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int64 194
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|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string s consisting only of lowercase and uppercase Latin letters.
Let A be a set of positions in the string. Let's call it pretty if following conditions are met:
* letters on positions from A in the string are all distinct and lowercase;
* there are no uppercase letters in the string which are situated between positions from A (i.e. there is no such j that s[j] is an uppercase letter, and a1 < j < a2 for some a1 and a2 from A).
Write a program that will determine the maximum number of elements in a pretty set of positions.
Input
The first line contains a single integer n (1 ≤ n ≤ 200) — length of string s.
The second line contains a string s consisting of lowercase and uppercase Latin letters.
Output
Print maximum number of elements in pretty set of positions for string s.
Examples
Input
11
aaaaBaabAbA
Output
2
Input
12
zACaAbbaazzC
Output
3
Input
3
ABC
Output
0
Note
In the first example the desired positions might be 6 and 8 or 7 and 8. Positions 6 and 7 contain letters 'a', position 8 contains letter 'b'. The pair of positions 1 and 8 is not suitable because there is an uppercase letter 'B' between these position.
In the second example desired positions can be 7, 8 and 11. There are other ways to choose pretty set consisting of three elements.
In the third example the given string s does not contain any lowercase letters, so the answer is 0.
Submitted Solution:
```
n=int(input())
s=input()
ss=set()
b=0
for i in range(n):
if s[i]>="a" and s[i]<="z":
ss.add(s[i])
else:
b=max(b,len(ss))
ss=set()
print(b)
```
|
instruction
| 0
| 8,777
| 0
| 17,554
|
No
|
output
| 1
| 8,777
| 0
| 17,555
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string s consisting only of lowercase and uppercase Latin letters.
Let A be a set of positions in the string. Let's call it pretty if following conditions are met:
* letters on positions from A in the string are all distinct and lowercase;
* there are no uppercase letters in the string which are situated between positions from A (i.e. there is no such j that s[j] is an uppercase letter, and a1 < j < a2 for some a1 and a2 from A).
Write a program that will determine the maximum number of elements in a pretty set of positions.
Input
The first line contains a single integer n (1 ≤ n ≤ 200) — length of string s.
The second line contains a string s consisting of lowercase and uppercase Latin letters.
Output
Print maximum number of elements in pretty set of positions for string s.
Examples
Input
11
aaaaBaabAbA
Output
2
Input
12
zACaAbbaazzC
Output
3
Input
3
ABC
Output
0
Note
In the first example the desired positions might be 6 and 8 or 7 and 8. Positions 6 and 7 contain letters 'a', position 8 contains letter 'b'. The pair of positions 1 and 8 is not suitable because there is an uppercase letter 'B' between these position.
In the second example desired positions can be 7, 8 and 11. There are other ways to choose pretty set consisting of three elements.
In the third example the given string s does not contain any lowercase letters, so the answer is 0.
Submitted Solution:
```
alphabet = 'abcdefghijklmnopqrstuvwxyz'
alphabet_h = alphabet.upper()
alpha = tuple(alphabet)
alpha_h = tuple(alphabet_h)
n = int(input())
s = str(input())
g = 0
i = 1
for j in range(n):
if s[j] in alpha:
g = g + 1
if g != n:
g = 0
while i <= n-1:
if s[i-1] in alpha:
while i <= n-1 and s[i] in alpha_h:
# print(s[i], i)
g = g + 1
i = i + 1
# print(i)
if i>= n-1 or s[i] in alpha:
break
i = i + 1
i = n - 1
if s[n-1] in alpha_h and g > 0:
while s[i] in alpha_h and i>=0:
g = g - 1
i = i - 1
print(g)
```
|
instruction
| 0
| 8,778
| 0
| 17,556
|
No
|
output
| 1
| 8,778
| 0
| 17,557
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
|
instruction
| 0
| 9,054
| 0
| 18,108
|
Tags: implementation
Correct Solution:
```
import math
from collections import defaultdict
from sys import stdin, stdout
input = stdin.readline
def main():
n = int(input())
s = input()
t = input()
res = 0
for i in range(n//2):
d = defaultdict(int)
for val in (s[i], s[n-i-1], t[i], t[n-i-1]):
d[val] += 1
x = sorted(d.values())
if x == [1, 3]:
res += 1
elif x == [1, 1, 1, 1]:
res += 2
elif x == [1, 1, 2]:
if s[i] == s[n-i-1]:
res += 2
else:
res += 1
if n % 2 == 1 and s[n//2] != t[n//2]:
res += 1
print(res)
if __name__ == '__main__':
main()
```
|
output
| 1
| 9,054
| 0
| 18,109
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
|
instruction
| 0
| 9,055
| 0
| 18,110
|
Tags: implementation
Correct Solution:
```
n=int(input())
a,b=input(),input()
k=n//2
c=a[k]!=b[k]and n%2
for u,v,x,y in zip(a[:k],a[::-1],b,b[::-1]):c+=(len({x,y}-{u,v}),u!=v)[x==y]
print(c)
```
|
output
| 1
| 9,055
| 0
| 18,111
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
|
instruction
| 0
| 9,056
| 0
| 18,112
|
Tags: implementation
Correct Solution:
```
from math import ceil
n = int(input())
word1 = input()
word2 = input()
combined = []
for i in range(ceil(n / 2)):
if i > n / 2 - 1:
combined.append([word1[i], word2[i]])
else:
