message stringlengths 2 23.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After learning some fancy algorithms about palindromes, Chouti found palindromes very interesting, so he wants to challenge you with this problem.
Chouti has got two strings A and B. Since he likes [palindromes](https://en.wikipedia.org/wiki/Palindrome), he would like to pick a as some non-empty palindromic substring of A and b as some non-empty palindromic substring of B. Concatenating them, he will get string ab.
Chouti thinks strings he could get this way are interesting, so he wants to know how many different strings he can get.
Input
The first line contains a single string A (1 ≤ |A| ≤ 2 ⋅ 10^5).
The second line contains a single string B (1 ≤ |B| ≤ 2 ⋅ 10^5).
Strings A and B contain only lowercase English letters.
Output
The first and only line should contain a single integer — the number of possible strings.
Examples
Input
aa
aba
Output
6
Input
aaba
abaa
Output
15
Note
In the first example, attainable strings are
* "a" + "a" = "aa",
* "aa" + "a" = "aaa",
* "aa" + "aba" = "aaaba",
* "aa" + "b" = "aab",
* "a" + "aba" = "aaba",
* "a" + "b" = "ab".
In the second example, attainable strings are "aa", "aaa", "aaaa", "aaaba", "aab", "aaba", "ab", "abaa", "abaaa", "abaaba", "abab", "ba", "baa", "baba", "bb".
Notice that though "a"+"aa"="aa"+"a"="aaa", "aaa" will only be counted once.
Submitted Solution:
```
s = input()
s2 = input()
a = s.split()
b = s2.split()
for i in range(len(s)):
for j in range(i + 1, len(s), 1):
stroka = s[i:j + 1]
if stroka[:] == stroka[::-1]:
a.append(stroka)
for i in range(len(s2)):
for j in range(i + 1, len(s2), 1):
stroka = s2[i:j + 1]
if stroka[:] == stroka[::-1]:
b.append(stroka)
r = set()
for i in range(len(a)):
for j in range(len(b)):
r.add(a[i] + b[j])
print(len(r))
``` | instruction | 0 | 18,588 | 0 | 37,176 |
No | output | 1 | 18,588 | 0 | 37,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After learning some fancy algorithms about palindromes, Chouti found palindromes very interesting, so he wants to challenge you with this problem.
Chouti has got two strings A and B. Since he likes [palindromes](https://en.wikipedia.org/wiki/Palindrome), he would like to pick a as some non-empty palindromic substring of A and b as some non-empty palindromic substring of B. Concatenating them, he will get string ab.
Chouti thinks strings he could get this way are interesting, so he wants to know how many different strings he can get.
Input
The first line contains a single string A (1 ≤ |A| ≤ 2 ⋅ 10^5).
The second line contains a single string B (1 ≤ |B| ≤ 2 ⋅ 10^5).
Strings A and B contain only lowercase English letters.
Output
The first and only line should contain a single integer — the number of possible strings.
Examples
Input
aa
aba
Output
6
Input
aaba
abaa
Output
15
Note
In the first example, attainable strings are
* "a" + "a" = "aa",
* "aa" + "a" = "aaa",
* "aa" + "aba" = "aaaba",
* "aa" + "b" = "aab",
* "a" + "aba" = "aaba",
* "a" + "b" = "ab".
In the second example, attainable strings are "aa", "aaa", "aaaa", "aaaba", "aab", "aaba", "ab", "abaa", "abaaa", "abaaba", "abab", "ba", "baa", "baba", "bb".
Notice that though "a"+"aa"="aa"+"a"="aaa", "aaa" will only be counted once.
Submitted Solution:
```
class Solution:
def countSubstrings(self, s):
l=[]
counter = 0
for i in range(len(s)):
for j in range(i+1,len(s)+1):
temp = s[i:j]
if temp == temp[::-1]:
counter+=1
l.append(temp)
return set(l)
ob1 = Solution()
l1=list(ob1.countSubstrings("abaa"))
l2=list(ob1.countSubstrings("aaba"))
x=[]
for i in range(len(l1)):
for j in range(len(l2)):
x.append(l1[i]+l2[j])
print(len(set(x)))
``` | instruction | 0 | 18,589 | 0 | 37,178 |
No | output | 1 | 18,589 | 0 | 37,179 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1 | instruction | 0 | 18,590 | 0 | 37,180 |
Tags: greedy, implementation
Correct Solution:
```
s = list(input())
for i in range(len(s)):
if s[i] == "[":
index = i
break
else:
print(-1)
exit()
s = s[index+1:]
# print(s)
for i in range(len(s)):
if s[i] == ":":
index = i
break
else:
print(-1)
exit()
s = s[index+1:]
s = s[::-1]
# print(s)
for i in range(len(s)):
if s[i] == "]":
index = i
break
else:
print(-1)
exit()
s = s[index+1:]
# print(s)
for i in range(len(s)):
if s[i] == ":":
index = i
break
else:
print(-1)
exit()
s = s[index+1:]
# print(s)
count = s.count("|")
print(count+4)
``` | output | 1 | 18,590 | 0 | 37,181 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1 | instruction | 0 | 18,591 | 0 | 37,182 |
Tags: greedy, implementation
Correct Solution:
```
accord = input()
open_bracket = accord.find("[")
accord=accord[open_bracket+1:]
open_colon = accord.find(":")
accord = accord[open_colon+1:]
close_bracket = accord.rfind("]")
close_colon = accord[:close_bracket].rfind(":")
if open_bracket == -1 or open_colon == -1 or close_bracket == -1 or close_colon == -1:
print(-1)
else:
res = accord[:close_colon].count("|")+4
print(res)
``` | output | 1 | 18,591 | 0 | 37,183 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1 | instruction | 0 | 18,592 | 0 | 37,184 |
Tags: greedy, implementation
Correct Solution:
```
import sys
s = sys.stdin.readline()
ls = len(s)
l1 = s.find('[')
r1 = ls - 1 - s[::-1].find(']')
if l1 == -1 or r1 == ls or l1 > r1:
print(-1)
exit(0)
l2 = s.find(':', l1+1, r1)
r2 = r1 - 1 - s[r1-1:l1:-1].find(':')
if l2 == -1 or r2 == ls or l2 == r2:
print(-1)
exit(0)
print(4 + s.count('|', l2+1, r2))
``` | output | 1 | 18,592 | 0 | 37,185 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1 | instruction | 0 | 18,593 | 0 | 37,186 |
Tags: greedy, implementation
Correct Solution:
```
s=input()
flag=False
ind=0
for i in range(len(s)-1,-1,-1):
if s[i]=="]":
flag=True
ind=i
break
if flag:
flag=False
for i in range(ind-1,-1,-1):
if s[i]==":":
flag=True
ind=i
break
if flag:
flag=False
for i in range(ind):
if s[i]=="[":
flag=True
ind2=i
break
if flag:
flag=False
for i in range(ind2+1,ind):
if s[i]==":":
flag=True
ind2=i
break
t=4
if flag:
for i in range(ind2+1,ind):
if s[i]=="|":
t+=1
print(t)
else:
print(-1)
else:
print(-1)
else:
print(-1)
else:
print(-1)
``` | output | 1 | 18,593 | 0 | 37,187 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1 | instruction | 0 | 18,594 | 0 | 37,188 |
Tags: greedy, implementation
Correct Solution:
```
s = input()
n = len(s)
open_b = -1
close_b = -1
open_d = -1
close_d = -1
for i in range(n):
if s[i] == '[':
open_b = i
break
for i in range(n):
if s[i] == ']':
close_b = i
