -n0-n2
hi = n1+n2
while True:
hi = bisect(-1, hi, lambda x: judge(lo, x))
lo = bisect(1, lo, lambda x: judge(x, hi))
if hi-lo >= best:
break
else:
best = hi-lo
hi += 1
lo = bisect(1, lo, lambda x: judge(x, hi))
return min(best, hi-lo)
# from itertools import product, accumulate
# from random import shuffle
# def naive(S):
# d = {'0':(-1,), '1':(1,), '?': (-1,1)}
# return min(max(0,max(accumulate(X)))-min(0,min(accumulate(X))) for X in product(*(d[c] for c in S)))
if __name__ == '__main__':
S = input()
print(solve(S))
# for _ in range(50):
# S = ['0']*15+['1']*15+['?']*14
# shuffle(S)
# S = ''.join(S)
# print(S, naive(S), solve(S))
``` | output | 1 | 57,773 | 0 | 115,547 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4 | instruction | 0 | 57,774 | 0 | 115,548 |
"Correct Solution:
```
S = list(input())
N = len(S)
Z = 0
SUM = 0
for s in S:
if s == "1":
SUM += 1
Z = max(Z,SUM)
else:
SUM -= 1
ruiseki_max = [0]
MAX = SUM
for s in S[1:][::-1]:
if s == "1":
SUM -= 1
else:
SUM += 1
MAX = max(MAX,SUM)
ruiseki_max.append(MAX-SUM)
def f(Z):
SUM = 0
MIN = 0
for i in range(N):
s = S[i]
if s == "1":
SUM += 1
elif s == "0":
SUM -= 1
MIN = min(MIN,SUM)
else:
if SUM + 1 + ruiseki_max[N-1-i] <= Z:
SUM += 1
else:
SUM -= 1
MIN = min(MIN,SUM)
return MIN
print(min(Z-f(Z),(Z+1-f(Z+1))))
``` | output | 1 | 57,774 | 0 | 115,549 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4 | instruction | 0 | 57,775 | 0 | 115,550 |
"Correct Solution:
```
S = input()
L = 0
U = len(S)
while U - L > 1:
M = (U + L) // 2
OK = False
for i in range(2):
A = i
B = M
if A % 2 != B % 2: B -= 1
NG = False
for s in S:
if s == '0':
A -= 1
B -= 1
elif s == '1':
A += 1
B += 1
else:
A -= 1
B += 1
if A < 0: A += 2
if B > M: B -= 2
if A > B:
NG = True
break
if not NG:
OK = True
break
if OK:
U = M
else:
L = M
print(U)
``` | output | 1 | 57,775 | 0 | 115,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4
Submitted Solution:
```
S = input()
max0 = 0
max1 = 0
amax0 = []
amax1 = []
mi0 = 0
mi1 = 0
h0 = 0
h1 = 0
for i, c in enumerate(S):
if c=="1":
h1 += 1
if h1-mi1 > max1:
max1 = h1-mi1
amax1 = [i]
elif h1-mi1 == max1:
amax1.append(i)
h0 -= 1
if h0 < mi0:
mi0 = h0
elif c=="0":
h1 -= 1
if h1 < mi1:
mi1 = h1
h0 += 1
if h0-mi0 > max0:
max0 = h0-mi0
amax0 = [i]
elif h0-mi0 == max0:
amax0.append(i)
else:
h1 -= 1
if h1 < mi1:
mi1 = h1
h0 -= 1
if h0 < mi0:
mi0 = h0
if max0 < max1:
if len(set([i%2 for i in amax1]))==1:
print(max1)
else:
print(max1+1)
elif max0 > max1:
if len(set([i%2 for i in amax0]))==1:
print(max0)
else:
print(max0+1)
else:
if len(set([i%2 for i in amax1]+[(max0+i)%2 for i in amax0]))==1:
print(max0)
else:
print(max0+1)
``` | instruction | 0 | 57,776 | 0 | 115,552 |
Yes | output | 1 | 57,776 | 0 | 115,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4
Submitted Solution:
```
"""
https://atcoder.jp/contests/agc045/tasks/agc045_b