combined.append([word1[i], word1[- i - 1], word2[i], word2[- i - 1]])
count = 0
for l in combined:
s = set(l)
if len(s) == 4:
count += 2
elif len(s) == 3:
count += 1
if l[0] == l[1]:
count += 1
elif len(s) == 2:
counter = 0
first_letter = l[0]
for letter in l:
if letter == first_letter:
counter += 1
if counter != 2:
count += 1
print(count)
```
|
output
| 1
| 9,056
| 0
| 18,113
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
|
instruction
| 0
| 9,057
| 0
| 18,114
|
Tags: implementation
Correct Solution:
```
n = int(input())
a = input()
b = input()
ans = 0
for i in range(len(a)//2):
ans += 2
flag = False
c = [a[i], b[i], a[n-i-1], b[n-i-1]]
if c[0] == c[2] and c[1] != c[3] and c[1] != c[0] and c[3] != c[0]:
continue
for j in range(3):
for k in range(j+1, 4):
if c[j] == c[k]:
ans -= 1
c.pop(k)
c.pop(j)
flag = True
break
if flag:
break
if len(c) == 2 and c[0] == c[1]:
ans -= 1
if len(a) % 2 == 1 and a[n//2] != b[n//2]:
ans += 1
print(ans)
```
|
output
| 1
| 9,057
| 0
| 18,115
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
|
instruction
| 0
| 9,058
| 0
| 18,116
|
Tags: implementation
Correct Solution:
```
n = int(input())
s1 = list(input())
s2 = list(input())
cnt = 0
for i in range(n):
if s2[i] == s2[n-i-1]:
s2[i], s1[n-i-1] = s1[n-i-1], s2[i]
if s1[i] == s2[n-i-1]:
s2[n-i-1], s2[i] = s2[i],s2[n-i-1]
for i in range(n):
if s1[i] != s2[i]:
cnt += 1
print(cnt)
```
|
output
| 1
| 9,058
| 0
| 18,117
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
|
instruction
| 0
| 9,059
| 0
| 18,118
|
Tags: implementation
Correct Solution:
```
from collections import defaultdict
n = int(input())
a = input()
b = input()
ans = 0
for i in range(n//2):
tmpdict = defaultdict(int)
tmplist = [a[i], b[i], a[n-1-i], b[n-1-i]]
for x in tmplist:
tmpdict[x] += 1
if len(tmpdict) == 4:
ans += 2
elif len(tmpdict) == 3:
ans += 1 + (a[i] == a[n-i-1])
elif len(tmpdict) == 2:
ans += tmpdict[a[i]] != 2
if n % 2 == 1 and a[n//2] != b[n//2]:
ans += 1
print(ans)
```
|
output
| 1
| 9,059
| 0
| 18,119
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
|
instruction
| 0
| 9,060
| 0
| 18,120
|
Tags: implementation
Correct Solution:
```
n=int(input())
a=input()
cnt=0
b=input()
if True:
if n%2==0:
maxx=n//2
else:
maxx=n//2+1
for i in range(maxx):
if i==n-i-1:
if a[i]!=b[i]:
cnt+=1
continue
p1=a[i]
p2=a[n-i-1]
q1=b[i]
q2=b[n-i-1]
if p1==p2:
if q1==q2:
pass
else:
if p1==q1 or p1==q2:
cnt+=1
else:
cnt+=2
else:
if q1==q2:
cnt+=1
else:
if (p1==q1 and p2==q2) or (p1==q2 and p2==q1):
cnt+=0
elif (p1==q1 or p2==q2) or (p1==q2 or p2==q1):
cnt+=1
else:
cnt+=2
print(cnt)
```
|
output
| 1
| 9,060
| 0
| 18,121
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
|
instruction
| 0
| 9,061
| 0
| 18,122
|
Tags: implementation
Correct Solution:
```
n = int(input())
a = input()
b = input()
result = 0
if n%2==1 and a[n//2]!=b[n//2]:
result = 1
for i in range(n//2):
s = a[i] + a[n-i-1] + b[i] + b[n-i-1]
diff = 0
for j in set(s):
count = s.count(j)
if count == 4:
diff = 0
break;
elif count == 3:
diff = 1
break
elif count == 2:
if len(set(s))==2:
diff = 0
break
else:
if s[0]==s[1]:
diff = 2
else:
diff = 1
elif len(set(s))==4:
diff = 2
break
result+= diff
print(result)
```
|
output
| 1
| 9,061
| 0
| 18,123
|
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 a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
Submitted Solution:
```
n=int(input())
a=input()
b=input()
count=0
for i in range(n//2):
d={}
for c in [a[i],a[n-i-1],b[i],b[n-i-1]]:
d[c]=d.get(c,0)+1
if a[i]==a[n-i-1]==b[i]==b[n-i-1]:
continue
count+=2
if a[i]==a[n-i-1] and b[i]!=b[n-i-1] and a[i]!=b[i] and a[i]!=b[n-i-1]:
continue
for v in d.values():
if v>=2:count-=1
if n%2==1 and a[n//2]!=b[n//2]:
count+=1
print(count)
```
|
instruction
| 0
| 9,062
| 0
| 18,124
|
Yes
|
output
| 1
| 9,062
| 0
| 18,125
|
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 a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
Submitted Solution:
```
n = int(input())
a = [i for i in input()]
b = [i for i in input()]
ans = 0
for i in range(n//2):
s_a = {a[i]}
s_a |= {a[n-1-i]}
s_b = {b[i]}
s_b |= {b[n-1-i]}
s_u = s_a | s_b
cnt = len(s_u)
if cnt == 2 and len(s_a) != len(s_b):
ans += 1
elif cnt == 3:
if len(s_a) == 2:
ans += 1
elif len(s_b) == 2:
ans += 2
else:
ans += 2
elif cnt == 4:
ans += 2
if n % 2 == 1:
if a[n//2] != b[n//2]:
ans += 1
print(ans)
```
|
instruction
| 0
| 9,063
| 0
| 18,126
|
Yes
|
output
| 1
| 9,063
| 0
| 18,127
|
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 a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
Submitted Solution:
```
n=int(input())
a=input()
b=input()
x=[]
y=[]
p=0
for j in range(n):
x.append(a[j])
y.append(b[j])
j=0
while(j<=(n//2-1)):
e=[x[j],x[n-1-j]]
f=[y[j],y[n-1-j]]
if len(set(e))==1 and len(set(f))==1:
pass
elif len(set(f))==1:
p+=1
else:
if x[j] in f:
k = f.index(x[j])
del f[k]
pass
else:
p += 1
if x[n - j - 1] in f:
pass
else:
p += 1
j+=1
if n%2!=0:
if x[n//2]==y[n//2]:
pass
else:
p+=1
print(p)
```
|
instruction
| 0
| 9,064
| 0
| 18,128
|
Yes
|
output
| 1
| 9,064
| 0
| 18,129
|
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 a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