if open_b == -1 or close_b == -1:
print(-1)
exit()
for i in range(open_b, close_b + 1):
if s[i] == ':':
open_d = i
break
if open_d == -1:
print(-1)
exit()
for i in range(open_d + 1, close_b + 1):
if s[i] == ':':
close_d = i
if close_d == -1 or open_d == close_d:
print(-1)
exit()
k = 0
for i in range(open_d, close_d + 1):
if s[i] == '|':
k += 1
print(4 + k)
``` | output | 1 | 18,594 | 0 | 37,189 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1 | instruction | 0 | 18,595 | 0 | 37,190 |
Tags: greedy, implementation
Correct Solution:
```
s=input()
ob=s.find('[')
cb=s.rfind(']')
fc=s[ob:cb].find(':')
sc=s[ob:cb].rfind(':')
if ob>cb or fc==sc: print(-1)
else: print(4+s[ob+fc:ob+sc].count('|'))
``` | output | 1 | 18,595 | 0 | 37,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1 | instruction | 0 | 18,596 | 0 | 37,192 |
Tags: greedy, implementation
Correct Solution:
```
s = input()
l,r = s.find('['), s.rfind(']')
if l == -1 or r == -1: print(-1)
else:
news = s[l:r+1]
l2,r2 = news.find(':'),news.rfind(':')
if l2 == r2: print(-1)
else:
v = news[l2:r2+1].count("|")
print(v+4)
``` | output | 1 | 18,596 | 0 | 37,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1 | instruction | 0 | 18,597 | 0 | 37,194 |
Tags: greedy, implementation
Correct Solution:
```
import re
s = input()
s = re.sub('[a-z]', '', s)
start = s.find('[')
if start > -1:
s = s[start:]
end = s.rfind(']')
if end > -1:
s = s[:end + 1]
start = s.find(':')
end = s.rfind(':')
if start != end:
s = s[start:(end + 1)]
print(s.count('|') + 4)
else:
print(-1)
else:
print(-1)
else:
print(-1)
``` | output | 1 | 18,597 | 0 | 37,195 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
# import time,random,resource
# sys.setrecursionlimit(10**6)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
mod2 = 998244353
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)
def IF(c, t, f): return t if c else f
def YES(c): return IF(c, "YES", "NO")
def Yes(c): return IF(c, "Yes", "No")
def main():
t = 1#I()
rr = []
for _ in range(t):
s = S()
l = len(s)
li = -1
for i in range(l):
if s[i] == '[':
li = i
break
if li < 0:
rr.append(-1)
continue
for i in range(li+1,l):
if s[i] == ':':
li = i
break
if s[li] != ':':
rr.append(-1)
continue
ri = l
for i in range(l-1,-1,-1):
if s[i] == ']':
ri = i
break
if ri == l:
rr.append(-1)
continue
for i in range(ri-1,-1,-1):
if s[i] == ':':
ri = i
break
if s[ri] != ':' or li >= ri:
rr.append(-1)
continue
if li == ri:
rr.append(l - 3)
else:
c = len([_ for _ in s[li+1:ri] if _ == '|'])
rr.append(c + 4)
return JA(rr, "\n")
print(main())
``` | instruction | 0 | 18,598 | 0 | 37,196 |
Yes | output | 1 | 18,598 | 0 | 37,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1
Submitted Solution:
```
s=input()
try:i=s.index(':',s.index('['))+1;r=s.count('|',i,s.rindex(':',i,s.rindex(']',i)))+4
except:r=-1
print(r)
``` | instruction | 0 | 18,599 | 0 | 37,198 |
Yes | output | 1 | 18,599 | 0 | 37,199 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1
Submitted Solution:
```
s = input()
if s.count('[')==0 or s.count(']')==0:
print(-1)
exit()
t = s[s.find('['):s.rfind(']')+1]
if t.count(':')<2:
print(-1)
exit()
t = t[t.find(':'):t.rfind(':')+1]
print(4+t.count('|'))
``` | instruction | 0 | 18,600 | 0 | 37,200 |
Yes | output | 1 | 18,600 | 0 | 37,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1
Submitted Solution:
```
s=input()
count=0
i=0
flag=0
for x in range(len(s)):
if s[x]=='[':
s=s[x+1:]
flag=1
break;
if flag==1:
flag=0
for x in range(len(s)):
if s[x]==':':
s=s[x+1:]
flag=1
break;
if flag==1:
flag=0
for x in range(len(s)-1,-1,-1):
if s[x]==']':
s=s[:x]
flag=1
break;
if flag==1:
flag=0
for x in range(len(s)-1,-1,-1):
if s[x]==':':
s=s[:x]
flag=1
break;
if flag==1:
for x in range(len(s)):
if(s[x]=='|'):
count+=1
else:
print(-1)
else:
print(-1)
else:
print(-1)
else:
print(-1)
if flag==1:
print(count+4)
``` | instruction | 0 | 18,601 | 0 | 37,202 |
Yes | output | 1 | 18,601 | 0 | 37,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1
Submitted Solution:
```
a=list(input())
n=len(a)
leftbraki=0
rightbraki=0
for i in range(n-1):
if(a[i]=='[' ):
leftbraki=i
break
for i in range(n-1,-1,-1):
if(a[i]==']'):
rightbraki=i
break
leftco=0
rightco=0
for i in range(leftbraki+1,n,1):
if(a[i]==':'):
leftco=i
break
for i in range(rightbraki-1,-1,-1):
if(a[i]==':'):
rightco=i
break
remo=0
danda=0
colo=0
#print(leftbraki,leftco,rightco,rightbraki)
for i in range(leftco+1,rightco,1):
if(a[i]=='|' ):
print("hello")
danda+=1
if(a[i]==':'):
colo+=1
if(leftbraki<leftco and leftco<rightco and rightco<rightbraki):
print(4+danda)
else:
print("-1")
``` | instruction | 0 | 18,602 | 0 | 37,204 |
No | output | 1 | 18,602 | 0 | 37,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1
Submitted Solution:
```
def main():
raw = input()
a = True
b = True
deadline_bracket = raw.rindex(']')
deadline_colon = raw.rindex(':')
while deadline_colon > deadline_bracket:
raw = raw[:deadline_colon]
try:
deadline_colon = raw.rindex(':')
except:
print(-1)
return
count = 0
if len(raw) < 4:
print(-1)
return
for x in range(0,len(raw)):
if a:
if raw[x] == '[':
a = False
count += 1
continue
elif b:
if raw[x] == ':':
if x == deadline_colon:
print(-1)
break
b = False
count += 1
continue
elif x < deadline_colon:
if raw[x] == '|':
count += 1
continue
else:
count += 2
print(count)
break
if a and b:
print(-1)
main()
``` | instruction | 0 | 18,603 | 0 | 37,206 |
No | output | 1 | 18,603 | 0 | 37,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1
Submitted Solution:
```
def main():
raw = input()
a = True
b = True
deadline_bracket = raw.rindex(']')
deadline_colon = raw.rindex(':')
end = len(raw)
while deadline_colon > deadline_bracket:
try:
for x in range(end,0,-1):
if raw[x] == ':':
deadline_colon = x
break
except:
print(-1)
return
count = 0
if len(raw) < 4:
print(-1)
return
for x in range(0,deadline_colon):
if a:
if raw[x] == '[':
a = False
count += 1
continue
elif b:
if raw[x] == ':':
b = False
count += 1
continue
elif x < deadline_colon:
if raw[x] == '|':
count += 1
continue
else:
count += 2
print(count)
break
if a or b:
print(-1)
main()
``` | instruction | 0 | 18,604 | 0 | 37,208 |
No | output | 1 | 18,604 | 0 | 37,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.
For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions.
You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?
Input
The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.
Output
If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.