交互が最高
アンバランス度の定義を変えたい
1の数 = 区間和
0の数 = 区間長さ-1の数
アンバランス度 = abs(区間長さ - 区間和 * 2)
→ abs( r-l+1 - (dp[r]-dp[l])*2 )
=abs( r-dp[r]*2 - (l-dp[l]*2) + 1 )
アンバランス度は2分探索すればよさそう
あとは、アンバランス度x以下を達成できるか？
の判定問題
連続してxこ並ぶ場所があると不可能
アンバランス度は、 (l-dp[l]*2) の最大・最小値さえ持っておけば計算できる
x以下を達成する方法は？
始めて?が来た時を考える
今までの (l-dp[l]*2) の最大・最小値 は持っている
なるべく最大・最小値は更新しない方がいい
1にする→1減らす
0にする→1増やす
1増やす・1減らすを選んで差を x-1以下にしたい問題
まず全部上にしておく
それ以降の最大値と最小値に影響する
下にすると、最大値を-2,最小値も-2できる
それ以前の最大値-それ以降の最小値 は2ふえる
それ以降の最大値-それ以前の最小値 は2へる
y=xを足してみる
====答えを見た=====
maspy氏の方針がわかりやすい
https://maspypy.com/atcoder-%E5%8F%82%E5%8A%A0%E6%84%9F%E6%83%B3-2020-06-07agc045
行ける点は連番になる(偶奇は違うので注意)
x以下が可能か → 0以上x以下の点からスタートして,維持できるか
なぜ偶奇の場合分けが必要？
→平行移動が不可能だからありえない答えを生み出す可能性があるため
ある数字以外不可能な時=幅が2になり、3つの数字の可能性が生まれるが
平行移動はできないため、存在しえない数字が生まれてしまう
偶奇をちゃんとすれば問題ない
"""
from sys import stdin
S = stdin.readline()
S = S[:-1]
N = len(S)
r = 2*10**6
l = 0
while r-l != 1:
m = (r+l)//2
flag = False
#even
nmax = m//2*2
nmin = 0
for i in range(N):
if S[i] == "0":
nmax = min(nmax+1,m)
nmin = min(nmin+1,m)
elif S[i] == "1":
nmax = max(nmax-1,0)
nmin = max(nmin-1,0)
else:
nmax = min(nmax+1,m)
nmin = max(nmin-1,0)
if i % 2 == 0:
if nmax % 2 == 0:
nmax -= 1
if nmin % 2 == 0:
nmin += 1
else:
if nmax % 2 == 1:
nmax -= 1
if nmin % 2 == 1:
nmin += 1
if nmax < nmin:
break
if nmax >= nmin:
flag = True
else:
#odd
nmax = (m-1)//2*2+1
nmin = 1
for i in range(N):
if S[i] == "0":
nmax = min(nmax+1,m)
nmin = min(nmin+1,m)
elif S[i] == "1":
nmax = max(nmax-1,0)
nmin = max(nmin-1,0)
else:
nmax = min(nmax+1,m)
nmin = max(nmin-1,0)
if i % 2 == 1:
if nmax % 2 == 0:
nmax -= 1
if nmin % 2 == 0:
nmin += 1
else:
if nmax % 2 == 1:
nmax -= 1
if nmin % 2 == 1:
nmin += 1
if nmax < nmin:
break
if nmax >= nmin:
flag = True
if flag:
r = m
else:
l = m
print (r)
``` | instruction | 0 | 57,777 | 0 | 115,554 |
Yes | output | 1 | 57,777 | 0 | 115,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4
Submitted Solution:
```
from itertools import accumulate
import copy
S=input()
ANS=[0]
for s in S:
if s=="0":
ANS.append(-1)
else:
ANS.append(1)
SUM=list(accumulate(ANS))
MIN=[SUM[-1]]
for s in SUM[::-1][1:]:
MIN.append(min(MIN[-1],s))
MIN.reverse()
MX=min(MIN)
ANS2=copy.deepcopy(ANS)
sa=0
for i in range(len(S)):
if S[i]=="?":
if MIN[i+1]-MX-sa>=2:
ANS2[i+1]=-1
sa+=2
ANS3=copy.deepcopy(ANS)
sa=0
MX-=1
for i in range(len(S)):
if S[i]=="?":
if MIN[i+1]-MX-sa>=2:
ANS3[i+1]=-1
sa+=2
S2=list(accumulate(ANS2))
S3=list(accumulate(ANS3))
print(min(max(S2)-min(S2),max(S3)-min(S3)))
``` | instruction | 0 | 57,778 | 0 | 115,556 |
Yes | output | 1 | 57,778 | 0 | 115,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4
Submitted Solution:
```
from itertools import accumulate
S = input()
N = len(S)
A = [0] + list(accumulate(1 if s == "1" else -1 for s in S))
ma = max(A)
cur = A[-1]
C = [ma - cur]
for a in reversed(A):
cur = max(a, cur)