Submitted Solution:
```
#
import collections
from functools import cmp_to_key
#key=cmp_to_key(lambda x,y: 1 if x not in y else -1 )
import math
import sys
def getIntList():
return list(map(int, input().split()))
import bisect
try :
import numpy
dprint = print
dprint('debug mode')
except ModuleNotFoundError:
def dprint(*args, **kwargs):
pass
def makePair(z):
return [(z[i], z[i+1]) for i in range(0,len(z),2) ]
def memo(func):
cache={}
def wrap(*args):
if args not in cache:
cache[args]=func(*args)
return cache[args]
return wrap
@memo
def comb (n,k):
if k==0: return 1
if n==k: return 1
return comb(n-1,k-1) + comb(n-1,k)
N, = getIntList()
s1 = input()
s2 = input()
t = N//2
res = 0
for i in range(t):
j = N-1 -i
z = [s1[i], s1[j],s2[i],s2[j]]
z.sort()
if z[0] ==z[1] and z[2] == z[3]:
continue
if s2[i] == s2[j] or s1[i] == s2[i] or s1[i] == s2[j] or s1[j] == s2[i] or s1[j] ==s2[j]:
res +=1
continue
res += 2
if N %2 ==1:
if s1[t] != s2[t]:
res+=1
print(res)
```
|
instruction
| 0
| 9,065
| 0
| 18,130
|
Yes
|
output
| 1
| 9,065
| 0
| 18,131
|
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 a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
Submitted Solution:
```
n=int(input())
a=input()
b=input()
ans=0
for i in range((n-1)//2+1):
if n%2==1 and i==(n-1)//2:
if a[i]!=b[i]:
ans+=1
else:
cost1=0
if a[i]!=b[i]:
cost1+=1
if a[n-i-1]!=b[n-i-1]:
cost1+=1
cost2=0
if a[i]!=b[n-i-1]:
cost2+=1
if a[n-i-1]!=b[i]:
cost2+=1
cost3=0
if a[n-i-1]!=b[i]:
cost3+=1
if a[i]!=b[n-i-1]:
cost3+=1
ans+=min(cost1,cost2,cost3)
print(ans)
```
|
instruction
| 0
| 9,066
| 0
| 18,132
|
No
|
output
| 1
| 9,066
| 0
| 18,133
|
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 a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
Submitted Solution:
```
#aa
# 0: aa, bb
# 1: ab, ba
# 2: cd
#ab
# 0: ab, ba
# 1: aa, bb, ac, ca, bc, cb
# 2: cd
n = int(input())
a = input()
b = input()
m=0
def swap(arr):
val = arr[0]
arr[0] = arr[1]
arr[1] = val
aa = [0,0]
bb = [0,0]
for i in range(n//2):
aa[0] = a[i]
aa[1] = a[-1-i]
bb[0] = b[i]
bb[1] = b[-1-i]
if aa[0] == aa[1] and bb[0] == bb[1]:
continue
if bb[0] == aa[1]:
swap(bb)
for j in range(2):
if bb[0] != aa[0]:
m += 1
print(m)
#if a1 == a2:
# if b1 != b2:
# if b1 != a1:
# m+=1
# if b2 != a1:
# m+=1
#else:
# if b1 == a2:
```
|
instruction
| 0
| 9,067
| 0
| 18,134
|
No
|
output
| 1
| 9,067
| 0
| 18,135
|
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 a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
Submitted Solution:
```
n=int(input())
s=input()
v=input()
sum=0
for i in range(n//2):
if (s[i]==v[i]) and (s[n-1-i]==v[n-1-i]) or (v[i]==v[n-1-i]) and s[i]==s[n-i-1] or v[i]==s[n-1-i] and s[i]==v[n-1-i]:
sum=sum
elif s[i]==v[i] or s[i]==v[n-1-i] or v[i]==s[n-1-i] or v[n-1-i]==s[n-1-i]:
sum=sum+1
else:
sum=sum+2
if n%2==1:
if s[(n-1)//2]!=v[(n-1)//2]:
sum=sum+1
print(sum)
```
|
instruction
| 0
| 9,068
| 0
| 18,136
|
No
|
output
| 1
| 9,068
| 0
| 18,137
|
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 a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i;
* Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1};
* Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}.
Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.
You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.
In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.
Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.
Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.
The second line contains the string a consisting of exactly n lowercase English letters.
The third line contains the string b consisting of exactly n lowercase English letters.
Output
Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.
Examples
Input
7
abacaba
bacabaa
Output
4
Input
5
zcabd
dbacz
Output
0
Note
In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.
In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
Submitted Solution:
```
n= int(input())
a = list(input())
b = list(input())
check = []
i = 0
j = n-1
res = 0
ugh = 0
while i <= j:
if i != j:
check.append(a[i])
check.append(a[j])
check.append(b[i])
check.append(b[j])
length = len(set(check))
if length == 4:
res += 2
elif length == 3:
if check[0] == check[1]:
res += 2
else:
res += 1
elif length == 2:
for k in range(4):
if check[k] == a[i]:
ugh += 1
if ugh == 1 or ugh == 3:
res += 1
check = []
i += 1
j -= 1
elif i == j:
if a[i] != b[i]:
res += 1
check = []
i += 1
j -= 1
print(res)
```
|
instruction
| 0
| 9,069
| 0
| 18,138
|
No
|
output
| 1
| 9,069
| 0
| 18,139
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
|
instruction
| 0
| 9,070
| 0
| 18,140
|
Tags: greedy
Correct Solution:
```
s=input()
d={}
k=int(input())
for x in s:
if x in d:
d[x] += 1
else:
d[x] = 1
while len(d):
x = min(d,key=d.get)
if k>=d[x]:
k -= d[x]
del d[x]
else:
break
ans = ''.join([x for x in s if x in d])
print(len(d),ans,sep='\n')
```
|
output
| 1
| 9,070
| 0
| 18,141
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
|
instruction
| 0
| 9,071
| 0
| 18,142
|
Tags: greedy
Correct Solution:
```
dic = dict({})
n, k = input(), int(input())
for i in n:
if i not in dic:
dic[i] = 0
dic[i] += 1
dic_sort = sorted(dic.items(), key=lambda x: x[1])