Examples
Input
|[a:b:|]
Output
4
Input
|]:[|:]
Output
-1
Submitted Solution:
```
s=str(input())
d={'[':[],']':[],':':[],'|':[]}
for i in range(len(s)):
if s[i]=='[' or s[i]==']' or s[i]==':' or s[i]=='|':
d[s[i]].append(i)
else:
pass
if len(d['['])>=1 and len(d[']'])>=1 and len(d[':'])>=2:
if d[':'][0]>d['['][0] and d[':'][1]<max(d[']']):
k=list(filter(lambda x:x<max(d[']']),d[':']))
m=list(filter(lambda x:x>min(d['[']),k))
start=min(m)
end=max(m)
m1=list(filter(lambda x:x<end,d['|']))
k1=list(filter(lambda x:x>start,m1))
print(len(k1)+4)
else:
print(-1)
else:
print(-1)
``` | instruction | 0 | 18,605 | 0 | 37,210 |
No | output | 1 | 18,605 | 0 | 37,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob play a game. They have a binary string s (a string such that each character in it is either 0 or 1). Alice moves first, then Bob, then Alice again, and so on.
During their move, the player can choose any number (not less than one) of consecutive equal characters in s and delete them.
For example, if the string is 10110, there are 6 possible moves (deleted characters are bold):
1. 10110 → 0110;
2. 10110 → 1110;
3. 10110 → 1010;
4. 10110 → 1010;
5. 10110 → 100;
6. 10110 → 1011.
After the characters are removed, the characters to the left and to the right of the removed block become adjacent. I. e. the following sequence of moves is valid: 10110 → 100 → 1.
The game ends when the string becomes empty, and the score of each player is the number of 1-characters deleted by them.
Each player wants to maximize their score. Calculate the resulting score of Alice.
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case contains exactly one line containing a binary string s (1 ≤ |s| ≤ 100).
Output
For each test case, print one integer — the resulting score of Alice (the number of 1-characters deleted by her).
Example
Input
5
01111001
0000
111111
101010101
011011110111
Output
4
0
6
3
6
Note
Questions about the optimal strategy will be ignored.
Submitted Solution:
```
# fin = open("a.in", "r")
# buf = fin.read()
# fin.close()
import sys
buf = sys.stdin.read()
nowbuf = 0
endbuf = len(buf)
def getint():
global nowbuf
valnow = 0
while buf[nowbuf] < '0' or buf[nowbuf] > '9':
nowbuf += 1
while nowbuf < endbuf and buf[nowbuf] >= '0' and buf[nowbuf] <= '9':
valnow = valnow * 10 + int(buf[nowbuf])
nowbuf += 1
return valnow
def getstr():
global nowbuf
strnow = ''
while buf[nowbuf] == ' ' or buf[nowbuf] == '\n':
nowbuf += 1
while nowbuf < endbuf and buf[nowbuf] != ' ' and buf[nowbuf] != '\n':
strnow += buf[nowbuf]
nowbuf += 1
return strnow
def solve(gamestr):
a = []
lenone = 0
ans = 0
i = 0
strlen = len(gamestr)
a.append(0)
for i in range(0, strlen):
if gamestr[i] == '1':
a[lenone] += 1
else:
lenone += 1
a.append(0)
a.sort(reverse = 1)
for i in range(0, lenone):
if i % 2 == 0:
ans += a[i]
print(ans)
return
t = getint()
for i in range(0, t):
solve(getstr())
``` | instruction | 0 | 18,716 | 0 | 37,432 |
No | output | 1 | 18,716 | 0 | 37,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob play a game. They have a binary string s (a string such that each character in it is either 0 or 1). Alice moves first, then Bob, then Alice again, and so on.
During their move, the player can choose any number (not less than one) of consecutive equal characters in s and delete them.
For example, if the string is 10110, there are 6 possible moves (deleted characters are bold):
1. 10110 → 0110;
2. 10110 → 1110;
3. 10110 → 1010;
4. 10110 → 1010;
5. 10110 → 100;
6. 10110 → 1011.
After the characters are removed, the characters to the left and to the right of the removed block become adjacent. I. e. the following sequence of moves is valid: 10110 → 100 → 1.
The game ends when the string becomes empty, and the score of each player is the number of 1-characters deleted by them.
Each player wants to maximize their score. Calculate the resulting score of Alice.
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case contains exactly one line containing a binary string s (1 ≤ |s| ≤ 100).
Output
For each test case, print one integer — the resulting score of Alice (the number of 1-characters deleted by her).
Example
Input
5
01111001
0000
111111
101010101
011011110111
Output
4
0
6
3
6
Note
Questions about the optimal strategy will be ignored.
Submitted Solution:
```
def solve(s):
res = ''
lst = []
for i in s:
if i == '1':
res += i
else:
if res:
lst.append(res)
res = ''
if res:
lst.append(res)
if not res:
return 0
lst.sort()
lst.reverse()
a = 0
b = 0
for i in range(len(lst)):
if i%2 == 0:
a += len(lst[i])
else:
b += len(lst[i])
return a
t = int(input())
for _ in range(t):
s = input()
print(solve(s))
``` | instruction | 0 | 18,717 | 0 | 37,434 |
No | output | 1 | 18,717 | 0 | 37,435 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required. | instruction | 0 | 18,720 | 0 | 37,440 |
Tags: constructive algorithms, strings
Correct Solution:
```
import math
import sys,bisect
from heapq import *
from itertools import *
from collections import *
sys.setrecursionlimit(10 ** 6)
I = lambda : int(sys.stdin.readline())
neo = lambda : map(int, sys.stdin.readline().split())
Neo = lambda : list(map(int, sys.stdin.readline().split()))
s = input()
if s == s[::-1]:
print(0)
exit()
print(4)
print('L',len(s)-1)
s = s[1:-1][::-1]+s
# print(s)
t = len(s)-1
print('L',t)
s = s[1:-1][::-1]+s
# print(s)
print('R',len(s)-t)
s = s+s[-t-1:-1][::-1]
#print(s)
t -= 1
print('R',len(s)-t)
s = s+s[-t-1:-1][::-1]
# print(s)
# if s == s[::-1]:
# print('y')
``` | output | 1 | 18,720 | 0 | 37,441 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required. | instruction | 0 | 18,721 | 0 | 37,442 |
Tags: constructive algorithms, strings
Correct Solution:
```
s = input()
n = len(s)
print(3)
pre = n-2
print('L',n-1)
print('R',n-2+1)
total = n + n-1 + n-2+1
print('R',total-pre-1)
``` | output | 1 | 18,721 | 0 | 37,443 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required. | instruction | 0 | 18,722 | 0 | 37,444 |
Tags: constructive algorithms, strings
Correct Solution:
```
x=input()
print(3)
print('L 2')
print('R 2')
print('R',2*len(x)-1)
``` | output | 1 | 18,722 | 0 | 37,445 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required. | instruction | 0 | 18,723 | 0 | 37,446 |
Tags: constructive algorithms, strings
Correct Solution:
```
s = input()
l = len(s)
ans = []
ans.append(['R',l-1])
l += 1
ans.append(['R',l-1])
l += 1
ans.append(['L',l-1])
l = l + l-2
ans.append(['L',2])
print(len(ans))
for g in ans:
print(*g)
``` | output | 1 | 18,723 | 0 | 37,447 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required. | instruction | 0 | 18,724 | 0 | 37,448 |
Tags: constructive algorithms, strings
Correct Solution:
```
s = input()
print(3)
print("L", 2)
print("R", 2)
k = 2 * len(s) - 1
print("R", k)
``` | output | 1 | 18,724 | 0 | 37,449 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required. | instruction | 0 | 18,725 | 0 | 37,450 |
Tags: constructive algorithms, strings
Correct Solution:
```
import sys
import math
from collections import Counter,defaultdict
LI=lambda:list(map(int,input().split()))
MAP=lambda:map(int,input().split())
IN=lambda:int(input())
S=lambda:input()
def case():
s = S()
n = len(s)
print(3)
n+=1
print("L",2)
n+=(n-2)
print("R",2)
print("R",n-1)
for _ in range(1):
case()
``` | output | 1 | 18,725 | 0 | 37,451 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required. | instruction | 0 | 18,726 | 0 | 37,452 |
Tags: constructive algorithms, strings
Correct Solution:
```
s = input()
n = len(s)
print(3)
print('L', n - 1)
print('R', n - 1)
print('R', 2 * n - 1)
``` | output | 1 | 18,726 | 0 | 37,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required. | instruction | 0 | 18,727 | 0 | 37,454 |
Tags: constructive algorithms, strings
Correct Solution:
```
a = input()
n = len(a)
print(3)
print("R",n-1)
print("L",n)
print("L",2)
``` | output | 1 | 18,727 | 0 | 37,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
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 RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def print_list(l):
print(' '.join(map(str,l)))
# import heapq as hq
# import bisect as bs
# from collections import deque as dq
# from collections import defaultdict as dc
# from math import ceil,floor,sqrt
# from collections import Counter
s = input().strip()
n = len(s)
print(3)
print('L',2)
print('R',2)
print('R',2*n-1)
``` | instruction | 0 | 18,728 | 0 | 37,456 |
Yes | output | 1 | 18,728 | 0 | 37,457 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required.
