message stringlengths 2 23.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Recall that string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly zero or all) characters. For example, for the string a="wowwo", the following strings are subsequences: "wowwo", "wowo", "oo", "wow", "", and others, but the following are not subsequences: "owoo", "owwwo", "ooo".
The wow factor of a string is the number of its subsequences equal to the word "wow". Bob wants to write a string that has a large wow factor. However, the "w" key on his keyboard is broken, so he types two "v"s instead.
Little did he realise that he may have introduced more "w"s than he thought. Consider for instance the string "ww". Bob would type it as "vvvv", but this string actually contains three occurrences of "w":
* "vvvv"
* "vvvv"
* "vvvv"
For example, the wow factor of the word "vvvovvv" equals to four because there are four wows:
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
Note that the subsequence "vvvovvv" does not count towards the wow factor, as the "v"s have to be consecutive.
For a given string s, compute and output its wow factor. Note that it is not guaranteed that it is possible to get s from another string replacing "w" with "vv". For example, s can be equal to "vov".
Input
The input contains a single non-empty string s, consisting only of characters "v" and "o". The length of s is at most 10^6.
Output
Output a single integer, the wow factor of s.
Examples
Input
vvvovvv
Output
4
Input
vvovooovovvovoovoovvvvovovvvov
Output
100
Note
The first example is explained in the legend. | instruction | 0 | 84,292 | 0 | 168,584 |
Tags: dp, strings
Correct Solution:
```
s=input()
n=len(s)
num=s.count('v')
left=[0]*n
right=[0]*n
#print(s.count('o'))
for i in range(1,n):
if s[i]=='v' and s[i-1]=='v':
left[i]=left[i-1]+1
else: left[i]=left[i-1]
for i in range(n-2,-1,-1):
if s[i]=='v' and s[i+1]=='v':
right[i]=right[i+1]+1
else: right[i]=right[i+1]
ans=0
#print(num)
for i in range(n):
if s[i]=='o':
ans+=(left[i]*right[i])
print(ans)
``` | output | 1 | 84,292 | 0 | 168,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recall that string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly zero or all) characters. For example, for the string a="wowwo", the following strings are subsequences: "wowwo", "wowo", "oo", "wow", "", and others, but the following are not subsequences: "owoo", "owwwo", "ooo".
The wow factor of a string is the number of its subsequences equal to the word "wow". Bob wants to write a string that has a large wow factor. However, the "w" key on his keyboard is broken, so he types two "v"s instead.
Little did he realise that he may have introduced more "w"s than he thought. Consider for instance the string "ww". Bob would type it as "vvvv", but this string actually contains three occurrences of "w":
* "vvvv"
* "vvvv"
* "vvvv"
For example, the wow factor of the word "vvvovvv" equals to four because there are four wows:
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
Note that the subsequence "vvvovvv" does not count towards the wow factor, as the "v"s have to be consecutive.
For a given string s, compute and output its wow factor. Note that it is not guaranteed that it is possible to get s from another string replacing "w" with "vv". For example, s can be equal to "vov".
Input
The input contains a single non-empty string s, consisting only of characters "v" and "o". The length of s is at most 10^6.
Output
Output a single integer, the wow factor of s.
Examples
Input
vvvovvv
Output
4
Input
vvovooovovvovoovoovvvvovovvvov
Output
100
Note
The first example is explained in the legend. | instruction | 0 | 84,293 | 0 | 168,586 |
Tags: dp, strings
Correct Solution:
```
def count( temp , n):
cw =0
co = 0
result = 0
C=0
for i in range(n):
if(temp[i]=='w'):
cw+=1
result+=C
continue
if(temp[i]=='o'):
co+=1
C+=cw
continue
else:
continue
return result
string = input()
temp = []
for i in range (len(string)-1):
if (string[i]=='v' and string[i+1]=='v'):
temp.append('w')
elif(string[i]=='o' ):
temp.append('o')
#print(temp)
print(count(temp,len(temp)))
``` | output | 1 | 84,293 | 0 | 168,587 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recall that string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly zero or all) characters. For example, for the string a="wowwo", the following strings are subsequences: "wowwo", "wowo", "oo", "wow", "", and others, but the following are not subsequences: "owoo", "owwwo", "ooo".
The wow factor of a string is the number of its subsequences equal to the word "wow". Bob wants to write a string that has a large wow factor. However, the "w" key on his keyboard is broken, so he types two "v"s instead.
Little did he realise that he may have introduced more "w"s than he thought. Consider for instance the string "ww". Bob would type it as "vvvv", but this string actually contains three occurrences of "w":
* "vvvv"
* "vvvv"
* "vvvv"
For example, the wow factor of the word "vvvovvv" equals to four because there are four wows:
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
Note that the subsequence "vvvovvv" does not count towards the wow factor, as the "v"s have to be consecutive.
For a given string s, compute and output its wow factor. Note that it is not guaranteed that it is possible to get s from another string replacing "w" with "vv". For example, s can be equal to "vov".
Input
The input contains a single non-empty string s, consisting only of characters "v" and "o". The length of s is at most 10^6.
Output
Output a single integer, the wow factor of s.
Examples
Input
vvvovvv
Output
4
Input
vvovooovovvovoovoovvvvovovvvov
Output
100
Note
The first example is explained in the legend. | instruction | 0 | 84,294 | 0 | 168,588 |
Tags: dp, strings
Correct Solution:
```
s = input()
w = o = total = 0
for i in range(len(s)):
if s[i] == 'o':
o += w
elif i > 0 and s[i - 1] == 'v':
w += 1
total += o
print(total)
``` | output | 1 | 84,294 | 0 | 168,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recall that string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly zero or all) characters. For example, for the string a="wowwo", the following strings are subsequences: "wowwo", "wowo", "oo", "wow", "", and others, but the following are not subsequences: "owoo", "owwwo", "ooo".
The wow factor of a string is the number of its subsequences equal to the word "wow". Bob wants to write a string that has a large wow factor. However, the "w" key on his keyboard is broken, so he types two "v"s instead.
Little did he realise that he may have introduced more "w"s than he thought. Consider for instance the string "ww". Bob would type it as "vvvv", but this string actually contains three occurrences of "w":
* "vvvv"
* "vvvv"
* "vvvv"
For example, the wow factor of the word "vvvovvv" equals to four because there are four wows:
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
Note that the subsequence "vvvovvv" does not count towards the wow factor, as the "v"s have to be consecutive.
For a given string s, compute and output its wow factor. Note that it is not guaranteed that it is possible to get s from another string replacing "w" with "vv". For example, s can be equal to "vov".
Input
The input contains a single non-empty string s, consisting only of characters "v" and "o". The length of s is at most 10^6.
Output
Output a single integer, the wow factor of s.