C.append(ma - cur)
d, e = 0, 0
D, E = A[:], A[:]
for i, (s, c) in enumerate(zip(S, reversed(C[:-1])), 1):
if s == '?' and c >= d + 2:
d += 2
if s == '?' and c >= e + 1:
e += 2
D[i] += d
E[i] += e
print(min(max(D) - min(D), max(E) - min(E)))
``` | instruction | 0 | 57,779 | 0 | 115,558 |
Yes | output | 1 | 57,779 | 0 | 115,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4
Submitted Solution:
```
# coding: utf-8
import sys
sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))
# 二分探索、左右からの貪欲法、?をスキップした状態で1の連続数と0の連続数
S = list(sr())
answer = 0
def solve(S):
global answer
one_seq = 0
one_max = 0
one_start = set()
zero_seq = 0
zero_max = 0
zero_start = set()
cur = 0
for i, s in enumerate(S):
if s == '1':
one_seq += 1
zero_seq -= 0
zero_seq = max(0, zero_seq)
cur += 1
elif s == '0':
zero_seq += 1
one_seq -= 0
one_seq = max(0, one_seq)
cur -= 1
elif s == '?':
one_seq -= 1
zero_seq -= 1
one_seq = max(0, one_seq)
zero_seq = max(0, zero_seq) # ここでは保留
if one_seq > 0 and one_seq == one_max:
one_start.add(i-one_seq+1)
elif one_seq > 0 and one_seq > one_max:
one_start = {i-one_seq+1}
one_max = one_seq
if zero_seq > 0 and zero_seq == zero_max:
zero_start.add(i-zero_seq+1)
elif zero_seq > 0 and zero_seq > zero_max:
zero_start = {i-zero_seq+1}
zero_max = zero_seq
answer = max(one_max, zero_max)
one_seq = 0
zero_seq = 0
cur = 0
for i, s in enumerate(S):
if s == '1':
one_seq += 1
zero_seq -= 0
zero_seq = max(0, zero_seq)
cur += 1
elif s == '0':
zero_seq += 1
one_seq -= 0
one_seq = max(0, one_seq)
cur -= 1
elif s == '?':
if zero_seq == answer and i+1 in one_start:
zero_max += 1
answer += 1
return None
elif one_seq == answer and i+1 in zero_start:
one_max += 1
answer += 1
return None
if i+1 in one_start:
S[i] = '0'
cur -= 1
elif i+1 in zero_start:
S[i] = '1'
cur += 1
else:
one_seq -= 1
zero_seq -= 1
one_seq = max(0, one_seq)
zero_seq = max(0, zero_seq) # ここでは保留
return S
for i in range(10):
ret = solve(S)
if ret:
S = ret
solve(S[::-1])
print(answer)
``` | instruction | 0 | 57,780 | 0 | 115,560 |
No | output | 1 | 57,780 | 0 | 115,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4
Submitted Solution:
```
def solve(S):
st = "" # debug
if S[0] == "?":
for i in range(len(S)):
if S[i] == "0":
if i % 2 == 0:
S[0] = "0"
else:
S[0] = "1"
break
elif S[i] == "1":
if i % 2 == 0:
S[0] = "1"
else:
S[0] = "0"
break
if S[0] == "?":
return 1
now_sum0 = 0
now_sum1 = 0
max_sum0 = 0
max_sum1 = 0
val = -1
for i in range(len(S)):
if S[i] == "0":
now_sum0 += 1
now_sum1 = max(now_sum1 - 1, 0)
val = -1
elif S[i] == "1":
now_sum0 = max(now_sum0 - 1, 0)
now_sum1 += 1
val = -1
else :
if now_sum0 > now_sum1:
if now_sum0 > now_sum1 + 1:
now_sum1 += 1
now_sum0 = max(now_sum0 - 1, 0)
st += "1" # debug
elif i + 1 < len(S):
if S[i+1] == "0":
now_sum0 = max(now_sum0 - 1, 0)
now_sum1 += 1
elif S[i+1] == "1":
now_sum0 += 1
now_sum1 = max(now_sum1 - 1, 0)