Submitted Solution:
```
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
string = input()
print(3)
print("L {}".format(len(string)-1))
print("R {}".format(len(string)-1))
print("R {}".format(2*len(string)-1))
#"{} {} {}".format(maxele,minele,minele)
# 0 1 1 0
# 1 0 1 1
# abcdefefe
# n
# 2n-2
# 3n-3-(n-2)
``` | instruction | 0 | 18,729 | 0 | 37,458 |
Yes | output | 1 | 18,729 | 0 | 37,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required.
Submitted Solution:
```
s=len(input())
print(3)
print("R {}".format(s-1))
print("L {}".format(s))
print("L {}".format(2))
``` | instruction | 0 | 18,730 | 0 | 37,460 |
Yes | output | 1 | 18,730 | 0 | 37,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required.
Submitted Solution:
```
def isPalin(s):
n = len(s)
for i in range(n>>1):
if(s[i] != s[n-i-1]):
return False
return True
s = input()
n = len(s)
if(isPalin(s)):
print(0)
else:
print(3)
print("L", 2)
print("R",2)
print("R", n+1+(n-1)-1)
``` | instruction | 0 | 18,731 | 0 | 37,462 |
Yes | output | 1 | 18,731 | 0 | 37,463 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required.
Submitted Solution:
```
def isPalin(st):
i,j=0,len(st)-1
while(i<=j):
if(st[i]==st[j]):
i+=1
j-=1
else:
return False
return True
def proC(st):
if(isPalin(st)):
print(0)
return
print('R',len(st)-1)
print('L',len(st))
print('L',2)
n=input()
proC(n)
``` | instruction | 0 | 18,732 | 0 | 37,464 |
No | output | 1 | 18,732 | 0 | 37,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required.
Submitted Solution:
```
s = input()
print('1')
print('L %d'%(len(s)-1))
``` | instruction | 0 | 18,733 | 0 | 37,466 |
No | output | 1 | 18,733 | 0 | 37,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required.
Submitted Solution:
```
s = input()
n = len(s)
print("R", n-1)
print("L", n)
print("L", 2)
``` | instruction | 0 | 18,734 | 0 | 37,468 |
No | output | 1 | 18,734 | 0 | 37,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
Note that characters in the string in this problem are indexed from 1.
For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
Input
The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
Output
The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
The length of the resulting palindrome must not exceed 10^6.
Examples
Input
abac
Output
2
R 2
R 5
Input
acccc
Output
2
L 4
L 2
Input
hannah
Output
0
Note
For the first example the following operations are performed:
abac → abacab → abacaba
The second sample performs the following operations: acccc → cccacccc → ccccacccc
The third example is already a palindrome so no operations are required.
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict,deque,Counter
from bisect import *
from math import sqrt,pi,ceil
import math
from itertools import permutations
from copy import deepcopy
def main():
s = input().rstrip()
b = s[::-1]
n = len(s)
if s == b:
print(0)
else:
print(4)
print("R", 2)
print("L", 2)
print("R",2*n-2)
print("L",2*n-2)
# region fastio
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 | 18,735 | 0 | 37,470 |
No | output | 1 | 18,735 | 0 | 37,471 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke has a string s. From this string, Anuke, Bnuke, and Cnuke obtained strings a, b, and c, respectively, as follows:
* Choose a non-empty (contiguous) substring of s (possibly s itself). Then, replace some characters (possibly all or none) in it with `?`s.
For example, if s is `mississippi`, we can choose the substring `ssissip` and replace its 1-st and 3-rd characters with `?` to obtain `?s?ssip`.
You are given the strings a, b, and c. Find the minimum possible length of s.
Constraints
* 1 \leq |a|, |b|, |c| \leq 2000
* a, b, and c consists of lowercase English letters and `?`s.
Input
Input is given from Standard Input in the following format:
a
b
c
Output
Print the minimum possible length of s.
Examples
Input
a?c
der
cod
Output
7
Input
atcoder
atcoder
???????
Output
7 | instruction | 0 | 19,138 | 0 | 38,276 |
"Correct Solution:
```
import sys
input=lambda: sys.stdin.readline().rstrip()
A=input()
B=input()
C=input()
a,b,c=len(A),len(B),len(C)
def f(X,Y):
XY=[True]*(10**4)
x,y=len(X),len(Y)
for i in range(5000-y+1,5000+x):
ret=True
if i<=5000:
for j in range(min(x,y+i-5000)):
if X[j]=="?" or Y[j-i+5000]=="?" or X[j]==Y[j-i+5000]:
continue
else:
ret=False
else:
for j in range(min(y,x-i+5000)):
if Y[j]=="?" or X[j+i-5000]=="?" or Y[j]==X[j+i-5000]:
continue
else:
ret=False
XY[i]=ret
return XY
AB=f(A,B)
BC=f(B,C)
AC=f(A,C)
ans=a+b+c
for i in range(5000-b-c,5000+a+c):
for j in range(5000-b-c,5000+a+b):
if -4000<=j-i<=4000:
if AB[i] and AC[j] and BC[j-i+5000]:
l=min(5000,i,j)
r=max(i+b,j+c,5000+a)
ans=min(ans,r-l)
else:
if AB[i] and AC[j]:
l=min(5000,i,j)
r=max(i+b,j+c,5000+a)
ans=min(ans,r-l)
print(ans)
``` | output | 1 | 19,138 | 0 | 38,277 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke has a string s. From this string, Anuke, Bnuke, and Cnuke obtained strings a, b, and c, respectively, as follows:
* Choose a non-empty (contiguous) substring of s (possibly s itself). Then, replace some characters (possibly all or none) in it with `?`s.
For example, if s is `mississippi`, we can choose the substring `ssissip` and replace its 1-st and 3-rd characters with `?` to obtain `?s?ssip`.
You are given the strings a, b, and c. Find the minimum possible length of s.
Constraints
* 1 \leq |a|, |b|, |c| \leq 2000
* a, b, and c consists of lowercase English letters and `?`s.
Input
Input is given from Standard Input in the following format:
a
b
c
Output
Print the minimum possible length of s.
Examples
Input
a?c
der
cod
Output
7
Input
atcoder
atcoder
???????