Examples
Input
vvvovvv
Output
4
Input
vvovooovovvovoovoovvvvovovvvov
Output
100
Note
The first example is explained in the legend. | instruction | 0 | 84,295 | 0 | 168,590 |
Tags: dp, strings
Correct Solution:
```
import sys
import math
#input = sys.stdin.readline
s=input()
count=0
if s[0]=="o":
count=0
else:
count=1
pre=[0]
for i in range(1,len(s)):
if s[i]=="v" and count>0:
pre.append(pre[-1]+1)
count+=1
elif s[i]=="v" and count==0:
pre.append(pre[i-1])
count=1
else:
pre.append(pre[i-1])
count=0
ans=0
for i in range(len(s)):
if s[i]=="o":
ans+=pre[i]*(pre[-1]-pre[i])
print(ans)
``` | output | 1 | 84,295 | 0 | 168,591 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recall that string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly zero or all) characters. For example, for the string a="wowwo", the following strings are subsequences: "wowwo", "wowo", "oo", "wow", "", and others, but the following are not subsequences: "owoo", "owwwo", "ooo".
The wow factor of a string is the number of its subsequences equal to the word "wow". Bob wants to write a string that has a large wow factor. However, the "w" key on his keyboard is broken, so he types two "v"s instead.
Little did he realise that he may have introduced more "w"s than he thought. Consider for instance the string "ww". Bob would type it as "vvvv", but this string actually contains three occurrences of "w":
* "vvvv"
* "vvvv"
* "vvvv"
For example, the wow factor of the word "vvvovvv" equals to four because there are four wows:
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
Note that the subsequence "vvvovvv" does not count towards the wow factor, as the "v"s have to be consecutive.
For a given string s, compute and output its wow factor. Note that it is not guaranteed that it is possible to get s from another string replacing "w" with "vv". For example, s can be equal to "vov".
Input
The input contains a single non-empty string s, consisting only of characters "v" and "o". The length of s is at most 10^6.
Output
Output a single integer, the wow factor of s.
Examples
Input
vvvovvv
Output
4
Input
vvovooovovvovoovoovvvvovovvvov
Output
100
Note
The first example is explained in the legend. | instruction | 0 | 84,296 | 0 | 168,592 |
Tags: dp, strings
Correct Solution:
```
s = input()
i = 0
count = [0, 0, 0]
ans = 0
while i < len(s):
j = i
while j + 1 < len(s) and s[j + 1] == s[i]:
j += 1
size = j - i + 1
if s[i] == 'v':
num = size - 1
count[2] += num * count[1]
count[0] += num
else:
count[1] += count[0] * size
#print(count)
i = j + 1
print(count[2])
``` | output | 1 | 84,296 | 0 | 168,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recall that string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly zero or all) characters. For example, for the string a="wowwo", the following strings are subsequences: "wowwo", "wowo", "oo", "wow", "", and others, but the following are not subsequences: "owoo", "owwwo", "ooo".
The wow factor of a string is the number of its subsequences equal to the word "wow". Bob wants to write a string that has a large wow factor. However, the "w" key on his keyboard is broken, so he types two "v"s instead.
Little did he realise that he may have introduced more "w"s than he thought. Consider for instance the string "ww". Bob would type it as "vvvv", but this string actually contains three occurrences of "w":
* "vvvv"
* "vvvv"
* "vvvv"
For example, the wow factor of the word "vvvovvv" equals to four because there are four wows:
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
Note that the subsequence "vvvovvv" does not count towards the wow factor, as the "v"s have to be consecutive.
For a given string s, compute and output its wow factor. Note that it is not guaranteed that it is possible to get s from another string replacing "w" with "vv". For example, s can be equal to "vov".
Input
The input contains a single non-empty string s, consisting only of characters "v" and "o". The length of s is at most 10^6.
Output
Output a single integer, the wow factor of s.
Examples
Input
vvvovvv
Output
4
Input
vvovooovovvovoovoovvvvovovvvov
Output
100
Note
The first example is explained in the legend.
Submitted Solution:
```
a = input()
#c = a.count('o')
n = len(a)
v,k,s = 0,0,0
l = []
l.append(0)
for i in range(n-1):
if a[i]=='v' and a[i+1]=='v': v+=1
if a[i]=='o':
k+=1
if v==s: l.append(v)
else: l.append(v+l[k-1])
s=v
l.append(v)
s=0
m = l[k+1]
#print(l,m)
for i in range(1,k+1):
s+=l[i]*(m-l[i])
print(s)
``` | instruction | 0 | 84,297 | 0 | 168,594 |
Yes | output | 1 | 84,297 | 0 | 168,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recall that string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly zero or all) characters. For example, for the string a="wowwo", the following strings are subsequences: "wowwo", "wowo", "oo", "wow", "", and others, but the following are not subsequences: "owoo", "owwwo", "ooo".
The wow factor of a string is the number of its subsequences equal to the word "wow". Bob wants to write a string that has a large wow factor. However, the "w" key on his keyboard is broken, so he types two "v"s instead.
Little did he realise that he may have introduced more "w"s than he thought. Consider for instance the string "ww". Bob would type it as "vvvv", but this string actually contains three occurrences of "w":
* "vvvv"
* "vvvv"
* "vvvv"
For example, the wow factor of the word "vvvovvv" equals to four because there are four wows:
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
Note that the subsequence "vvvovvv" does not count towards the wow factor, as the "v"s have to be consecutive.
For a given string s, compute and output its wow factor. Note that it is not guaranteed that it is possible to get s from another string replacing "w" with "vv". For example, s can be equal to "vov".
Input
The input contains a single non-empty string s, consisting only of characters "v" and "o". The length of s is at most 10^6.
Output
Output a single integer, the wow factor of s.
Examples
Input
vvvovvv
Output
4
Input
vvovooovovvovoovoovvvvovovvvov
Output
100
Note
The first example is explained in the legend.
Submitted Solution:
```
st = input()
n = len(st)
left = [0 for k in range(n) ]
right = [0 for k in range(n)]
for k in range(1,n) :
if st[k] == 'v' and (st[k] == st[k-1]) :
left[k] = left[k-1] + 1
else :
left[k] = left[k-1]
for k in range(2,n+1) :
if st[n-k] == 'v' and (st[n-k] == st[n-k+1]) :
right[n-k] = right[n-k +1] + 1
else :
right[n-k] = right[n-k+1]
c = 0
for k in range(n) :
if st[k] == 'o' :
c += left[k]*right[k]
print(c)
``` | instruction | 0 | 84,298 | 0 | 168,596 |
Yes | output | 1 | 84,298 | 0 | 168,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recall that string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly zero or all) characters. For example, for the string a="wowwo", the following strings are subsequences: "wowwo", "wowo", "oo", "wow", "", and others, but the following are not subsequences: "owoo", "owwwo", "ooo".
The wow factor of a string is the number of its subsequences equal to the word "wow". Bob wants to write a string that has a large wow factor. However, the "w" key on his keyboard is broken, so he types two "v"s instead.
Little did he realise that he may have introduced more "w"s than he thought. Consider for instance the string "ww". Bob would type it as "vvvv", but this string actually contains three occurrences of "w":
* "vvvv"
* "vvvv"
* "vvvv"
For example, the wow factor of the word "vvvovvv" equals to four because there are four wows:
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
Note that the subsequence "vvvovvv" does not count towards the wow factor, as the "v"s have to be consecutive.