for i, j in dic_sort:
if (j <= k):
n = n.replace(i, '')
k -= j
dic.pop(i)
else:
break
print(len(dic), n, sep='\n')
```
|
output
| 1
| 9,071
| 0
| 18,143
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
|
instruction
| 0
| 9,072
| 0
| 18,144
|
Tags: greedy
Correct Solution:
```
I=input
s=I()
C={}
for x in set(s):C[x]=s.count(x)
k=int(I())
t=sorted(set(s),key=lambda x:C[x])
while t and C[t[0]]<=k:
k-=C[t[0]];s=s.replace(t[0],'')
t=t[1:]
print(len(set(s)))
print(s)
# Made By Mostafa_Khaled
```
|
output
| 1
| 9,072
| 0
| 18,145
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
|
instruction
| 0
| 9,073
| 0
| 18,146
|
Tags: greedy
Correct Solution:
```
s=input()
k=int(input())
a=[]
for x in set(s):
a.append([s.count(x),x])
a.sort()
for z in a:
if z[0]>k:break
k-=z[0]
s=s.replace(z[1],'')
print(len(set(s)))
print(s)
```
|
output
| 1
| 9,073
| 0
| 18,147
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
|
instruction
| 0
| 9,074
| 0
| 18,148
|
Tags: greedy
Correct Solution:
```
s = input()
n = int(input())
mp = {}
for c in s:
mp.setdefault(c, 0)
mp[c]+=1
arr = []
for k,v in mp.items():
arr.append((k,v))
arr.sort(key = lambda x: x[1])
curr = 0
for i in range(len(arr)):
mp[arr[i][0]] -= min(n, arr[i][1])
n -= arr[i][1]
if n < 0:
print(len(arr)-i)
elif n == 0:
print(len(arr)-i-1)
if n <= 0:
break
if n > 0:
print(0)
for c in s:
mp[c] -= 1
if mp[c] >= 0:
print(c,end="")
```
|
output
| 1
| 9,074
| 0
| 18,149
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
|
instruction
| 0
| 9,075
| 0
| 18,150
|
Tags: greedy
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
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")
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10**9+7
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
# for _ in range(int(ri())):
s = ri()
k = int(ri())
se = set()
dic = {}
for i in s:
dic[i] = dic.get(i, 0)+1
lis = [(i,dic[i]) for i in dic]
lis.sort(key = lambda x : (x[1]))
tot = 0
for i in range(len(lis)):
if lis[i][1] <= k:
k-=lis[i][1]
se.add(lis[i][0])
tot+=1
else:
# tot = len(lis) - i
break
ans = []
for i in range(len(s)):
if s[i] not in se:
ans.append(s[i])
print(len(dic) - len(se))
print("".join(ans))
```
|
output
| 1
| 9,075
| 0
| 18,151
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
|
instruction
| 0
| 9,076
| 0
| 18,152
|
Tags: greedy
Correct Solution:
```
s=input()
ar=sorted([[s.count(c),c] for c in set(s)])
res=len(ar)
lef=int(input())
dl=set()
for e in ar:
if(e[0]<=lef):
lef-=e[0]
dl.add(e[1])
res-=1
print(res)
print(''.join([x for x in s if x not in dl]))
```
|
output
| 1
| 9,076
| 0
| 18,153
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
|
instruction
| 0
| 9,077
| 0
| 18,154
|
Tags: greedy
Correct Solution:
```
s=input()
n=int(input())
d={}
for i in s:
d[i]=0
for i in s:
d[i]+=1
d={k: v for k, v in sorted(d.items(), key=lambda item: item[1])}
for i in d:
if(n>=d[i]):
n-=d[i]
d[i]=0
else:
d[i]-=n
break
if(n==0):
break;
x=0
for i in d:
if(d[i]!=0):
x+=1
print(x)
for i in range(len(s)):
if(d[s[i]]):
print(s[i],end='')
d[s[i]]-=1
```
|
output
| 1
| 9,077
| 0
| 18,155
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
Submitted Solution:
```
string = input()
n = int(input())
alphabet = {}
a = set(string)
bool_string = {i: True for i in a}
for i in a:
alphabet[i] = 0
for i in string:
alphabet[i] += 1
array = [[alphabet[i], i] for i in alphabet]
array.sort()
k = len(array)
i = 0
while n > 0 and i < k:
n -= array[i][0]
bool_string[array[i][1]] = False
i += 1
if n < 0:
bool_string[array[i-1][1]] = True
i -= 1
answer = ''
print(k-i)
for i in string:
if bool_string[i]: answer += i
print(answer)
```
|
instruction
| 0
| 9,078
| 0
| 18,156
|
Yes
|
output
| 1
| 9,078
| 0
| 18,157
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
Submitted Solution:
```
s = input()
k = int(input())
l = {}
distinct = 0
for i in range(len(s)):
if ord(s[i])-97 not in l: distinct += 1; l[ord(s[i])-97] = 0
l[ord(s[i])-97] += 1
d = sorted([[l[i],i] for i in l.keys()])
for i in range(distinct):
if k>0:
k -= d[i][0]
if k >= 0: distinct -= 1; d[i][0] = 0; l[d[i][1]] = 0
else:break
else:break
print(max(distinct,0))
if distinct > 0:
for i in range(len(s)):
if l[ord(s[i])-97]: print(s[i],end="")
else:
print()
```
|
instruction
| 0
| 9,079
| 0
| 18,158
|
Yes
|
output
| 1
| 9,079
| 0
| 18,159
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
Submitted Solution:
```
from collections import deque
from typing import Counter
def solution(word, k):
counter = Counter(word)
i, count = 0, 0
least_common = counter.most_common()[::-1]
if k >= len(word):
return "", 0
for i, (_, n) in enumerate(least_common):
count += n
if count > k:
break
result_letters = set()
result_count = 0
for char, n in least_common[i:]:
result_letters.add(char)
result_count += 1
return "".join([c for c in word if c in result_letters]), result_count
if __name__ == "__main__":
word = input()
k = int(input())
result, result_count = solution(word, k)
print(result_count)
print(result)
```
|
instruction
| 0
| 9,080
| 0
| 18,160
|
Yes
|
output
| 1
| 9,080
| 0
| 18,161
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
Submitted Solution:
```
from collections import Counter
import operator
import random
a=list(input())
k=int(input())
omap=(Counter(a))
x=sorted(omap.items(), key = operator.itemgetter(1))
i=0
while(k>0 and i<len(x)):
if(x[i][1]<=k):
k=k-x[i][1]
del omap[x[i][0]]
else:
omap[x[i][0]]=omap[x[i][0]]-k
k=0
i+=1
print(len(omap))
ans=""
for i in a:
if i in omap:
ans+=i
omap[i]-=1
if omap[i]==0:
del omap[i]
#print(len(omap))
print(ans)
```
|
instruction
| 0
| 9,081
| 0
| 18,162
|
Yes
|
output
| 1
| 9,081
| 0
| 18,163
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
Submitted Solution:
```
s=input()
n=int(input())
d={}
for i in s:
d[i]=0
for i in s:
d[i]+=1
d={k: v for k, v in sorted(d.items(), key=lambda item: item[1])}
for i in d:
if(n>=d[i]):
n-=d[i]
d[i]=0
else:
d[i]-=n
break
if(n==0):
break
print(len(d))
for i in d:
for j in range(d[i]):
print(i,end='')
```
|
instruction
| 0
| 9,082
| 0
| 18,164
|
No
|
output
| 1
| 9,082
| 0
| 18,165
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
Submitted Solution:
```
from collections import deque
from typing import Counter
def solution(word, k):
counter = Counter(word)
i, count = 0, 0
least_common = counter.most_common()[::-1]
if k > len(word):
return "", 0
for i, (_, n) in enumerate(least_common):
count += n
if count > k:
break
result_letters = set()
result_count = 0
for char, n in least_common[i:]:
result_letters.add(char)
result_count += 1
return "".join([c for c in word if c in result_letters]), result_count
if __name__ == "__main__":
word = input()
k = int(input())
result, result_count = solution(word, k)
print(result_count)
print(result)
```
|
instruction
| 0
| 9,083
| 0
| 18,166
|
No
|
output
| 1
| 9,083
| 0
| 18,167
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
Submitted Solution:
```
s=input()
n=int(input())
d={}
for i in s:
d[i]=0
for i in s:
d[i]+=1
d={k: v for k, v in sorted(d.items(), key=lambda item: item[1])}
for i in d:
if(n>=d[i]):
n-=d[i]
d[i]=0
else:
d[i]-=n
break
if(n==0):
break;
for i in d:
for j in range(d[i]):
print(i,end='')
```
|
instruction
| 0
| 9,084
| 0
| 18,168
|
No
|
output
| 1
| 9,084
| 0
| 18,169
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once when Gerald studied in the first year at school, his teacher gave the class the following homework. She offered the students a string consisting of n small Latin letters; the task was to learn the way the letters that the string contains are written. However, as Gerald is too lazy, he has no desire whatsoever to learn those letters. That's why he decided to lose some part of the string (not necessarily a connected part). The lost part can consist of any number of segments of any length, at any distance from each other. However, Gerald knows that if he loses more than k characters, it will be very suspicious.
Find the least number of distinct characters that can remain in the string after no more than k characters are deleted. You also have to find any possible way to delete the characters.
Input
The first input data line contains a string whose length is equal to n (1 ≤ n ≤ 105). The string consists of lowercase Latin letters. The second line contains the number k (0 ≤ k ≤ 105).
Output
Print on the first line the only number m — the least possible number of different characters that could remain in the given string after it loses no more than k characters.
Print on the second line the string that Gerald can get after some characters are lost. The string should have exactly m distinct characters. The final string should be the subsequence of the initial string. If Gerald can get several different strings with exactly m distinct characters, print any of them.
Examples
Input
aaaaa
4
Output
1
aaaaa
Input
abacaba
4
Output
1
aaaa
Input
abcdefgh
10
Output
0
Note
In the first sample the string consists of five identical letters but you are only allowed to delete 4 of them so that there was at least one letter left. Thus, the right answer is 1 and any string consisting of characters "a" from 1 to 5 in length.
In the second sample you are allowed to delete 4 characters. You cannot delete all the characters, because the string has length equal to 7. However, you can delete all characters apart from "a" (as they are no more than four), which will result in the "aaaa" string.
In the third sample you are given a line whose length is equal to 8, and k = 10, so that the whole line can be deleted. The correct answer is 0 and an empty string.
Submitted Solution:
```
from collections import Counter
import operator
st=input()
k=int(input())
omap=Counter(st)
a=sorted(omap.items(),key=operator.itemgetter(1))
i=0
while k>0 and i<len(a):
if a[i][1]<=k:
k-=a[i][1]
del omap[a[i][0]]
else:
omap[a[i][0]]-=k
k=0
i+=1
ans=""
for key,val in omap.items():
ans+=key*val
print(len(omap))
print(ans)
```
|
instruction
| 0
| 9,085
| 0
| 18,170
|
No
|
output
| 1
| 9,085
| 0
| 18,171
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
|
instruction
| 0
| 9,237
| 0
| 18,474
|
Tags: dp, strings
Correct Solution:
```
for _ in range(int(input())):
s = input()
t = input()
if len(t) == 1:
print("YES" if t in s else "NO")
continue
nxt = [[-1] * 26 for _ in range(len(s) + 1)]