else:
now_sum1 += 1
now_sum0 = max(now_sum0 - 1, 0)
elif now_sum1 > now_sum0:
if now_sum1 > now_sum0 + 1:
now_sum0 += 1
now_sum1 = max(now_sum1 - 1, 0)
elif i + 1 < len(S):
if S[i+1] == "0":
now_sum0 = max(now_sum0 - 1, 0)
now_sum1 += 1
elif S[i+1] == "1":
now_sum0 += 1
now_sum1 = max(now_sum1 - 1, 0)
else:
now_sum0 += 1
now_sum1 = max(now_sum1 - 1, 0)
else:
if i + 1 < len(S):
if S[i+1] == "0":
now_sum0 = max(now_sum0 - 1, 0)
now_sum1 += 1
elif S[i+1] == "1":
now_sum0 += 1
now_sum1 = max(now_sum1 - 1, 0)
else:
if val == -1:
for j in range(i, len(S)):
if S[j] == "0":
if (j - i) % 2 == 0:
val = 0
else:
val = 1
break
elif S[j] == "1":
if (j - i) % 2 == 0:
val = 1
else:
val = 0
break
if val == 1:
now_sum0 = max(now_sum0 - 1, 0)
now_sum1 += 1
elif val == 0:
now_sum0 += 1
now_sum1 = max(now_sum1 - 1, 0)
else:
return max(max_sum0, max_sum1)
else:
return max(max_sum0, max_sum1)
max_sum0 = max(max_sum0, now_sum0)
max_sum1 = max(max_sum1, now_sum1)
return max(max_sum0, max_sum1)
S = list(input())
print(min(solve(S), solve(S[::-1])))
``` | instruction | 0 | 57,781 | 0 | 115,562 |
No | output | 1 | 57,781 | 0 | 115,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4
Submitted Solution:
```
z=0
n=0
s=input()
l=[]
q=0
for i in range(len(s)):
if s[i]=="0":
z+=1
elif s[i]=="1":
n+=1
else:
if z>n:
q="1"
n+=1
elif z<n:
q="0"
z+=1
else:
if s[i-1]=="1":
z+=1
elif s[i-1]=="?":
if q=="0":
n+=1
else:
z+=1
else:
n+=1
l.append(abs(z-n))
print(max(l))
``` | instruction | 0 | 57,782 | 0 | 115,564 |
No | output | 1 | 57,782 | 0 | 115,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a string S, where each character is `0`, `1`, or `?`.
Consider making a string S' by replacing each occurrence of `?` with `0` or `1` (we can choose the character for each `?` independently). Let us define the unbalancedness of S' as follows:
* (The unbalancedness of S') = \max \\{ The absolute difference between the number of occurrences of `0` and `1` between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\\}
Find the minimum possible unbalancedness of S'.
Constraints
* 1 \leq |S| \leq 10^6
* Each character of S is `0`, `1`, or `?`.
Input
Input is given from Standard Input in the following format:
S
Output
Print the minimum possible unbalancedness of S'.
Examples
Input
0??
Output
1
Input
0??0
Output
2
Input
??00????0??0????0?0??00??1???11?1?1???1?11?111???1
Output
4
Submitted Solution:
```
z=0
n=0
s=input()
l=[]
for i in range(len(s)):
if s[i]=="0":
z+=1
elif s[i]=="1":
n+=1
else:
if z>n:
n+=1
elif z<n:
z+=1
else:
if s[i-1]=="1":
z+=1
elif s[i-1]=="?":
t1=s[max(l):i].count("1")
t2=s[max(l):i].count("0")
if t1>=t2:
z+=1
else:
n+=1
else:
n+=1
l.append(abs(z-n))
print(max(l))
``` | instruction | 0 | 57,783 | 0 | 115,566 |
No | output | 1 | 57,783 | 0 | 115,567 |
Provide a correct Python 3 solution for this coding contest problem.