For a given string s, compute and output its wow factor. Note that it is not guaranteed that it is possible to get s from another string replacing "w" with "vv". For example, s can be equal to "vov".
Input
The input contains a single non-empty string s, consisting only of characters "v" and "o". The length of s is at most 10^6.
Output
Output a single integer, the wow factor of s.
Examples
Input
vvvovvv
Output
4
Input
vvovooovovvovoovoovvvvovovvvov
Output
100
Note
The first example is explained in the legend.
Submitted Solution:
```
s=list(input().rstrip())
n=len(s)
i=c=0
d=[0 for i in range(n+1)]
while i<n:
c=0
while i<n and s[i]!='o':
d[i]=d[i-1]
c+=1
i+=1
d[i]=d[i-1]+(c-1)*(c>=1)
i+=1
ans=0
if d[-1]<d[-2]:
d[-1]=d[-2]
#print(d)
for i in range(n):
if s[i]=='o':
ans+=(d[i]-d[0])*(d[-1]-d[i+1])
print(ans)
``` | instruction | 0 | 84,299 | 0 | 168,598 |
Yes | output | 1 | 84,299 | 0 | 168,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recall that string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly zero or all) characters. For example, for the string a="wowwo", the following strings are subsequences: "wowwo", "wowo", "oo", "wow", "", and others, but the following are not subsequences: "owoo", "owwwo", "ooo".
The wow factor of a string is the number of its subsequences equal to the word "wow". Bob wants to write a string that has a large wow factor. However, the "w" key on his keyboard is broken, so he types two "v"s instead.
Little did he realise that he may have introduced more "w"s than he thought. Consider for instance the string "ww". Bob would type it as "vvvv", but this string actually contains three occurrences of "w":
* "vvvv"
* "vvvv"
* "vvvv"
For example, the wow factor of the word "vvvovvv" equals to four because there are four wows:
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
Note that the subsequence "vvvovvv" does not count towards the wow factor, as the "v"s have to be consecutive.
For a given string s, compute and output its wow factor. Note that it is not guaranteed that it is possible to get s from another string replacing "w" with "vv". For example, s can be equal to "vov".
Input
The input contains a single non-empty string s, consisting only of characters "v" and "o". The length of s is at most 10^6.
Output
Output a single integer, the wow factor of s.
Examples
Input
vvvovvv
Output
4
Input
vvovooovovvovoovoovvvvovovvvov
Output
100
Note
The first example is explained in the legend.
Submitted Solution:
```
s=input()
cnt=0
for i in range(len(s)-1):
if(s[i]=='v' and s[i+1]=='v'):
cnt+=1
cc=0
ans=0
for i in range(len(s)-2):
if(s[i]=='o'):
ans+=cc*(cnt-cc)
elif(s[i]==s[i+1] and s[i+1]=='v'):
cc+=1
print(ans)
``` | instruction | 0 | 84,300 | 0 | 168,600 |
Yes | output | 1 | 84,300 | 0 | 168,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recall that string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly zero or all) characters. For example, for the string a="wowwo", the following strings are subsequences: "wowwo", "wowo", "oo", "wow", "", and others, but the following are not subsequences: "owoo", "owwwo", "ooo".
The wow factor of a string is the number of its subsequences equal to the word "wow". Bob wants to write a string that has a large wow factor. However, the "w" key on his keyboard is broken, so he types two "v"s instead.
Little did he realise that he may have introduced more "w"s than he thought. Consider for instance the string "ww". Bob would type it as "vvvv", but this string actually contains three occurrences of "w":
* "vvvv"
* "vvvv"
* "vvvv"
For example, the wow factor of the word "vvvovvv" equals to four because there are four wows:
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
Note that the subsequence "vvvovvv" does not count towards the wow factor, as the "v"s have to be consecutive.
For a given string s, compute and output its wow factor. Note that it is not guaranteed that it is possible to get s from another string replacing "w" with "vv". For example, s can be equal to "vov".
Input
The input contains a single non-empty string s, consisting only of characters "v" and "o". The length of s is at most 10^6.
Output
Output a single integer, the wow factor of s.
Examples
Input
vvvovvv
Output
4
Input
vvovooovovvovoovoovvvvovovvvov
Output
100
Note
The first example is explained in the legend.
Submitted Solution:
```
import sys
input=sys.stdin.readline
s=list(input().rstrip())
n=len(s)
i=c=0
d=[0 for i in range(n+1)]
while i<n:
c=0
while i<n and s[i]!='o':
d[i]=d[i-1]
c+=1
i+=1
d[i]=d[i-1]+(c-1)*(c>=1)
i+=1
ans=0
#print(d)
for i in range(n):
if s[i]=='o':
ans+=(d[i]-d[0])*(d[-1]-d[i+1])
print(ans)
``` | instruction | 0 | 84,301 | 0 | 168,602 |
No | output | 1 | 84,301 | 0 | 168,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recall that string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly zero or all) characters. For example, for the string a="wowwo", the following strings are subsequences: "wowwo", "wowo", "oo", "wow", "", and others, but the following are not subsequences: "owoo", "owwwo", "ooo".
The wow factor of a string is the number of its subsequences equal to the word "wow". Bob wants to write a string that has a large wow factor. However, the "w" key on his keyboard is broken, so he types two "v"s instead.
Little did he realise that he may have introduced more "w"s than he thought. Consider for instance the string "ww". Bob would type it as "vvvv", but this string actually contains three occurrences of "w":
* "vvvv"
* "vvvv"
* "vvvv"
For example, the wow factor of the word "vvvovvv" equals to four because there are four wows:
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
Note that the subsequence "vvvovvv" does not count towards the wow factor, as the "v"s have to be consecutive.
For a given string s, compute and output its wow factor. Note that it is not guaranteed that it is possible to get s from another string replacing "w" with "vv". For example, s can be equal to "vov".
Input
The input contains a single non-empty string s, consisting only of characters "v" and "o". The length of s is at most 10^6.
Output
Output a single integer, the wow factor of s.
Examples
Input
vvvovvv
Output
4
Input
vvovooovovvovoovoovvvvovovvvov
Output
100
Note
The first example is explained in the legend.
Submitted Solution:
```
s=input()
v=0
o=0
for i in range(0,len(s)-1):
if(s[i]=='v' and s[i]==s[i+1]):
v=v+1
if(s[i]=='o'):
o=o+1
p=o*v
print(p)
``` | instruction | 0 | 84,302 | 0 | 168,604 |
No | output | 1 | 84,302 | 0 | 168,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recall that string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly zero or all) characters. For example, for the string a="wowwo", the following strings are subsequences: "wowwo", "wowo", "oo", "wow", "", and others, but the following are not subsequences: "owoo", "owwwo", "ooo".
The wow factor of a string is the number of its subsequences equal to the word "wow". Bob wants to write a string that has a large wow factor. However, the "w" key on his keyboard is broken, so he types two "v"s instead.
