Balanced Substring CodeForces - 873B (思维+前缀和)

Balanced Substring

CodeForces - 873B

You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string s_ls_l + 1s_l + 2... s_r, 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

8
11010111

Output

4

Input

3
111

Output

0

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.

题目大意：字符串是由0和1组成，求字符串的最长子串，使0和1的数目相同

思路：维护一个前缀和，初始化sum为n，如果是0，--，如果是1就++，如果这个sum值之前出现过了，那么说明最早出现这个值的位置和现在位置的长度就是当前构成01相等字符串的长度，每次更新ans。

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int INF = 0x3f3f3f3f;
int Min[200005];//Min[sum]表示前缀和sum的最早出现位置
char s[100005];
int main(){
    int n,sum,len,ans;
    while(scanf("%d",&n) != EOF){
        scanf("%s",s+1);
        sum = n;//防止sum为负数导致Min数组越界初始化为n
        ans = 0;
        len = strlen(s+1);
        memset(Min,INF,sizeof(Min));
        Min[sum] = 0;//初始Min[n]=0,如果整个串01个数相同就是原长度
        for(int i = 1; i <= len; i++){
            if(s[i] == '0') sum--;
            else sum++;
            Min[sum] = min(Min[sum],i);
            ans = max(ans,i - Min[sum]);
        }
        printf("%d\n",ans);
    }
    return 0;
}