We have a string X, which has an even number of characters. Half the characters are `S`, and the other half are `T`.
Takahashi, who hates the string `ST`, will perform the following operation 10^{10000} times:
* Among the occurrences of `ST` in X as (contiguous) substrings, remove the leftmost one. If there is no occurrence, do nothing.
Find the eventual length of X.
Constraints
* 2 ≦ |X| ≦ 200,000
* The length of X is even.
* Half the characters in X are `S`, and the other half are `T`.
Input
The input is given from Standard Input in the following format:
X
Output
Print the eventual length of X.
Examples
Input
TSTTSS
Output
4
Input
SSTTST
Output
0
Input
TSSTTTSS
Output
4 | instruction | 0 | 57,883 | 0 | 115,766 |
"Correct Solution:
```
x=input()
s_cnt=0
t_cnt=0
for i in x:
if i=="S":
s_cnt+=1
elif s_cnt>0:
s_cnt-=1
else:
t_cnt+=1
print(s_cnt+t_cnt)
``` | output | 1 | 57,883 | 0 | 115,767 |
Provide a correct Python 3 solution for this coding contest problem.
We have a string X, which has an even number of characters. Half the characters are `S`, and the other half are `T`.
Takahashi, who hates the string `ST`, will perform the following operation 10^{10000} times:
* Among the occurrences of `ST` in X as (contiguous) substrings, remove the leftmost one. If there is no occurrence, do nothing.
Find the eventual length of X.
Constraints
* 2 ≦ |X| ≦ 200,000
* The length of X is even.
* Half the characters in X are `S`, and the other half are `T`.
Input
The input is given from Standard Input in the following format:
X
Output
Print the eventual length of X.
Examples
Input
TSTTSS
Output
4
Input
SSTTST
Output
0
Input
TSSTTTSS
Output
4 | instruction | 0 | 57,884 | 0 | 115,768 |
"Correct Solution:
```
s=input()
ls=len(s)
c=0
ans=0
for i in range(ls):
if s[i]=='S':
c+=1
elif c>0:
ans+=2
c-=1
print(ls-ans)
``` | output | 1 | 57,884 | 0 | 115,769 |
Provide a correct Python 3 solution for this coding contest problem.
We have a string X, which has an even number of characters. Half the characters are `S`, and the other half are `T`.
Takahashi, who hates the string `ST`, will perform the following operation 10^{10000} times:
* Among the occurrences of `ST` in X as (contiguous) substrings, remove the leftmost one. If there is no occurrence, do nothing.
Find the eventual length of X.
Constraints
* 2 ≦ |X| ≦ 200,000
* The length of X is even.
* Half the characters in X are `S`, and the other half are `T`.
Input
The input is given from Standard Input in the following format:
X
Output
Print the eventual length of X.
Examples
Input
TSTTSS
Output
4
Input
SSTTST
Output
0
Input
TSSTTTSS
Output
4 | instruction | 0 | 57,885 | 0 | 115,770 |
"Correct Solution:
```
s = input().strip()
x=0
r=0
for c in s:
if c == "S":
x += 1
else:
if x != 0:
x -= 1
r += 2
print(len(s)-r)
``` | output | 1 | 57,885 | 0 | 115,771 |
Provide a correct Python 3 solution for this coding contest problem.
We have a string X, which has an even number of characters. Half the characters are `S`, and the other half are `T`.
Takahashi, who hates the string `ST`, will perform the following operation 10^{10000} times:
* Among the occurrences of `ST` in X as (contiguous) substrings, remove the leftmost one. If there is no occurrence, do nothing.
Find the eventual length of X.
Constraints
* 2 ≦ |X| ≦ 200,000
* The length of X is even.
* Half the characters in X are `S`, and the other half are `T`.
Input
The input is given from Standard Input in the following format:
X
Output
Print the eventual length of X.
Examples
Input
TSTTSS
Output
4
Input
SSTTST
Output
0
Input
TSSTTTSS
Output
4 | instruction | 0 | 57,886 | 0 | 115,772 |
"Correct Solution:
```
s = input()
count = 0
sc = 0
for c in s:
if c=='S':
sc+=1
else:
if sc > 0:
sc -= 1
count += 1
print(len(s) - count*2)
``` | output | 1 | 57,886 | 0 | 115,773 |
Provide a correct Python 3 solution for this coding contest problem.