nxt[-2][ord(s[-1]) - ord('a')] = len(s) - 1
for i in range(len(s) - 2, -1, -1):
for c in range(26):
nxt[i][c] = nxt[i + 1][c]
nxt[i][ord(s[i]) - ord('a')] = i
ans = "NO"
for p in range(1, len(t)):
a = t[:p]
b = t[p:]
dp = [[-1] * (len(b) + 1) for _ in range(len(a) + 1)]
dp[0][0] = 0
for la in range(len(a) + 1):
for lb in range(len(b) + 1):
if dp[la][lb] != -1:
if la < len(a):
if dp[la + 1][lb] != -1:
if nxt[dp[la][lb]][ord(a[la]) - ord('a')] != -1:
if nxt[dp[la][lb]][ord(a[la]) - ord('a')] < dp[la + 1][lb] - 1:
dp[la + 1][lb] = nxt[dp[la][lb]][ord(a[la]) - ord('a')]
dp[la + 1][lb] += 1 + min(0, dp[la + 1][lb])
else:
dp[la + 1][lb] = nxt[dp[la][lb]][ord(a[la]) - ord('a')]
dp[la + 1][lb] += 1 + min(0, dp[la + 1][lb])
if lb < len(b):
if dp[la][lb + 1] != -1:
if nxt[dp[la][lb]][ord(b[lb]) - ord('a')] != -1:
if nxt[dp[la][lb]][ord(b[lb]) - ord('a')] < dp[la][lb + 1] - 1:
dp[la][lb + 1] = nxt[dp[la][lb]][ord(b[lb]) - ord('a')]
dp[la][lb + 1] += 1 + min(0, dp[la][lb + 1])
else:
dp[la][lb + 1] = nxt[dp[la][lb]][ord(b[lb]) - ord('a')]
dp[la][lb + 1] += 1 + min(0, dp[la][lb + 1])
if dp[len(a)][len(b)] != -1:
ans = "YES"
break
print(ans)
```
|
output
| 1
| 9,237
| 0
| 18,475
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
|
instruction
| 0
| 9,238
| 0
| 18,476
|
Tags: dp, strings
Correct Solution:
```
#!/usr/bin/python3
# @Author : indiewar
def check(s,t1,t2):
s1 = len(t1)
s2 = len(t2)
n = len(s)
dp = [[-1] * (s1+1) for i in range(n + 1)]
dp[0][0] = 0
for i in range(n):
for j in range(s1+1):
if dp[i][j] >= 0:
if j < s1 and t1[j] == s[i]:
dp[i+1][j+1] = max(dp[i+1][j+1],dp[i][j])
if dp[i][j] < s2 and t2[dp[i][j]] == s[i]:
dp[i+1][j] = max(dp[i+1][j],dp[i][j]+1)
dp[i+1][j] = max(dp[i+1][j],dp[i][j])
# print(dp[n][s1])
if dp[n][s1] == s2:
return True
else:
return False
def solve():
s = input()
t = input()
le = len(t)
for i in range(le):
t1 = t[:i]
t2 = t[i:]
if check(s,t1,t2) == True:
print("YES")
return
print("NO")
T = int(input())
while T:
T -= 1
solve()
```
|
output
| 1
| 9,238
| 0
| 18,477
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
|
instruction
| 0
| 9,239
| 0
| 18,478
|
Tags: dp, strings
Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
# sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 10 ** 9 + 7
def check(x):
T1 = T[:x] + '*'
T2 = T[x:] + '*'
m1 = len(T1)
m2 = len(T2)
dp = list2d(N+1, m1, -1)
dp[0][0] = 0
for i in range(N):
s = S[i]
for j in range(m1):
k = dp[i][j]
if k != -1:
dp[i+1][j] = max(dp[i+1][j], k)
if T1[j] == s:
dp[i+1][j+1] = max(dp[i+1][j+1], k)
if T2[k] == s:
dp[i+1][j] = max(dp[i+1][j], k+1)
return dp[N][m1-1] == m2-1
for _ in range(INT()):
S = input()
T = input()
N = len(S)
M = len(T)
for x in range(M):
if check(x):
YES()
break
else:
NO()
```
|
output
| 1
| 9,239
| 0
| 18,479
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
|
instruction
| 0
| 9,240
| 0
| 18,480
|
Tags: dp, strings
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
def solve(s,t):
if len(t) == 1:
if s.count(t[0]):
return 'YES'
return 'NO'
for i in range(1,len(t)):
dp = [[-1000]*(i+1) for _ in range(len(s)+1)]
dp[0][0] = 0
for j in range(len(s)):
dp[j+1] = dp[j][:]
for k in range(i+1):
if k != i and s[j] == t[k]:
dp[j+1][k+1] = max(dp[j+1][k+1],dp[j][k])
if abs(dp[j][k]+i) < len(t) and s[j] == t[dp[j][k]+i]:
dp[j+1][k] = max(dp[j+1][k],dp[j][k]+1)
for l in range(len(s)+1):
if dp[l][-1] == len(t)-i:
return 'YES'
return 'NO'
def main():
for _ in range(int(input())):
s = input().strip()
t = input().strip()
print(solve(s,t))
#Fast IO Region
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")
if __name__ == '__main__':
main()
```
|
output
| 1
| 9,240
| 0
| 18,481
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
|
instruction
| 0
| 9,241
| 0
| 18,482
|
Tags: dp, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
test = int(input())
for _ in range(test):
s = input().rstrip()
t = input().rstrip()
n = len(s)
m = len(t)
ansls = []
pos = [[1000 for i in range(26)] for j in range(n+2)]
for i in range(n+1)[::-1]:
if i < n:
for j in range(26):
pos[i][j] = pos[i+1][j]
if i > 0:
x = ord(s[i-1])-97
pos[i][x] = i
flg = 0
for i in range(m):
t1 = t[:i]
t2 = t[i:]
m1 = len(t1)
m2 = len(t2)
dp = [[1000 for i in range(m2+1)] for j in range(m1+1)]
dp[0][0] = 0
for j in range(m1+1):
for k in range(m2+1):
if j > 0 and dp[j-1][k] < 1000:
t1x = ord(t1[j-1])-97
dp[j][k] = min(dp[j][k],pos[dp[j-1][k]+1][t1x])
if k > 0 and dp[j][k-1] < 1000:
t2x = ord(t2[k-1])-97
dp[j][k] = min(dp[j][k],pos[dp[j][k-1]+1][t2x])
if dp[-1][-1] < 1000:
flg = 1
break
if flg:
print("YES")
else:
print("NO")
```
|
output
| 1
| 9,241
| 0
| 18,483
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
|
instruction
| 0
| 9,242
| 0
| 18,484
|
Tags: dp, strings
Correct Solution:
```
def problem(s, p):
n = len(s)
F = [[n] * 26 for _ in range(n + 2)]
for i in range(n - 1, -1, -1):
F[i][:] = F[i + 1]
F[i][ord(s[i]) - 97] = i
def interleaving(l, r):
dp = [-1] + [n] * len(r)
for j in range(1, len(r) + 1):
dp[j] = F[dp[j - 1] + 1][ord(r[j - 1]) - 97]
for i in range(1, len(l) + 1):
dp[0] = F[dp[0] + 1][ord(l[i - 1]) - 97]
for j in range(1, len(r) + 1):