Little did he realise that he may have introduced more "w"s than he thought. Consider for instance the string "ww". Bob would type it as "vvvv", but this string actually contains three occurrences of "w":
* "vvvv"
* "vvvv"
* "vvvv"
For example, the wow factor of the word "vvvovvv" equals to four because there are four wows:
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
Note that the subsequence "vvvovvv" does not count towards the wow factor, as the "v"s have to be consecutive.
For a given string s, compute and output its wow factor. Note that it is not guaranteed that it is possible to get s from another string replacing "w" with "vv". For example, s can be equal to "vov".
Input
The input contains a single non-empty string s, consisting only of characters "v" and "o". The length of s is at most 10^6.
Output
Output a single integer, the wow factor of s.
Examples
Input
vvvovvv
Output
4
Input
vvovooovovvovoovoovvvvovovvvov
Output
100
Note
The first example is explained in the legend.
Submitted Solution:
```
s = input()
string = s
n = len(s)
a = []
i = 0
while i < n:
j = 1
while i < n-1 and s[i] == s[i+1]:
i += 1
j += 1
a.append(j)
i += 1
s = 0
b = []
n = len(a)
if string[0] == 'o':
a = [0]+a
n += 1
for i in range(n):
if i%2 == 0:
a[i] -= 1
for i in range(n-1,-1,-1):
if i%2 == 0:
s += a[i]
b.append(s)
b.reverse()
b = list(b)
#print(b)
s = 0
ans = 0
for i in range(n):
if i%2:
ans += s*b[i]*a[i]
else:
s += a[i]
print(ans)
``` | instruction | 0 | 84,303 | 0 | 168,606 |
No | output | 1 | 84,303 | 0 | 168,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recall that string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly zero or all) characters. For example, for the string a="wowwo", the following strings are subsequences: "wowwo", "wowo", "oo", "wow", "", and others, but the following are not subsequences: "owoo", "owwwo", "ooo".
The wow factor of a string is the number of its subsequences equal to the word "wow". Bob wants to write a string that has a large wow factor. However, the "w" key on his keyboard is broken, so he types two "v"s instead.
Little did he realise that he may have introduced more "w"s than he thought. Consider for instance the string "ww". Bob would type it as "vvvv", but this string actually contains three occurrences of "w":
* "vvvv"
* "vvvv"
* "vvvv"
For example, the wow factor of the word "vvvovvv" equals to four because there are four wows:
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
* "vvvovvv"
Note that the subsequence "vvvovvv" does not count towards the wow factor, as the "v"s have to be consecutive.
For a given string s, compute and output its wow factor. Note that it is not guaranteed that it is possible to get s from another string replacing "w" with "vv". For example, s can be equal to "vov".
Input
The input contains a single non-empty string s, consisting only of characters "v" and "o". The length of s is at most 10^6.
Output
Output a single integer, the wow factor of s.
Examples
Input
vvvovvv
Output
4
Input
vvovooovovvovoovoovvvvovovvvov
Output
100
Note
The first example is explained in the legend.
Submitted Solution:
```
a = [0]
for x in input().split('o'):
a += a[-1]+max(0, len(x)-1),
print(a)
print(sum(i*(a[-1]-i)for i in a))
``` | instruction | 0 | 84,304 | 0 | 168,608 |
No | output | 1 | 84,304 | 0 | 168,609 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1 | instruction | 0 | 85,119 | 0 | 170,238 |
Tags: data structures, dp
Correct Solution:
```
import io
import os
from collections import Counter, defaultdict, deque
def solve(N, S):
# Same color must be already sorted since they can't be swapped with each other
# Greedily build increasing subsequences
indices = [[0]]
# last value -> which list
for i, x in enumerate(S[1:], 1):
for l in indices:
if S[l[-1]] <= x:
l.append(i)
break
else:
indices.append([i])
ans = [None for i in range(N)]
for color, l in enumerate(indices):
for i in l:
ans[i] = color
# Format for E2
possible = str(len(indices))
return possible + "\n" + " ".join(str(x + 1) for x in ans)
# Format for E1
if len(indices) <= 2:
return "YES\n" + "".join(map(str, ans))
else:
return 'NO'
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
(N,) = [int(x) for x in input().split()]
S = input().decode().rstrip()
ans = solve(N, S)
print(ans)
``` | output | 1 | 85,119 | 0 | 170,239 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1 | instruction | 0 | 85,120 | 0 | 170,240 |
Tags: data structures, dp
Correct Solution:
```
n = int(input())
s = str(input())
lit = ['Z']*26
wyn = ''
m = -1
for x in s:
for y in range(26):
if lit[y] <= x:
wyn += str(y + 1)+ ' '
lit[y] = x
if y+1 >= m:
m = y+1
break
print(m)
print(wyn)
``` | output | 1 | 85,120 | 0 | 170,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1 | instruction | 0 | 85,121 | 0 | 170,242 |
Tags: data structures, dp
Correct Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def main():
n=II()
s=SI()
col=[0]*27
ans=[0]*n
a=ord("a")
for i,c in enumerate(s):
code=ord(c)-a
cur=max(col[code+1:])+1
ans[i]=cur
col[code]=cur
print(max(col))
print(*ans)
main()
``` | output | 1 | 85,121 | 0 | 170,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1 | instruction | 0 | 85,122 | 0 | 170,244 |
Tags: data structures, dp
Correct Solution:
```
class RangeMinimumQuery:
def __init__(self, n, func=min, inf=float("inf")):
self.n0 = 2**(n-1).bit_length()
self.op = func
self.inf = inf
self.data = [self.inf]*(2*self.n0)
def query(self, l,r):
l += self.n0
r += self.n0
res = self.inf
while l < r:
if r&1:
r -= 1
res = self.op(res, self.data[r-1])
if l&1:
res = self.op(res, self.data[l-1])
l += 1
l >>=1
r >>=1
return res
def update(self, i, x):
i += self.n0-1
self.data[i] = x
while i:
i = ~-i//2
self.data[i] = self.op(self.data[2*i+1], self.data[2*i+2])
n = int(input())
s = input()
a = [(c,i) for i, c in enumerate(s)]
a.sort()
RMQ = RangeMinimumQuery(n,max,0)
col = [0]*n
for _, i in a:
c = RMQ.query(i, n)+1
col[i] = c
RMQ.update(i, c)
max_col = RMQ.query(0, n)
print(max_col)
print(*col)
``` | output | 1 | 85,122 | 0 | 170,245 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1 | instruction | 0 | 85,123 | 0 | 170,246 |
Tags: data structures, dp
Correct Solution:
```
n = int(input())
a = list(map(lambda c: ord(c)-97, input()))
color = [0]*26
ans = [0]*n
last = -1
for i, c in enumerate(a):
col = 0
if last <= c:
last = c
if color[c] == 0:
col = 1
else:
col = color[c] & (-color[c])
else:
col = 1
for j in range(last, c, -1):
while col & color[j]:
col <<= 1
color[c] |= col
ans[i] = len(bin(col)) - 2
print(max(ans))
print(*ans)
``` | output | 1 | 85,123 | 0 | 170,247 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1 | instruction | 0 | 85,124 | 0 | 170,248 |
Tags: data structures, dp
Correct Solution:
```
input()
arr = input().strip()
ans = []
color = 0
mx = [0 for i in range(26)]
for i in arr:
c = ord(i) - 97
_max = 0
for j in range(c+1,26):
_max = max(_max,mx[j])
ans.append(_max + 1)
color = max(color,ans[-1])
mx[c] = max(mx[c],_max+1)
print(color)
for i in ans:
print(i,end=' ')
``` | output | 1 | 85,124 | 0 | 170,249 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1 | instruction | 0 | 85,125 | 0 | 170,250 |
Tags: data structures, dp
Correct Solution:
```
from sys import stdin
input = stdin.readline
n = int(input()) ; s = input().strip()
tmp = [s[0]] + ['' for i in range(25)] ; ans = [1]
for i in range(1,len(s)):
for j in range(26):
if tmp[j] <= s[i]:
ans.append(j+1)
tmp[j] = s[i]
break
print(max(ans)) ; print(*ans)