Output
7 | instruction | 0 | 19,139 | 0 | 38,278 |
"Correct Solution:
```
from itertools import chain, permutations, dropwhile
from heapq import merge
from functools import wraps
def listify(fn):
@wraps(fn)
def wraper(*args, **kwargs):
return list(fn(*args, **kwargs))
return wraper
def comp_wild(a,b):
return all(x==y or "?" in (x,y) for x,y in zip(a, b))
@listify
def includers(a,b):
for i in range(len(a)-len(b)+1):
if comp_wild(a[i:], b):
yield i
@listify
def connectors(a,b):
la = len(a)
for i in range(max(la-len(b)+1, 1), la):
if comp_wild(a[i-la:], b):
yield i
yield la
class c_int(int):
def __new__(cls, *args, **kwargs):
cond = kwargs.pop("cond", lambda:True)
i = int.__new__(cls, *args, **kwargs)
i.cond = cond
return i
def iter_from(it, fm):
return dropwhile(fm.__gt__, it)
def solve(a, b, c):
keys = sorted((a,b,c), key=len, reverse=True)
len_keys = list(map(len, keys))
inc = { (i,j):includers(keys[i],keys[j]) for i,j in permutations(range(3), 2) }
con = { (i,j):connectors(keys[i],keys[j]) for i,j in permutations(range(3), 2) }
for i in inc[0,1]:
for j in inc[0,2]:
if (
j+len_keys[2] <= i or i+len_keys[2] <= j or
j-i in inc[1,2] or j-i in con[1,2] or i-j in con[2,1]
):
return len_keys[0]
def candicates(x,y,z):
return chain(
( c_int(
i+len_keys[z],
cond=lambda:(
(inc[x,y] and inc[x,y][0]+len_keys[y] <= i) or
(inc[z,y] and len_keys[x] <= i+inc[z,y][-1]) or
any(
(i-j in con[y,z] if i>j else j-i in inc[z,y])
for j in iter_from(inc[x,y]+con[x,y], i-len_keys[y])
)
)
) for i in iter_from(con[x,z], len_keys[y]-len_keys[z]+1)
),
( c_int(
min( i+next(iter_from(q, len_keys[x]-i+1))+len_keys[z] for i in p )
) for (p, q) in [sorted((con[x,y], con[y,z]), key=len)]
),
)
for c in merge(*( candicates(*xyz) for xyz in permutations(range(3), 3) )):
if c.cond():
return c
return sum(len_keys)
print(solve(*(input() for _ in range(3))))
``` | output | 1 | 19,139 | 0 | 38,279 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke has a string s. From this string, Anuke, Bnuke, and Cnuke obtained strings a, b, and c, respectively, as follows:
* Choose a non-empty (contiguous) substring of s (possibly s itself). Then, replace some characters (possibly all or none) in it with `?`s.
For example, if s is `mississippi`, we can choose the substring `ssissip` and replace its 1-st and 3-rd characters with `?` to obtain `?s?ssip`.
You are given the strings a, b, and c. Find the minimum possible length of s.
Constraints
* 1 \leq |a|, |b|, |c| \leq 2000
* a, b, and c consists of lowercase English letters and `?`s.
Input
Input is given from Standard Input in the following format:
a
b
c
Output
Print the minimum possible length of s.
Examples
Input
a?c
der
cod
Output
7
Input
atcoder
atcoder
???????
Output
7 | instruction | 0 | 19,140 | 0 | 38,280 |
"Correct Solution:
```
def match(x, y):
"""
１文字x, yが一致するか少なくとも一方が？であればTrueを返す。
"""
return x == y or x == "?" or y == "?"
def func(a, b, c, la, lb, lc):
"""
文字列の並びがa->b->cの場合を仮定して考える。
aに対するb, cの配置が高々2000 * 2000通りあるので、文字列が一致するかの判定はO(1)でしないとTLE。
これは先にaとb、bとc、aとcに対して文字列の一致を前処理確認しておけばOK <= 思いつかなかったので反省。
ab[i] = (b[0]をa[i]から重ねたとき、合成文字列は一致するか)
bc[i] = (c[0]をb[i]から重ねたとき、合成文字列は一致するか)
ac[i] = (c[0]をa[i]から重ねたとき、合成文字列は一致するか)
とおく。後ろに[True] * 2000を付け加えているのはaとbがdisjointな場合を考慮するため。
"""
ab = [True] * (la + 4000)
bc = [True] * (lb + 4000)
ac = [True] * (la + 4000)
for i in range(la):
for j in range(min(la - i, lb)):
if not match(a[i + j], b[j]):
ab[i] = False
break
for i in range(lb):
for j in range(min(lb - i, lc)):
if not match(b[i + j], c[j]):
bc[i] = False
break
for i in range(la):
for j in range(min(la - i, lc)):
if not match(a[i + j], c[j]):
ac[i] = False
break
#print("ab={}".format(ab[:la]))
#print("bc={}".format(bc[:lb]))
#print("ac={}".format(ac[:la]))
"""
前処理終わり。これより実際の比較を行う。
"""
ans = la + lb + lc
for i in range(la + 1):
for j in range(max(lb + 1, la - i + 1)): # A上でBとCがdisjointに乗っかっている場合、j <= lbだと不十分(ex: A = "qaawbbe", B = "aa", C = "bb")
if ab[i] and bc[j] and ac[i + j]:
preans = max(la, i + lb, i + j + lc) # ここ(i + lb)入れないと、Bが一番長い場合が抜け落ちる
ans = min(ans, preans)
#print("{}->{}->{}: ans={}".format(a, b, c, ans))
return ans
def main():
a = input()
b = input()
c = input()
la = len(a)
lb = len(b)
lc = len(c)
ans = func(a, b, c, la, lb, lc)
ans = min(ans, func(a, c, b, la, lc, lb))
ans = min(ans, func(b, a, c, lb, la, lc))
ans = min(ans, func(b, c, a, lb, lc, la))
ans = min(ans, func(c, a, b, lc, la, lb))
ans = min(ans, func(c, b, a, lc, lb, la))
print(ans)
if __name__ == "__main__":
main()
``` | output | 1 | 19,140 | 0 | 38,281 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke has a string s. From this string, Anuke, Bnuke, and Cnuke obtained strings a, b, and c, respectively, as follows:
* Choose a non-empty (contiguous) substring of s (possibly s itself). Then, replace some characters (possibly all or none) in it with `?`s.
For example, if s is `mississippi`, we can choose the substring `ssissip` and replace its 1-st and 3-rd characters with `?` to obtain `?s?ssip`.
You are given the strings a, b, and c. Find the minimum possible length of s.
Constraints
* 1 \leq |a|, |b|, |c| \leq 2000
* a, b, and c consists of lowercase English letters and `?`s.
Input
Input is given from Standard Input in the following format:
a
b
c
Output
Print the minimum possible length of s.
Examples
Input
a?c
der
cod
Output
7
Input
atcoder
atcoder
???????
Output
7 | instruction | 0 | 19,142 | 0 | 38,284 |
"Correct Solution:
```
from itertools import chain, permutations, dropwhile
from heapq import merge
from functools import wraps
def comp_wild(a,b):
return all(x==y or "?" in (x,y) for x,y in zip(a, b))
def listify(fn):
@wraps(fn)
def wraper(*args, **kwargs):
return list(fn(*args, **kwargs))
return wraper
@listify
def includers(a,b):
for i in range(len(a)-len(b)+1):
if comp_wild(a[i:],b):
yield i
@listify
def connectors(a,b):
la = len(a)
for i in range(max(la-len(b)+1, 1), la):
if comp_wild(b, a[i-la:]):
yield i
yield la
class K(int):
def __new__(cls, *args, **kwargs):
cb = kwargs.pop("cb", lambda:True)
i = int.__new__(cls, *args, **kwargs)
i.cb = cb
return i
def iter_from(it, fm):
return dropwhile(fm.__gt__, it)
def solve(a,b,c):
keys = sorted((a,b,c), key=len, reverse=True)
len_keys = list(map(len, keys))
inc = {(i,j):includers(keys[i],keys[j]) for i,j in permutations(range(3), 2)}
con = {(i,j):connectors(keys[i],keys[j]) for i,j in permutations(range(3), 2)}
for i in inc[0,1]:
for j in inc[0,2]:
if j+len_keys[2] <= i or i+len_keys[2] <= j or j-i in inc[1,2] or j-i in con[1,2] or i-j in con[2,1]:
return len_keys[0]
def candicates(x,y,z):
return chain(
(K(i+len_keys[z], cb=lambda:(
(inc[x,y] and inc[x,y][0]+len_keys[y] <= i) or
(inc[z,y] and len_keys[x] <= i+inc[z,y][-1]) or
any((i-j in con[y,z] if i>j else j-i in inc[z,y]) for j in iter_from(inc[x,y]+con[x,y], i-len_keys[y]))
)) for i in iter_from(con[x,z], len_keys[y]-len_keys[z]+1)),
(K(min(i+next(iter_from(q, len_keys[x]-i+1))+len_keys[z] for i in p)) for (p, q) in [sorted((con[x,y], con[y,z]), key=len)]),
)
can = merge(*(candicates(*xyz) for xyz in permutations(range(3),3)))
for k in can:
if k.cb():
return k
return sum(len_keys)
print(solve(*(input() for _ in range(3))))
``` | output | 1 | 19,142 | 0 | 38,285 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke has a string s. From this string, Anuke, Bnuke, and Cnuke obtained strings a, b, and c, respectively, as follows:
* Choose a non-empty (contiguous) substring of s (possibly s itself). Then, replace some characters (possibly all or none) in it with `?`s.