We have a string X, which has an even number of characters. Half the characters are `S`, and the other half are `T`.
Takahashi, who hates the string `ST`, will perform the following operation 10^{10000} times:
* Among the occurrences of `ST` in X as (contiguous) substrings, remove the leftmost one. If there is no occurrence, do nothing.
Find the eventual length of X.
Constraints
* 2 ≦ |X| ≦ 200,000
* The length of X is even.
* Half the characters in X are `S`, and the other half are `T`.
Input
The input is given from Standard Input in the following format:
X
Output
Print the eventual length of X.
Examples
Input
TSTTSS
Output
4
Input
SSTTST
Output
0
Input
TSSTTTSS
Output
4 | instruction | 0 | 57,887 | 0 | 115,774 |
"Correct Solution:
```
S=input()
t=0
u=0
for v in S:
if v=="S":
t+=1
else:
if t>=1:
u+=2
t-=1
print(len(S)-u)
``` | output | 1 | 57,887 | 0 | 115,775 |
Provide a correct Python 3 solution for this coding contest problem.
We have a string X, which has an even number of characters. Half the characters are `S`, and the other half are `T`.
Takahashi, who hates the string `ST`, will perform the following operation 10^{10000} times:
* Among the occurrences of `ST` in X as (contiguous) substrings, remove the leftmost one. If there is no occurrence, do nothing.
Find the eventual length of X.
Constraints
* 2 ≦ |X| ≦ 200,000
* The length of X is even.
* Half the characters in X are `S`, and the other half are `T`.
Input
The input is given from Standard Input in the following format:
X
Output
Print the eventual length of X.
Examples
Input
TSTTSS
Output
4
Input
SSTTST
Output
0
Input
TSSTTTSS
Output
4 | instruction | 0 | 57,888 | 0 | 115,776 |
"Correct Solution:
```
X = input()
s, t = 0, 0
for x in X:
if x == "S":
s += 1
else:
if s == 0:
t += 1
else:
s -= 1
print(s+t)
``` | output | 1 | 57,888 | 0 | 115,777 |
Provide a correct Python 3 solution for this coding contest problem.
We have a string X, which has an even number of characters. Half the characters are `S`, and the other half are `T`.
Takahashi, who hates the string `ST`, will perform the following operation 10^{10000} times:
* Among the occurrences of `ST` in X as (contiguous) substrings, remove the leftmost one. If there is no occurrence, do nothing.
Find the eventual length of X.
Constraints
* 2 ≦ |X| ≦ 200,000
* The length of X is even.
* Half the characters in X are `S`, and the other half are `T`.
Input
The input is given from Standard Input in the following format:
X
Output
Print the eventual length of X.
Examples
Input
TSTTSS
Output
4
Input
SSTTST
Output
0
Input
TSSTTTSS
Output
4 | instruction | 0 | 57,889 | 0 | 115,778 |
"Correct Solution:
```
x = input()
s=0
ans=0
for i in x:
if i == "S":
s+=1
else:
if s==0:
ans +=1
else:
s-=1
print(ans+s)
``` | output | 1 | 57,889 | 0 | 115,779 |
Provide a correct Python 3 solution for this coding contest problem.
We have a string X, which has an even number of characters. Half the characters are `S`, and the other half are `T`.
Takahashi, who hates the string `ST`, will perform the following operation 10^{10000} times:
* Among the occurrences of `ST` in X as (contiguous) substrings, remove the leftmost one. If there is no occurrence, do nothing.
Find the eventual length of X.
Constraints
* 2 ≦ |X| ≦ 200,000
* The length of X is even.
* Half the characters in X are `S`, and the other half are `T`.
Input
The input is given from Standard Input in the following format:
X
Output
Print the eventual length of X.
Examples
Input
TSTTSS
Output
4
Input
SSTTST
Output
0
Input
TSSTTTSS
Output
4 | instruction | 0 | 57,890 | 0 | 115,780 |
"Correct Solution:
```
x = input()
n = len(x)
res = 0
tres = 0
for c in x:
if c == 'T':
tres += 1
else:
tres -= 1
res = max(tres,res)
print(res * 2)
``` | output | 1 | 57,890 | 0 | 115,781 |
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