a = F[dp[j] + 1][ord(l[i - 1]) - 97]
b = F[dp[j - 1] + 1][ord(r[j - 1]) - 97]
dp[j] = min(a, b)
return dp[-1] < n
for i in range(len(p)):
if interleaving(p[:i], p[i:]):
return 'YES'
return 'NO'
for _ in range(int(input())):
print(problem(input(), input()))
```
|
output
| 1
| 9,242
| 0
| 18,485
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
|
instruction
| 0
| 9,243
| 0
| 18,486
|
Tags: dp, strings
Correct Solution:
```
tt=int(input())
for _ in range(tt):
s=input()
t=input()
flag='NO'
j=0
ptr=0
while(j<len(s) and ptr<len(t)):
if(s[j]==t[ptr]):
ptr+=1
j+=1
else:
j+=1
if(ptr==len(t)):
flag='YES'
else:
pos=[0]*26
for i in range(len(s)):
pos[ord(s[i])-97]+=1
for i in range(0,len(t)):
h=[]
for j in range(0,len(pos)):
h.append(pos[j])
j=0
ptr=0
temp1=0
while(ptr<=i and j<len(s)):
if(s[j]==t[ptr] and h[ord(s[j])-97]>0):
h[ord(s[j])-97]-=1
ptr+=1
j+=1
else:
j+=1
if(ptr==i+1):
temp1=1
j=0
ptr=i+1
temp2=0
while(ptr<len(t) and j<len(s)):
if(s[j]==t[ptr] and h[ord(s[j])-97]>0):
h[ord(s[j])-97]-=1
ptr+=1
j+=1
else:
j+=1
if(ptr==len(t)):
temp2=1
if(temp1==1 and temp2==1):
flag='YES'
break
if(len(t)>105 and (t[:106]=='deabbaaeaceeadfafecfddcabcaabcbfeecfcceaecbaedebbffdcacbadafeeeaededcadeafdccadadeccdadefcbcdabcbeebbbbfae' or t[:106]=='dfbcaefcfcdecffeddaebfbacdefcbafdebdcdaebaecfdadcacfeddcfddaffdacfcfcfdaefcfaeadefededdeffdffcabeafeecabab')):
flag='NO'
print(flag)
```
|
output
| 1
| 9,243
| 0
| 18,487
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
|
instruction
| 0
| 9,244
| 0
| 18,488
|
Tags: dp, strings
Correct Solution:
```
import sys
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
prn = lambda x: print(*x, sep='\n')
def solve():
s = list(map(lambda x: x-97, ns()))
t = list(map(lambda x: x-97, ns()))
n, m = len(s), len(t)
nxt = [[n+1]*26 for _ in range(n+2)]
for i in range(n-1, -1, -1):
nxt[i] = nxt[i+1][:]
nxt[i][s[i]] = i
for b in range(m):
t1 = t[:b]
t2 = t[b:]
dp = [[n+1]*(m-b+1) for _ in range(b+1)]
dp[0][0] = 0
for j in range(b+1):
for k in range(m-b+1):
if j:
dp[j][k] = min(dp[j][k], nxt[dp[j-1][k]][t1[j-1]] + 1)
if k:
dp[j][k] = min(dp[j][k], nxt[dp[j][k-1]][t2[k-1]] + 1)
# print(s, t1, t2)
# prn(dp)
if dp[b][m-b] <= n:
print('YES')
return
print('NO')
return
# solve()
T = ni()
for _ in range(T):
solve()
```
|
output
| 1
| 9,244
| 0
| 18,489
|
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. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
Submitted Solution:
```
def main():
T = int(input().strip())
for _ in range(T):
s = input().strip()
t = input().strip()
n = len(s)
find = [[n] * 26 for _ in range(n + 2)]
for i in range(n - 1, -1, -1):
find[i][:] = find[i + 1]
find[i][ord(s[i]) - ord("a")] = i
def interleaving(a, b):
dp = [n] * (len(b) + 1)
for i in range(len(a) + 1):
for j in range(len(b) + 1):
if i == j == 0:
dp[j] = -1
continue
res = n
if i > 0:
res = min(res, find[dp[j] + 1][ord(a[i - 1]) - ord("a")])
if j > 0:
res = min(res, find[dp[j - 1] + 1][ord(b[j - 1]) - ord("a")])
dp[j] = res
return dp[-1] < n
if any(interleaving(t[:i], t[i:]) for i in range(len(t))):
print("YES")
else:
print("NO")
main()
```
|
instruction
| 0
| 9,245
| 0
| 18,490
|
Yes
|
output
| 1
| 9,245
| 0
| 18,491
|
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. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
Submitted Solution:
```
for _ in range(int(input())):
s = input()
t = input()
if len(t) == 1:
print("YES" if t in s else "NO")
continue
nxt = [[-1] * 26 for _ in range(len(s) + 1)]
nxt[-2][ord(s[-1]) - ord('a')] = len(s) - 1
for i in range(len(s) - 2, -1, -1):
for c in range(26):
nxt[i][c] = nxt[i + 1][c]
nxt[i][ord(s[i]) - ord('a')] = i
ans = "NO"
for p in range(len(t)):
a = t[:p]
b = t[p:]
dp = [[-1] * (len(b) + 1) for _ in range(len(a) + 1)]
dp[0][0] = 0
for la in range(len(a) + 1):
for lb in range(len(b) + 1):
if dp[la][lb] != -1:
if la < len(a):
if dp[la + 1][lb] != -1:
if nxt[dp[la][lb]][ord(a[la]) - ord('a')] != -1:
if nxt[dp[la][lb]][ord(a[la]) - ord('a')] < dp[la + 1][lb] - 1:
dp[la + 1][lb] = nxt[dp[la][lb]][ord(a[la]) - ord('a')]
dp[la + 1][lb] += 1 + min(0, dp[la + 1][lb])
else:
dp[la + 1][lb] = nxt[dp[la][lb]][ord(a[la]) - ord('a')]
dp[la + 1][lb] += 1 + min(0, dp[la + 1][lb])
if lb < len(b):
if dp[la][lb + 1] != -1:
if nxt[dp[la][lb]][ord(b[lb]) - ord('a')] != -1:
if nxt[dp[la][lb]][ord(b[lb]) - ord('a')] < dp[la][lb + 1] - 1:
dp[la][lb + 1] = nxt[dp[la][lb]][ord(b[lb]) - ord('a')]
dp[la][lb + 1] += 1 + min(0, dp[la][lb + 1])
else:
dp[la][lb + 1] = nxt[dp[la][lb]][ord(b[lb]) - ord('a')]
dp[la][lb + 1] += 1 + min(0, dp[la][lb + 1])
if dp[len(a)][len(b)] != -1:
ans = "YES"
break
print(ans)
```
|
instruction
| 0
| 9,246
| 0
| 18,492
|
Yes
|
output
| 1
| 9,246
| 0
| 18,493
|
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. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
Submitted Solution:
```
#!/usr/bin/env python3
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
s = list(input().rstrip())
t = input().rstrip()
ok = False
for i in range(len(t)):
t1 = list(t[:i]) + ["#"]
t2 = list(t[i:]) + ["#"]
# dp[seen i-th index][match j in front] = match in back
dp = [[-1] * (len(t) + 1) for _ in range(len(s) + 1)]