``` | output | 1 | 85,125 | 0 | 170,251 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1 | instruction | 0 | 85,126 | 0 | 170,252 |
Tags: data structures, dp
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
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")
#-------------------------------------------------------------------------
#mod = 9223372036854775807
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a+b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=10**6, func=lambda a, b: min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
MOD=10**9+7
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
mod=10**9+7
omod=998244353
#-------------------------------------------------------------------------
prime = [True for i in range(10)]
pp=[0]*10
def SieveOfEratosthenes(n=10):
p = 2
c=0
while (p * p <= n):
if (prime[p] == True):
c+=1
for i in range(p, n+1, p):
pp[i]+=1
prime[i] = False
p += 1
#---------------------------------Binary Search------------------------------------------
def binarySearch(arr, n, key):
left = 0
right = n-1
mid = 0
res=arr[n-1]
while (left <= right):
mid = (right + left)//2
if (arr[mid] >= key):
res=arr[mid]
right = mid-1
else:
left = mid + 1
return res
def binarySearch1(arr, n, key):
left = 0
right = n-1
mid = 0
res=arr[0]
while (left <= right):
mid = (right + left)//2
if (arr[mid] > key):
right = mid-1
else:
res=arr[mid]
left = mid + 1
return res
#---------------------------------running code------------------------------------------
n=int(input())
s=input()
curr=['']*100
m=1
res=[]
for i in range (n):
for j in range (100):
if s[i]>=curr[j]:
curr[j]=s[i]
res.append(j+1)
break
print(max(res))
print(*res)
``` | output | 1 | 85,126 | 0 | 170,253 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
#threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
#sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**30, func=lambda a, b: min(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b:a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] > k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
#-----------------------------------------trie---------------------------------
class Node:
def __init__(self, data):
self.data = data
self.count=0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count+=1
if not val:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += 1
self.temp.data = pre_xor
def query(self, xor):
self.temp = self.root
for i in range(31, -1, -1):
val = xor & (1 << i)
if not val:
if self.temp.left and self.temp.left.count>0:
self.temp = self.temp.left
elif self.temp.right:
self.temp = self.temp.right
else:
if self.temp.right and self.temp.right.count>0:
self.temp = self.temp.right
elif self.temp.left:
self.temp = self.temp.left
self.temp.count-=1
return xor ^ self.temp.data
#-------------------------bin trie-------------------------------------------
n=int(input())
d=dict()
for i in range(97,124):
d[chr(i)]=0
l=input()
d[l[0]]+=1
ans=[0]*n
ans[0]=d[l[0]]
for i in range(1,n):
for j in sorted(d.keys(),reverse=True):
if j>l[i]:
ans[i]=max(ans[i],d[j]+1)
else:
break
d[l[i]]=ans[i]
print(max(ans))
print(*ans,sep=" ")
``` | instruction | 0 | 85,127 | 0 | 170,254 |
Yes | output | 1 | 85,127 | 0 | 170,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1
Submitted Solution:
```
n=int(input())
s=input()
rama=[]
for i in range(n):
rama.append(ord(s[i])-96)
visit=[1 for i in range(27)]
a=[]
maxi=0
for i in range(n):
p=visit[rama[i]]
a.append(p)
maxi=max(maxi,p)
for j in range(rama[i]):
visit[j]=max(visit[j],p+1)
print(maxi)
for i in a:
print(i,end=' ')
``` | instruction | 0 | 85,128 | 0 | 170,256 |
Yes | output | 1 | 85,128 | 0 | 170,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1
Submitted Solution:
```
import sys
# inf = open('input.txt', 'r')
# reader = (line.rstrip() for line in inf)
reader = (line.rstrip() for line in sys.stdin)
input = reader.__next__
def ceil(tails, L, R, key):
while L + 1 < R:
m = (L + R) // 2
if key < tails[m]:
L = m
else:
R = m
return R
def LIS(a, n):
tails = [0] * (n + 1)
tails[0] = a[0]
seq_len = 1 # LIS for a[:1]
for i in range(1, n):
if (a[i] > tails[0]): # edit for other order
tails[0] = a[i] # new LIS start
ans.append(1)
elif (a[i] < tails[seq_len - 1]): # edit for other order
tails[seq_len] = a[i] # extend existing LIS
seq_len += 1
ans.append(seq_len)
else: # find LIS that ends in a[i] and update tail value for it
pos = ceil(tails, -1, seq_len - 1, a[i])
tails[pos] = a[i]
ans.append(pos + 1)
return seq_len
n = int(input())
s = input()
ans = [1]
res = LIS(s, n)
print(res)
print(*ans)
# inf.close()
``` | instruction | 0 | 85,129 | 0 | 170,258 |
Yes | output | 1 | 85,129 | 0 | 170,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1
Submitted Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from math import gcd, ceil
def prod(a, mod=10**9+7):
ans = 1
for each in a:
ans = (ans * each) % mod
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if not True else 1):
n = int(input())
#n, k = map(int, input().split())
#a, b = map(int, input().split())
#c, d = map(int, input().split())
#a = list(map(int, input().split()))
#b = list(map(int, input().split()))
s = input()
ans = [0]*n
colors = ['a']*50
for i in range(n):
for j in range(50):
if ord(s[i]) >= ord(colors[j]):
colors[j] = s[i]
ans[i] = j+1
break
print(max(ans))
print(*ans)
``` | instruction | 0 | 85,130 | 0 | 170,260 |
Yes | output | 1 | 85,130 | 0 | 170,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1
Submitted Solution:
```
def opp(c):
if(c=='0'):
return '1'
return '0'
n=int(input())
s=input()
t=s[0]
pos=[0]
for i in range(1,len(s)):
if(s[i]<t):
pos.append(1)
else:
pos.append(0)
t=max(t,s[i])
temp=0
col=[0]*n
for i in range(1,len(s)):
if(ord(s[i])>ord(s[i-1])):
if(pos[i]==0):
if(col[i-1]==0):
col[i]=0
else:
col[i]=col[i-1]-1
else:
col[i]=col[i-1]
elif(ord(s[i])==ord(s[i-1])):
col[i]=col[i-1]
else:
col[i]=col[i-1]+1
for i in range(0,len(col)):
col[i]=(col[i])+1
print(max(col))
print(*col)
``` | instruction | 0 | 85,131 | 0 | 170,262 |
No | output | 1 | 85,131 | 0 | 170,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1
Submitted Solution:
```
n=int(input())
s=input()
rama=[]
for i in range(n):
rama.append(ord(s[i])-96)
visit=[1 for i in range(27)]
a=[]
for i in range(n):
p=visit[rama[i]]
a.append(p)
for j in range(rama[i]):
visit[j]=max(visit[j],p+1)
for i in a:
print(i,end='')
``` | instruction | 0 | 85,132 | 0 | 170,264 |
No | output | 1 | 85,132 | 0 | 170,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1
Submitted Solution:
```
from sys import stdin
from collections import deque
mod = 10**9 + 7
import sys
# sys.setrecursionlimit(10**6)
from queue import PriorityQueue
# def rl():
# return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
import heapq
# input = lambda: sys.stdin.readline().rstrip()
input = lambda : sys.stdin.readline().rstrip()
from sys import stdin, stdout
from heapq import heapify, heappush, heappop
n = int(input())
s = input().strip()
l = [97 - (ord(i)-96) for i in s]
dp = [float("inf")]*n
ans = []
for i in l:
posn = bisect_left(dp,i)
print(posn)
ans.append(posn+1)
dp[posn] = i
print(max(ans))
print(*ans)
``` | instruction | 0 | 85,133 | 0 | 170,266 |
No | output | 1 | 85,133 | 0 | 170,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters the minimum number of colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
In the first line print one integer res (1 ≤ res ≤ n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.
