codeforces676C Vasya and String

Vasya and String

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.

Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve?

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n) — the length of the string and the maximum number of characters to change.

The second line contains the string, consisting of letters 'a' and 'b' only.

Output

Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than k characters.

Examples

input

4 2abba

output

4

input

8 1aabaabaa

output

5

Note

In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".

In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".

题意：

给出一个长度为n的字符串，只有字符'a'和'b'。最多能改变k个字符，即把'a'变成'b'或把'b'变成'a'。

问改变后的最长连续相同字符的字串长度为多少。

话说开始我想dp来着，但是数据太大。。。后来看到别人写的题解才恍然大悟

思路：用两个指针，i=j=1，保持它们之间的a个数是k，算出最大值，初始化（i，j）然后保持它们之间b个数是k，算出最大值。

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<string>#include<vector>#include<stack>#include<set>#include<map>#include<queue>#include<algorithm>using namespace std;char c[100005];int pa[100005],pb[100005];int main(){int n,k;int i,j;scanf("%d %d",&n,&k);int left,right;getchar();for(i=1;i<=n;i++){scanf("%c",&c[i]);if(c[i]=='a')pa[i]=pa[i-1]+1;elsepa[i]=pa[i-1];if(c[i]=='b')pb[i]=pb[i-1]+1;elsepb[i]=pb[i-1];}left=1;right=1;int ans=0;while(right<=n){while(pa[right]-pa[left-1]<=k&&right<=n){right++;}ans=max(ans,right-left);left++;}left=right=1;while(right<=n){while(pb[right]-pb[left-1]<=k&&right<=n){right++;}ans=max(ans,right-left);left++;}printf("%d\n",ans);    return 0;}




                                             
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