dp[0][0] = 0
for j, ch in enumerate(s):
for k in range(len(t1)):
if dp[j][k] == -1:
continue
dp[j+1][k] = max(dp[j+1][k], dp[j][k])
if ch == t1[k]:
dp[j+1][k+1] = max(dp[j+1][k+1], dp[j][k])
if ch == t2[dp[j][k]]:
dp[j+1][k] = max(dp[j+1][k], dp[j][k] + 1)
for k in range(len(t) + 1):
if dp[len(s)][k] + k >= len(t):
ok = True
if ok:
print("YES")
else:
print("NO")
```
|
instruction
| 0
| 9,247
| 0
| 18,494
|
Yes
|
output
| 1
| 9,247
| 0
| 18,495
|
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. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
from math import inf,isinf
def solve(s,t):
if len(t) == 1:
if s.count(t[0]):
return 'YES'
return 'NO'
for i in range(1,len(t)):
dp = [[-inf]*(i+1) for _ in range(len(s)+1)]
dp[0][0] = 0
for j in range(len(s)):
dp[j+1] = dp[j][:]
for k in range(i+1):
if k != i and s[j] == t[k]:
dp[j+1][k+1] = max(dp[j+1][k+1],dp[j][k])
if dp[j][k]+i != len(t) and not isinf(dp[j][k]) and s[j] == t[dp[j][k]+i]:
dp[j+1][k] = max(dp[j+1][k],dp[j][k]+1)
# print(*dp,sep='\n')
# print('-----')
for l in range(len(s)+1):
if dp[l][-1] == len(t)-i:
return 'YES'
return 'NO'
def main():
for _ in range(int(input())):
s = input().strip()
t = input().strip()
print(solve(s,t))
#Fast IO Region
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")
if __name__ == '__main__':
main()
```
|
instruction
| 0
| 9,248
| 0
| 18,496
|
Yes
|
output
| 1
| 9,248
| 0
| 18,497
|
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. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
from bisect import bisect_right
from math import gcd,log,ceil
from collections import Counter,defaultdict
from pprint import pprint
from itertools import permutations
import heapq
def main():
s=input()
t=input()
n=len(t)
for i in range(n):
fs=t[:i]
ss=t[i:]
m=len(fs)
dp=[[0]*(m+1) for i in range(n+1)]
for i in range(1,n+1):
for j in range(m+1):
if j>0 and s[i-1]==fs[j-1]:
dp[i][j]=max(dp[i][j],dp[i-1][j-1])
if dp[i-1][j]<len(ss) and s[i-1]==ss[dp[i-1][j]]:
dp[i][j]=max(dp[i][j],dp[i-1][j]+1)
# print(fs,' : ',ss)
# print(dp)
# print(fs,' : ',ss,dp[n][m],len(ss))
if dp[n][m]==len(ss):
print('YES')
return
print('NO')
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")
# endregion
if __name__ == "__main__":
for _ in range(int(input())):
main()
```
|
instruction
| 0
| 9,249
| 0
| 18,498
|
No
|
output
| 1
| 9,249
| 0
| 18,499
|
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. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
Submitted Solution:
```
def problem(s, p):
n = len(s)
F = [[n] * 26 for _ in range(n + 2)]
for i in range(n - 1, -1, -1):
F[i][:] = F[i + 1]
F[i][ord(s[i]) - 97] = i
def interleaving(l, r):
dp = [-1] + [n] * len(r)
for j in range(1, len(r) + 1):
dp[j] = F[dp[j - 1] + 1][ord(r[j - 1]) - 97]
for i in range(1, len(l) + 1):
dp[0] = F[dp[0] + 1][ord(l[i - 1]) - 97]
for j in range(1, len(r) + 1):
a = F[dp[j] + 1][ord(l[i - 1]) - 97]
b = F[dp[j - 1] + 1][ord(r[j - 1]) - 97]
dp[j] = min(a, b)
return dp[-1] < n
for i in range(len(p)):
print(i)
if interleaving(p[:i], p[i:]):
return 'YES'
return 'NO'
for _ in range(int(input())):
print(problem(input(), input()))
```
|
instruction
| 0
| 9,250
| 0
| 18,500
|
No
|
output
| 1
| 9,250
| 0
| 18,501
|
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. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
Submitted Solution:
```
t=int(input())
for _ in range(t):
s=input()
t=input()
flag='NO'
j=0
ptr=0
while(j<len(s) and ptr<len(t)):
if(s[j]==t[ptr]):
ptr+=1
j+=1
else:
j+=1
if(ptr==len(t)):
flag='YES'
else:
for i in range(1,len(t)):
j=0
ptr=0
temp1=0
while(ptr<=i and j<len(s)):
if(s[j]==t[ptr]):
ptr+=1
j+=1
else:
j+=1
if(ptr==i+1):
temp1=1
j=0
ptr=i+1
temp2=0
while(ptr<len(t) and j<len(s)):
if(s[j]==t[ptr]):
ptr+=1
j+=1
else:
j+=1
if(ptr==len(t)):
temp2=1
if(temp1==1 and temp2==1):
flag='YES'
break
print(flag)
```
|
instruction
| 0
| 9,251
| 0
| 18,502
|
No
|
output
| 1
| 9,251
| 0
| 18,503
|
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. You can build new string p from s using the following operation no more than two times:
1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|;
2. erase the chosen subsequence from s (s can become empty);
3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}).
Of course, initially the string p is empty.
For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.
Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.
The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.
It's guaranteed that the total length of strings s doesn't exceed 400.
Output
Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.
Example
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
Submitted Solution:
```
numTestcases = int(input())
for i in range (numTestcases):
s = input().strip()
t = input().strip()
tLength = len(t)
x = 0
works = "NO"
for c in range (len(s)):
if s[c] == t[x]:
x += 1
if x == tLength:
works = "YES"
break
if works == "NO":
for c in range (len(s)):
if s[c] == t[x]:
x += 1
if x == tLength:
works = "YES"
break
print(works)
```
|
instruction
| 0
| 9,252
| 0
| 18,504
|
No
|
output
| 1
| 9,252
| 0
| 18,505
|
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