In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array c of length n, where 1 ≤ c_i ≤ res and c_i means the color of the i-th character.
Examples
Input
9
abacbecfd
Output
2
1 1 2 1 2 1 2 1 2
Input
8
aaabbcbb
Output
2
1 2 1 2 1 2 1 1
Input
7
abcdedc
Output
3
1 1 1 1 1 2 3
Input
5
abcde
Output
1
1 1 1 1 1
Submitted Solution:
```
n = int(input())
s = list(str(input()))
letters = [[y for y in range(26)] for x in range(26)]
wyn = ''
wyn_int = -1
col = 1
for x in range(0, len(s)):
curr_let = ord(s[x])-97
if x != 0:
prev_let = ord(s[x-1])-97
if prev_let == curr_let:
wyn += str(wyn_int+1)+ ' '
continue
curr_col = -1
m_color = 0
for y in range(0, len(letters[curr_let])):
if letters[curr_let][y] != -1:
wyn += str(letters[curr_let][y]+1) + ' '
for z in range(curr_let-1, -1, -1):
letters[z][letters[curr_let][y]] = -1
wyn_int = letters[curr_let][y]
letters[curr_let][y] = -1
break
print(max(wyn))
print(wyn)
``` | instruction | 0 | 85,134 | 0 | 170,268 |
No | output | 1 | 85,134 | 0 | 170,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 85,167 | 0 | 170,334 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
def stringflip(l,s):
news=""
for i in range(l):
if s[l-i-1]=="0":
news+="1"
else:
news+="0"
if l!=n:
news+=s[l:]
return news
def flip(n,a,b):
if a==b:
print(0)
return
#a to b
moves=0
array=[]
for i in range(n):
if a[0]==b[n-i-1]:
array.append(1)
array.append(n-i)
moves+=2
stringflip(1,a)
a=stringflip(n-i,a)
else:
array.append(n-i)
moves+=1
a=stringflip(n-i,a)
print(moves,end=" ")
for x in array:
print(x,end=" ")
print()
return
t=int(input())
a=[]
for j in range(t):
n=int(input())
d=input()
e=input()
a.append([n,d,e])
for x in a:
flip(*x)
``` | output | 1 | 85,167 | 0 | 170,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 85,168 | 0 | 170,336 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
from sys import stdin
# input=stdin.buffer.readline
input=lambda : stdin.readline().strip()
lin=lambda :list(map(int,input().split()))
iin=lambda :int(input())
main=lambda :map(int,input().split())
from math import ceil,sqrt,factorial,log
from collections import deque
from bisect import bisect_left
mod=998244353
mod=1000000007
def gcd(a,b):
a,b=max(a,b),min(a,b)
while a%b!=0:
a,b=b,a%b
return b
def moduloinverse(a):
return(pow(a,mod-2,mod))
def solve(we):
n=iin()
a=list(input())
b=list(input())
t=-1
ans=[]
a=a+['0']
for i in range(n+1):
if a[i]=='0':
if t!=-1 and t<i-1:
ans.append(t+1)
ans.append(i)
elif t==-1 and i!=0:
ans.append(i)
t=i
t='0'
for i in range(n-1,-1,-1):
if b[i]!=t:
ans.append(i+1)
if t=='1':
t='0'
else:
t='1'
print(str(len(ans)),*ans)
qwe=1
qwe=iin()
for _ in range(qwe):
solve(_+1)
``` | output | 1 | 85,168 | 0 | 170,337 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 85,169 | 0 | 170,338 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
#dt = {} for i in x: dt[i] = dt.get(i,0)+1
import sys;input = sys.stdin.readline
inp,ip = lambda :int(input()),lambda :[int(w) for w in input().split()]
for _ in range(inp()):
n = inp()
a = input().strip()
b = input().strip()
ans = []
for i in range(n-1,-1,-1):
if a[i] != b[i]:
ans.extend([i+1,1,i+1])
print(len(ans),*ans)
``` | output | 1 | 85,169 | 0 | 170,339 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 85,170 | 0 | 170,340 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
# code by RAJ BHAVSAR
for _ in range(int(input())):
n = int(input())
a = list(str(input()))
b = list(str(input()))
res = []
for i in range(n):
if(a[i] != b[i]):
res += [i+1,1,i+1]
print(len(res),*res)
``` | output | 1 | 85,170 | 0 | 170,341 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 85,171 | 0 | 170,342 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
import sys
ii = lambda: sys.stdin.readline().strip()
idata = lambda: [int(x) for x in ii().split()]
def solve():
n = int(ii())
data_a = ii()
data_b = ii()
ans = []
for i in range(n - 1):
if data_a[i] != data_a[i + 1]:
ans += [i + 1]
if data_a[-1] == '1':
ans += [n]
d = 0
for i in range(n - 1, -1, -1):
if d % 2 != int(data_b[i]):
d = (d + 1) % 2
ans += [i + 1]
print(len(ans), *ans)
return
for t in range(int(ii())):
solve()
``` | output | 1 | 85,171 | 0 | 170,343 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 85,172 | 0 | 170,344 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
t1=int(input())
for _ in range(t1):
n=int(input())
a2=input()
a=[]
for i in range(n):
a.append(a2[i])
b=input()
ans=[]
for i in range(n-1,-1,-1):
if a[i]!=b[i]:
temp=[]
for j in range(i+1):
temp.append(a[j])
temp2=[]
for j in range(i,-1,-1):
if temp[j]=='0':
temp2.append('1')
else:
temp2.append('0')
if a[0]!=b[i]:
ans.append(str(i+1))
for j in range(i+1):
a[j]=temp2[j]
else:
ans.append('1')
ans.append(str(i+1))
for j in range(i+1):
a[j]=temp2[j]
print(len(ans))
if len(ans)>0:
print(' '.join(ans))
``` | output | 1 | 85,172 | 0 | 170,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 85,173 | 0 | 170,346 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
res=[]
a=input()
b=input()
i=n-1
while i>0:
if a[i]!=b[i]:
res.append(str(i+1))
res.append("1")
res.append(str(i + 1))
i-=1
if a[0]!=b[0]:
res.append("1")
print(len(res),end=" ")
for x in res:
print(x,end=" ")
print()
``` | output | 1 | 85,173 | 0 | 170,347 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged. | instruction | 0 | 85,174 | 0 | 170,348 |
Tags: constructive algorithms, data structures, strings
Correct Solution:
```
from math import *
from collections import *
from random import *
from bisect import *
import sys
input=sys.stdin.readline
d={'1':'0','0':'1'}
t=int(input())
while(t):
t-=1
n=int(input())
a=input().rstrip("\n")
b=input().rstrip("\n")
a=list(a)
b=list(b)
r=[]
lp=n-1
tu=0
while(lp>=0):
if(a[lp]==b[lp]):
lp-=1
continue
if(a[0]==b[lp]):
r.append(1)
a[0]=d[a[0]]
#print(a)
tu+=1
continue
else:
r.append(lp+1)
for i in range(lp+1):
a[i]=d[a[i]]
a=a[:lp+1][-1::-1]+a[lp+1:]
#print(a,lp)
lp-=1
#print(a,b)
print(len(r),end=" ")
print(*r)
``` | output | 1 | 85,174 | 0 | 170,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
import math
import sys
#input=sys.stdin.readline
t=int(input())
#t=1
for _ in range(t):
n=int(input())
#n,m=map(int,input().split())
#l1=list(map(int,input().split()))
a=input()
a+='0'
b=input()
b+='0'
ans=[]
ans1=[]
for i in range(n):
if a[i]=='1' and a[i+1]=='0' :
ans.append(i+1)
if a[i]=='0' and a[i+1]=='1':
ans.append(i+1)
if b[i]=='1' and b[i+1]=='0' :
ans1.append(i+1)
if b[i]=='0' and b[i+1]=='1':
ans1.append(i+1)
if len(ans)+len(ans1)==0:
print(0)
else:
print(len(ans)+len(ans1),*ans,*ans1[::-1])
``` | instruction | 0 | 85,175 | 0 | 170,350 |
Yes | output | 1 | 85,175 | 0 | 170,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
s=input()
target=input()
ans=[]
for i in range(n-1,-1,-1):
if s[i]==target[i]:
continue
ans.append(i+1)
ans.append(1)
ans.append(i+1)
print(len(ans),end=' ')
print(*ans)
``` | instruction | 0 | 85,176 | 0 | 170,352 |
Yes | output | 1 | 85,176 | 0 | 170,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
a = input()
b = input()
k = 0
ans = []
for i in range(n, 0, -1):
if a[i - 1] == b[i - 1]:
continue
k += 3
ans.append(i)
ans.append(1)
ans.append(i)
ans = [k] + ans
print(*ans)
``` | instruction | 0 | 85,177 | 0 | 170,354 |
Yes | output | 1 | 85,177 | 0 | 170,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
def do_reverse(s):
s1 = ''
for i in s:
if i=='0':
s1='1'+s1
else:
s1='0'+s1
return s1
for _ in range(int(input())):
n = int(input())
a = input()
b = input()
if a==b:
print(0)
continue
l = []
for i in range(n-1,-1,-1):
if a[i]==b[i]:
continue
else:
if a[i]==a[0]:
a = do_reverse(a[:i+1])+a[i+1:]
l.append(i+1)
else:
l.append(1)
a = a[i]+a[1:]
a = do_reverse(a[:i+1])+a[i+1:]
l.append(i+1)
print(len(l),*l)
``` | instruction | 0 | 85,178 | 0 | 170,356 |
Yes | output | 1 | 85,178 | 0 | 170,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
# a="1011001"
# b=a[5:0:-1]
# print(a)
# print(b)
t=int(input())
for i in range(t):
n=int(input())
a=input()
b=input()
li=[]
for i in range(n-1,-1,-1):
if a[i]==b[i]:
continue
else:
if b[i]!=a[0]:
li.append(i+1)
c=""
for j in range(i,-1,-1):
if(a[i]=='1'):
c+='0'
else: c+='1'
c+=a[i+1:n]
a=c
else:
c=""
li.append(1)
if(a[0]=='0') : c+='1'
else: c+='0'
c+=a[1:n]
a=c
li.append(i+1)
c=""
for j in range(i,-1,-1):
if(a[i]=='1'):
c+='0'
else: c+='1'
c+=a[i+1:n]
a=c
print(len(li),end=" ")
for i in range(0,len(li)):
print(li[i],end=" ")
print()
``` | instruction | 0 | 85,179 | 0 | 170,358 |
No | output | 1 | 85,179 | 0 | 170,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
from sys import stdin
input = stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
a = input().rstrip()
b = input().rstrip()
out = []
for i in range(n - 1):
if a[n - i - 1] == b[n - i - 1]:
continue
if b[n - i - 1] == a[i]:
out.append(1)
out.append(n - i)
out.append(n - i - 1)
else:
out.append(n - i)
out.append(n - i - 1)
if a[-1] != b[0]:
out.append(1)
print(len(out), *out)
``` | instruction | 0 | 85,180 | 0 | 170,360 |
No | output | 1 | 85,180 | 0 | 170,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
from sys import stdin,stdout
nmbr=lambda:int(stdin.readline())
lst=lambda:list(map(int, stdin.readline().split()))
for _ in range(nmbr()):
n=nmbr()
a=input()
b=input()
f=a[0]
ans=[]
for i in range(n-1,-1,-1):
if a[i]==b[i]:continue
if b[i]==f:
ans+=[1,i+1]
f='1' if a[i]=='0' else '0'
else:
ans+=[1+i]
f = '1' if a[i] == '0' else '0'
print(len(ans))
if ans:print(*ans)
``` | instruction | 0 | 85,181 | 0 | 170,362 |
No | output | 1 | 85,181 | 0 | 170,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.
There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.
For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.
Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.
Input
The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.
The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.
The next two lines contain two binary strings a and b of length n.
It is guaranteed that the sum of n across all test cases does not exceed 1000.
Output
For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.
Example
Input
5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1
Output
3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1
Note
In the first test case, we have 01→ 11→ 00→ 10.
In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.