For example, if s is `mississippi`, we can choose the substring `ssissip` and replace its 1-st and 3-rd characters with `?` to obtain `?s?ssip`.
You are given the strings a, b, and c. Find the minimum possible length of s.
Constraints
* 1 \leq |a|, |b|, |c| \leq 2000
* a, b, and c consists of lowercase English letters and `?`s.
Input
Input is given from Standard Input in the following format:
a
b
c
Output
Print the minimum possible length of s.
Examples
Input
a?c
der
cod
Output
7
Input
atcoder
atcoder
???????
Output
7 | instruction | 0 | 19,143 | 0 | 38,286 |
"Correct Solution:
```
import sys
input=lambda: sys.stdin.readline().rstrip()
A=input()
B=input()
C=input()
a,b,c=len(A),len(B),len(C)
def f(X,Y):
XY=[True]*(10**4)
x,y=len(X),len(Y)
for i in range(5000-y+1,5000+x):
ret=True
if i<=5000:
for j in range(min(x,y+i-5000)):
if X[j]=="?" or Y[j-i+5000]=="?" or X[j]==Y[j-i+5000]:
continue
else:
ret=False
else:
for j in range(min(y,x-i+5000)):
if Y[j]=="?" or X[j+i-5000]=="?" or Y[j]==X[j+i-5000]:
continue
else:
ret=False
XY[i]=ret
return XY
AB=f(A,B)
BC=f(B,C)
AC=f(A,C)
ans=a+b+c
for i in range(5000-b-c,5000+a+c):
if AB[i]:
for j in range(5000-b-c,5000+a+b):
if -4000<=j-i<=4000:
if AC[j] and BC[j-i+5000]:
l=min(5000,i,j)
r=max(i+b,j+c,5000+a)
ans=min(ans,r-l)
else:
if AC[j]:
l=min(5000,i,j)
r=max(i+b,j+c,5000+a)
ans=min(ans,r-l)
print(ans)
``` | output | 1 | 19,143 | 0 | 38,287 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke has a string s. From this string, Anuke, Bnuke, and Cnuke obtained strings a, b, and c, respectively, as follows:
* Choose a non-empty (contiguous) substring of s (possibly s itself). Then, replace some characters (possibly all or none) in it with `?`s.
For example, if s is `mississippi`, we can choose the substring `ssissip` and replace its 1-st and 3-rd characters with `?` to obtain `?s?ssip`.
You are given the strings a, b, and c. Find the minimum possible length of s.
Constraints
* 1 \leq |a|, |b|, |c| \leq 2000
* a, b, and c consists of lowercase English letters and `?`s.
Input
Input is given from Standard Input in the following format:
a
b
c
Output
Print the minimum possible length of s.
Examples
Input
a?c
der
cod
Output
7
Input
atcoder
atcoder
???????
Output
7 | instruction | 0 | 19,144 | 0 | 38,288 |
"Correct Solution:
```
R=range
def F(a,b):A=len(a);r=[all([len(set([a[i+j],b[j],'?']))<3for j in R(min(A-i,len(b)))])for i in R(A)];return r+[1]
def Z(X):
i,j,k=X;U,V,W=M[i][j],M[j][k],M[i][k];A,B,C=map(len,[S[i]for i in X]);q=A+B+C
for l in R(A+1):
if U[l]:
for r in R(l,A+B+1):
if(B<r-l or V[r-l])*W[min(A,r)]:q=min(q,max(A,l+B,r+C))
return q
S=[input()for i in'abc']
T=R(3)
M=[[F(S[i],S[j])for j in T]for i in T]
from itertools import *;print(min([Z(i)for i in permutations(T)]))
``` | output | 1 | 19,144 | 0 | 38,289 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke has a string s. From this string, Anuke, Bnuke, and Cnuke obtained strings a, b, and c, respectively, as follows:
* Choose a non-empty (contiguous) substring of s (possibly s itself). Then, replace some characters (possibly all or none) in it with `?`s.
For example, if s is `mississippi`, we can choose the substring `ssissip` and replace its 1-st and 3-rd characters with `?` to obtain `?s?ssip`.
You are given the strings a, b, and c. Find the minimum possible length of s.
Constraints
* 1 \leq |a|, |b|, |c| \leq 2000
* a, b, and c consists of lowercase English letters and `?`s.
Input
Input is given from Standard Input in the following format:
a
b
c
Output
Print the minimum possible length of s.
Examples
Input
a?c
der
cod
Output
7
Input
atcoder
atcoder
???????
Output
7 | instruction | 0 | 19,145 | 0 | 38,290 |
"Correct Solution:
```
from itertools import *
from heapq import *
def comp(a,b):
return all(x==y or "?" in (x,y) for x,y in zip(a, b))
def comp3(a,b,c):
return
def comp3(i,j,k,a,b,c):
m = max(i,j,k)
n = min(i+len(a),j+len(b),k+len(c))
return all(len({x,y,z}-{"?"}) <= 1 for x,y,z in zip(a[m-i:n-i],b[m-j:n-j],c[m-k:n-k]))
def _merge_mid(a,b):
la, lb = map(len,[a,b])
for i in range(la-lb+1):
if comp(a[i:],b):
yield i
def merge_mid(a,b):
return list(_merge_mid(a,b))
def _merge_right(a,b):
la, lb = map(len,[a,b])
for i in range(max(la-lb+1, 1),la):
if comp(b, a[i-la:]):
yield i
yield la
def merge_right(a,b):
return list(_merge_right(a,b))
class K(int):
def __new__(cls, *args, **kwargs):
cb = kwargs.pop("cb")
i = int.__new__(cls, *args, **kwargs)
i.cb = cb
return i
def min_merge(a,b,c):
keys = sorted([a,b,c],key=len,reverse=True)
len_keys = list(map(len, keys))
mid = {(i,j):merge_mid(keys[i],keys[j]) for i,j in permutations(range(3), 2)}
right = {(i,j):merge_right(keys[i],keys[j]) for i,j in permutations(range(3), 2)}
for i in mid[0,1]:
for j in mid[0,2]:
if j+len_keys[2] <= i or i+len_keys[2] <= j or (j-i in mid[1,2] or j-i in right[1,2] or i-j in right[2,1]):
return len_keys[0]
def candicates(a,b,c):
return chain(
(K(i+len_keys[b], cb=lambda:(
(mid[a,c] and mid[a,c][0]+len_keys[c] <= i) or (mid[b,c] and len_keys[a] <= i+mid[b,c][-1]) or
any((i-j in right[c,b] if i>j else j-i in mid[b,c]) and comp3(0,i,j,keys[a],keys[b],keys[c]) for j in dropwhile((i-len_keys[c]).__gt__, mid[a,c]+right[a,c]))
)) for i in dropwhile((len_keys[c]-len_keys[b]+1).__gt__, right[a,b])),
(K(min(i+next(dropwhile((len_keys[a]-i+1).__gt__, y))+len_keys[b] for i in x), cb=lambda:True) for (x, y) in [sorted((right[a,c], right[c,b]), key=len)]),
)
can = merge(*(candicates(x,y,z) for x,y,z in permutations(range(3),3)))
for k in can:
if k.cb():
return k
return sum(len_keys)
a=input()
b=input()
c=input()
print(min_merge(a,b,c))
``` | output | 1 | 19,145 | 0 | 38,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has a string s. From this string, Anuke, Bnuke, and Cnuke obtained strings a, b, and c, respectively, as follows:
* Choose a non-empty (contiguous) substring of s (possibly s itself). Then, replace some characters (possibly all or none) in it with `?`s.
