CodeForces - 873B 01串

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本文介绍了一种算法，用于在由0和1组成的字符串中找到最长的平衡子串（即0和1数量相等的子串）。通过将0映射为-1、1映射为+1，并记录下相同数值的最远位置来解决此问题。

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B. Balanced Substring

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.

You have to determine the length of the longest balanced substring of s.

Input

The first line contains n (1 ≤ n ≤ 100000) — the number of characters in s.

The second line contains a string s consisting of exactly n characters. Only characters 0 and 1 can appear in s.

Output

If there is no non-empty balanced substring in s, print 0. Otherwise, print the length of the longest balanced substring.

Examples

input

Copy

8
11010111

output

Copy

input

Copy

3
111

output

Copy

Note

In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.

In the second example it's impossible to find a non-empty balanced substring.

思路： 1为+1,0为-1，寻找值一样的最长间隔

#include <cstring>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string.h>
using namespace std;

int main(){
    int n;
    string s;
    cin>>n>>s;
    int sum=0;
    int idx[100005],iidx[100005];
    const int inf=0x3f3f3f3f;
    memset(idx,inf,sizeof(idx));
    memset(iidx,inf,sizeof(iidx));
    int ans=0;
    for(int i=0,j=1;i<n;i++,j++){
        if(s[i]=='0') sum--;
        else sum++;
        if(sum>0&&idx[sum]==inf) idx[sum]=j;
        else if(sum<0&&iidx[-sum]==inf) iidx[-sum]=j;
        if(sum>0)
            ans=max(ans,j-idx[sum]);
        else if(sum==0) ans=max(ans,j);
        else ans=max(ans,j-iidx[-sum]);
        //cout<<i<<" "<<sum<<endl;
    }
    cout<<ans<<endl;
    return 0;
}
/*
 11010111  0
012121234  3
 0123456789 10
   0 0 1 1 1 1 0 0
  -1-2-1 0 1 2 1 0
   1 2 3 4 5 6 7 8 9
8
00111100
*/