In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
Submitted Solution:
```
#codeforces round 658 div 2 Problem C1
import sys
def input():
return sys.stdin.readline()[:-1]
def getInt():
return int(input())
def getIntIter():
return map(int, input().split())
def getIntList():
return list(getIntIter())
for _ in range(getInt()):
n = getInt()
a = list(x=="1" for x in input())
b = list(x=="1" for x in input())
aLast = a[-1]
bLast = b[-1]
while aLast == bLast and len(a) > 1:
a = a[:-1]
b = b[:-1]
aLast = a[-1]
bLast = b[-1]
outList = []
bit = 0
while len(a) > 1:
if a[0] ^ bit == b[-1]:
outList.append("1")
outList.append(str(len(a)))
b = b[:-1]
a = a[-1:0:-1]
bit = not bit
if a != b:
outList.append("1")
print(len(outList),' '.join(outList))
``` | instruction | 0 | 85,182 | 0 | 170,364 |
No | output | 1 | 85,182 | 0 | 170,365 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dreamoon has a string s and a pattern string p. He first removes exactly x characters from s obtaining string s' as a result. Then he calculates <image> that is defined as the maximal number of non-overlapping substrings equal to p that can be found in s'. He wants to make this number as big as possible.
More formally, let's define <image> as maximum value of <image> over all s' that can be obtained by removing exactly x characters from s. Dreamoon wants to know <image> for all x from 0 to |s| where |s| denotes the length of string s.
Input
The first line of the input contains the string s (1 ≤ |s| ≤ 2 000).
The second line of the input contains the string p (1 ≤ |p| ≤ 500).
Both strings will only consist of lower case English letters.
Output
Print |s| + 1 space-separated integers in a single line representing the <image> for all x from 0 to |s|.
Examples
Input
aaaaa
aa
Output
2 2 1 1 0 0
Input
axbaxxb
ab
Output
0 1 1 2 1 1 0 0
Note
For the first sample, the corresponding optimal values of s' after removal 0 through |s| = 5 characters from s are {"aaaaa", "aaaa", "aaa", "aa", "a", ""}.
For the second sample, possible corresponding optimal values of s' are {"axbaxxb", "abaxxb", "axbab", "abab", "aba", "ab", "a", ""}. | instruction | 0 | 85,328 | 0 | 170,656 |
Tags: dp, strings
Correct Solution:
```
s, p = input(), input()
n, m = len(s) + 1, len(p)
d = [[0] * n for t in range(n)]
for x in range(1, n):
i, j = x, m
while i and j:
j -= s[i - 1] == p[j - 1]
i -= 1
if not j:
for y in range(i + 1): d[x][y + x - i - m] = d[i][y] + 1
for y in range(x): d[x][y] = max(d[x][y], d[x - 1][y])
print(*d[-1])
``` | output | 1 | 85,328 | 0 | 170,657 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given string s. Your task is to determine if the given string s contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order).
Input
The only line of input contains a string s of length between 1 and 105 consisting of uppercase Latin letters.
Output
Print "YES" (without the quotes), if string s contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise.
Examples
Input
ABA
Output
NO
Input
BACFAB
Output
YES
Input
AXBYBXA
Output
NO
Note
In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".
In the second sample test there are the following occurrences of the substrings: BACFAB.
In the third sample test there is no substring "AB" nor substring "BA". | instruction | 0 | 85,337 | 0 | 170,674 |
Tags: brute force, dp, greedy, implementation, strings
Correct Solution:
```
s = input()
pA = -10
pB = -10
al = []
bl = []
for i in range(len(s)):
if s[i] == 'A' and pA == -10:
pA = i
if s[i] == 'B' and pB == -10:
pB = i
if s[i] == 'A':
if (i - pB == 1):
bl.append(i)
pA = i
if s[i] == 'B':
if (i - pA == 1):
al.append(i)
pB = i
#print(al, bl)
br = False
for i in bl:
if br:
break
tS = set(al[:])
if i-1 in tS:
tS.remove(i-1)
if i+1 in tS:
tS.remove(i+1)
if len(tS) > 0:
br = True
break
if br:
print('YES')
else:
print('NO')
``` | output | 1 | 85,337 | 0 | 170,675 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given string s. Your task is to determine if the given string s contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order).
Input
The only line of input contains a string s of length between 1 and 105 consisting of uppercase Latin letters.
Output
Print "YES" (without the quotes), if string s contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise.
Examples
Input
ABA
Output
NO
Input
BACFAB
Output
YES
Input
AXBYBXA
Output
NO
Note
In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".
In the second sample test there are the following occurrences of the substrings: BACFAB.
In the third sample test there is no substring "AB" nor substring "BA". | instruction | 0 | 85,338 | 0 | 170,676 |
Tags: brute force, dp, greedy, implementation, strings
Correct Solution:
```
import re
mystr = input()
r1 = mystr.find('AB')
r2 = mystr.rfind('BA')
if r1 != -1 and r2 != -1 and r1 != r2-1 and r2 != r1-1:
print('YES')
else:
r1 = mystr.find('BA')
r2 = mystr.rfind('AB')
if r1 != -1 and r2 != -1 and r1 != r2-1 and r2 != r1-1:
print('YES')
else:
print('NO')
``` | output | 1 | 85,338 | 0 | 170,677 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given string s. Your task is to determine if the given string s contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order).
Input
The only line of input contains a string s of length between 1 and 105 consisting of uppercase Latin letters.
Output
Print "YES" (without the quotes), if string s contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise.
Examples
Input
ABA
Output
NO
Input
BACFAB
Output
YES
Input
AXBYBXA
Output
NO
Note
In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".
In the second sample test there are the following occurrences of the substrings: BACFAB.
In the third sample test there is no substring "AB" nor substring "BA". | instruction | 0 | 85,339 | 0 | 170,678 |
Tags: brute force, dp, greedy, implementation, strings
Correct Solution:
```
s = input()
ab, ba = [], []
for i in range(len(s) - 1):
if s[i:i+2] == 'AB': ab.append(i)
elif s[i:i+2] == 'BA': ba.append(i)
for u in ab:
for v in ba:
if abs(u-v) > 1:
print('YES')
exit()
print('NO')
``` | output | 1 | 85,339 | 0 | 170,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given string s. Your task is to determine if the given string s contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order).
Input
The only line of input contains a string s of length between 1 and 105 consisting of uppercase Latin letters.
Output
Print "YES" (without the quotes), if string s contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise.
Examples
Input
ABA
Output
NO
Input
BACFAB
Output
YES
Input
AXBYBXA
Output
NO
Note
In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".
In the second sample test there are the following occurrences of the substrings: BACFAB.
In the third sample test there is no substring "AB" nor substring "BA". | instruction | 0 | 85,340 | 0 | 170,680 |
Tags: brute force, dp, greedy, implementation, strings
Correct Solution:
```
s = input()
matches = {'AB': [], 'BA': []}
for i in range(len(s) - 1):
substring = s[i:i + 2]
if substring in matches:
matches[substring].append(i)
if not matches['AB'] or not matches['BA']:
print('NO')
elif abs(max(matches['AB']) - min(matches['BA'])) > 1:
print('YES')
elif abs(min(matches['AB']) - max(matches['BA'])) > 1:
print('YES')
else:
print('NO')
``` | output | 1 | 85,340 | 0 | 170,681 |
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