For example, if s is `mississippi`, we can choose the substring `ssissip` and replace its 1-st and 3-rd characters with `?` to obtain `?s?ssip`.
You are given the strings a, b, and c. Find the minimum possible length of s.
Constraints
* 1 \leq |a|, |b|, |c| \leq 2000
* a, b, and c consists of lowercase English letters and `?`s.
Input
Input is given from Standard Input in the following format:
a
b
c
Output
Print the minimum possible length of s.
Examples
Input
a?c
der
cod
Output
7
Input
atcoder
atcoder
???????
Output
7
Submitted Solution:
```
printn = lambda x: print(x,end='')
inn = lambda : int(input())
inl = lambda: list(map(int, input().split()))
inm = lambda: map(int, input().split())
ins = lambda : input().strip()
DBG = True and False
BIG = 10**18
R = 10**9 + 7
def ddprint(x):
if DBG:
print(x)
def cmp1(a,b):
l = []
for i in range(len(a)):
ok = True
for j in range(min(len(b),len(a)-i)):
#ddprint("a {} b {} i {} j {}".format(a,b,i,j))
#ddprint("aij {} bj {}".format(a[i+j],b[j]))
if a[i+j]!='?' and b[j]!='?' and a[i+j]!=b[j]:
ok = False
break
if ok:
l.append(i)
#ddprint("cmp1 a {} b {} l {}".format(a,b,l))
return l
def chk3(mn,a,b,c,ab,ac,bc):
for i in ab:
v = max(len(a),len(b)+i)+len(c)
mn = min(mn,v)
for j in bc:
if len(a)-i<=j or (j+i) in ac:
v = max([len(a),len(b)+i,len(c)+i+j])
mn = min(mn,v)
for j in ac:
if len(b)+i<=j:
v = max(len(a),len(c)+j)
mn = min(mn,v)
for i in bc:
v = max(len(b),len(c)+i)+len(a)
mn = min(mn,v)
return mn
def solve(a,b,c):
ab = cmp1(a,b)
ac = cmp1(a,c)
bc = cmp1(b,c)
ba = cmp1(b,a)
ca = cmp1(c,a)
cb = cmp1(c,b)
mn = len(a)+len(b)+len(c)
ddprint(mn)
mn = chk3(mn,a,b,c,ab,ac,bc)
ddprint(mn)
mn = chk3(mn,a,c,b,ac,ab,cb)
ddprint(mn)
mn = chk3(mn,b,a,c,ba,bc,ac)
ddprint(mn)
mn = chk3(mn,b,c,a,bc,ba,ca)
ddprint(mn)
mn = chk3(mn,c,a,b,ca,cb,ab)
ddprint(mn)
mn = chk3(mn,c,b,a,cb,ca,ba)
return mn
a = ins()
b = ins()
c = ins()
mn = solve(a,b,c)
print(mn)
``` | instruction | 0 | 19,146 | 0 | 38,292 |
Yes | output | 1 | 19,146 | 0 | 38,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has a string s. From this string, Anuke, Bnuke, and Cnuke obtained strings a, b, and c, respectively, as follows:
* Choose a non-empty (contiguous) substring of s (possibly s itself). Then, replace some characters (possibly all or none) in it with `?`s.
For example, if s is `mississippi`, we can choose the substring `ssissip` and replace its 1-st and 3-rd characters with `?` to obtain `?s?ssip`.
You are given the strings a, b, and c. Find the minimum possible length of s.
Constraints
* 1 \leq |a|, |b|, |c| \leq 2000
* a, b, and c consists of lowercase English letters and `?`s.
Input
Input is given from Standard Input in the following format:
a
b
c
Output
Print the minimum possible length of s.
Examples
Input
a?c
der
cod
Output
7
Input
atcoder
atcoder
???????
Output
7
Submitted Solution:
```
def F(a,b):
A=len(a)
r=[all([len(set([a[i+j],b[j],'?']))<3for j in range(min(A-i,len(b)))])for i in range(A)]
return r+[1]
def Z(ix):
i1,i2,i3=ix;ab,bc,ac=M[i1][i2],M[i2][i3],M[i1][i3];A,B,C=map(len,[S[i]for i in ix]);q=A+B+C
for l in range(A+1):
if ab[l]!=0:
for r in range(l,A+B+1):
if (B<r-l or bc[r-l]==1) and ac[min(A,r)]==1:q=min(q,max(A,l+B,r+C))
return q
S=[input()for i in'abc']
T=range(3);M=[[F(S[i],S[j])for j in T]for i in T]
print(min([Z(i)for i in[(0,1,2),(0,2,1),(1,0,2),(1,2,0),(2,1,0),(2,0,1)]]))
``` | instruction | 0 | 19,147 | 0 | 38,294 |
Yes | output | 1 | 19,147 | 0 | 38,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has a string s. From this string, Anuke, Bnuke, and Cnuke obtained strings a, b, and c, respectively, as follows:
* Choose a non-empty (contiguous) substring of s (possibly s itself). Then, replace some characters (possibly all or none) in it with `?`s.
For example, if s is `mississippi`, we can choose the substring `ssissip` and replace its 1-st and 3-rd characters with `?` to obtain `?s?ssip`.
You are given the strings a, b, and c. Find the minimum possible length of s.
Constraints
* 1 \leq |a|, |b|, |c| \leq 2000
* a, b, and c consists of lowercase English letters and `?`s.
Input
Input is given from Standard Input in the following format:
a
b
c
Output
Print the minimum possible length of s.
Examples
Input
a?c
der
cod
Output
7
Input
atcoder
atcoder
???????
Output
7
Submitted Solution:
```
a = input()
b = input()
c = input()
La = len(a)
Lb = len(b)
Lc = len(c)
def func(a, b):
ret = set()
for i in range(-len(b)+1, len(a)):
ok = True
for j in range(max(0, i), min(len(a), len(b)+i)):
if a[j] != '?' and b[j-i] != '?' and a[j] != b[j-i]:
ok = False
break
if ok:
ret.add(i)
return ret
L1 = func(a, b)
L2 = func(a, c)
L3 = func(b, c)
L1.add(-Lb)
L1.add(La)
L2.add(-Lc)
L2.add(La)
L3.add(-Lc)
L3.add(Lb)
ans = 10000
for i in L1:
for j in L2:
if j - i in L3 or j - i <= -Lc or j - i >= Lb:
lo = min(0, i, j)
hi = max(len(a), i+len(b), j+len(c))
ans = min(ans, hi-lo)
for k in L3:
j = i + k
if j >= La or j <= -Lc:
lo = min(0, i, j)
hi = max(len(a), i+len(b), j+len(c))
ans = min(ans, hi-lo)
for j in L2:
for k in L3:
i = j - k
if i >= La or i <= -Lb:
lo = min(0, i, j)
hi = max(len(a), i+len(b), j+len(c))
ans = min(ans, hi-lo)
print(ans)
``` | instruction | 0 | 19,148 | 0 | 38,296 |
Yes | output | 1 | 19,148 | 0 | 38,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has a string s. From this string, Anuke, Bnuke, and Cnuke obtained strings a, b, and c, respectively, as follows:
* Choose a non-empty (contiguous) substring of s (possibly s itself). Then, replace some characters (possibly all or none) in it with `?`s.
For example, if s is `mississippi`, we can choose the substring `ssissip` and replace its 1-st and 3-rd characters with `?` to obtain `?s?ssip`.
You are given the strings a, b, and c. Find the minimum possible length of s.
Constraints
* 1 \leq |a|, |b|, |c| \leq 2000
* a, b, and c consists of lowercase English letters and `?`s.
Input
Input is given from Standard Input in the following format:
a
b
c
Output
Print the minimum possible length of s.
Examples
Input
a?c
der
cod
Output
7
Input
atcoder
atcoder
???????
Output
7
Submitted Solution:
```
from itertools import permutations
ans = 6000
for a, b, c in permutations(input() for _ in range(3)):
ab, bc, ac = [], [], []
for s, t, st in [(a, b, ab), (b, c, bc), (a, c, ac)]:
for i in range(len(s) + 1):
for si, ti in zip(s[i:], t):
if si != '?' and ti != '?' and si != ti:
st.append(False)
break
else:
st.append(True)
for i in range(len(a) + 1):
for j in range(max(len(b), len(a) - i) + 1):
if ab[i] and (j > len(b) or bc[j]) and (i+j > len(a) or ac[i+j]):
ans = min(ans, max(len(a), i+len(b), i+j+len(c)))
print(ans)
``` | instruction | 0 | 19,149 | 0 | 38,298 |
Yes | output | 1 | 19,149 | 0 | 38,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has a string s. From this string, Anuke, Bnuke, and Cnuke obtained strings a, b, and c, respectively, as follows:
* Choose a non-empty (contiguous) substring of s (possibly s itself). Then, replace some characters (possibly all or none) in it with `?`s.
For example, if s is `mississippi`, we can choose the substring `ssissip` and replace its 1-st and 3-rd characters with `?` to obtain `?s?ssip`.
You are given the strings a, b, and c. Find the minimum possible length of s.
Constraints
* 1 \leq |a|, |b|, |c| \leq 2000
* a, b, and c consists of lowercase English letters and `?`s.
Input
Input is given from Standard Input in the following format:
a
b
c
Output
Print the minimum possible length of s.
Examples
Input
a?c
der
cod
Output
7
Input
atcoder
atcoder
???????
Output
7
Submitted Solution:
```
a = input()
b = input()
c = input()
A, B, C = len(a), len(b), len(c)
ab, bc, ac = [1] * 120000, [1] * 120000, [1] * 120000
for i in range(A):
if a[i] == '?':
continue
for j in range(B):
if b[j] == '?':
continue
if a[i] != b[j]:
ab[i-j+6000]= 0
for i in range(B):
if b[i] == '?':
continue
for j in range(C):
if c[j] == '?':
continue
if b[i] != c[j]:
bc[i-j+6000]= 0
for i in range(A):
if a[i] == '?':
continue
for j in range(C):
if c[j] == '?':
continue
if a[i] != c[j]:
ac[i-j+6000]= 0
ans = 6000
for i in range(-2000, 2000):
for j in range(-2000, 2000):
if ab[i+6000] and bc[j-i+6000] and ac[j+6000]:
L = min(0, min(i, j))
R = max(A, max(B+i, C+j))
ans = min(ans, R-L)
print(ans)
``` | instruction | 0 | 19,150 | 0 | 38,300 |
No | output | 1 | 19,150 | 0 | 38,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has a string s. From this string, Anuke, Bnuke, and Cnuke obtained strings a, b, and c, respectively, as follows:
* Choose a non-empty (contiguous) substring of s (possibly s itself). Then, replace some characters (possibly all or none) in it with `?`s.
For example, if s is `mississippi`, we can choose the substring `ssissip` and replace its 1-st and 3-rd characters with `?` to obtain `?s?ssip`.
You are given the strings a, b, and c. Find the minimum possible length of s.
Constraints
* 1 \leq |a|, |b|, |c| \leq 2000
* a, b, and c consists of lowercase English letters and `?`s.
Input
Input is given from Standard Input in the following format:
a
b
c
Output
Print the minimum possible length of s.
Examples
Input
a?c
der
cod
Output
7
Input
atcoder
atcoder
???????
Output
7
Submitted Solution:
```
import sys
input = sys.stdin.readline
A = input().rstrip()
B = input().rstrip()
C = input().rstrip()
def check(P1, P2, P3):
L1 = len(P1)
L2 = len(P2)
L3 = len(P3)
L = max([L1, L2, L3])*2+3
P12 = [True]*L
for i2 in range(-L2, L1+1):
for l in range(max(i2, 0), min(i2+L2, L1)):
if P1[l] != "?" and P2[l-i2] != "?" and P1[l] != P2[l-i2]:
P12[i2] = False
break
P13 = [True]*L
for i3 in range(-L3, L1+1):
for l in range(max(i3, 0), min(i3+L3, L1)):
if P1[l] != "?" and P3[l-i3] != "?" and P1[l] != P3[l-i3]:
P13[i3] = False
break
P23 = [True]*L
for i3 in range(-L3, L2+1):
for l in range(max(i3, 0), min(i3+L3, L2)):
if P2[l] != "?" and P3[l-i3] != "?" and P2[l] != P3[l-i3]:
P23[i3] = False
break
ans = L1 + L2 + L3
for i2 in range(-L2, L1+1):
for i3 in range(-L3, L1+1):
if P12[i2] and P13[i3] and P23[i3-i2]:
ans = min(ans, max([L1, L2+i2, L3+i3]) - min([0, i2, i3]))
return ans
print(min([check(A, B, C), check(B, A, C), check(C, A, B)]))
``` | instruction | 0 | 19,151 | 0 | 38,302 |
No | output | 1 | 19,151 | 0 | 38,303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A non-empty string is called palindrome, if it reads the same from the left to the right and from the right to the left. For example, "abcba", "a", and "abba" are palindromes, while "abab" and "xy" are not.
A string is called a substring of another string, if it can be obtained from that string by dropping some (possibly zero) number of characters from the beginning and from the end of it. For example, "abc", "ab", and "c" are substrings of the string "abc", while "ac" and "d" are not.
Let's define a palindromic count of the string as the number of its substrings that are palindromes. For example, the palindromic count of the string "aaa" is 6 because all its substrings are palindromes, and the palindromic count of the string "abc" is 3 because only its substrings of length 1 are palindromes.
You are given a string s. You can arbitrarily rearrange its characters. You goal is to obtain a string with the maximum possible value of palindromic count.
Input
The first line contains an integer n (1 ≤ n ≤ 100 000) — the length of string s.
The second line contains string s that consists of exactly n lowercase characters of Latin alphabet.
Output
Print string t, which consists of the same set of characters (and each characters appears exactly the same number of times) as string s. Moreover, t should have the maximum possible value of palindromic count among all such strings strings.
If there are multiple such strings, print any of them.
Examples
Input
5
oolol
Output
ololo
Input
16
gagadbcgghhchbdf
Output
abccbaghghghgdfd
Note
In the first example, string "ololo" has 9 palindromic substrings: "o", "l", "o", "l", "o", "olo", "lol", "olo", "ololo". Note, that even though some substrings coincide, they are counted as many times as they appear in the resulting string.
In the second example, the palindromic count of string "abccbaghghghgdfd" is 29. | instruction | 0 | 19,437 | 0 | 38,874 |
Tags: constructive algorithms, strings
Correct Solution:
```
n = int(input())
s = [x for x in input().strip()]
s.sort()
print(*s, sep='')
``` | output | 1 | 19,437 | 0 | 38,875 |
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