837B. Balanced Substring

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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

inputCopy

8

11010111

outputCopy

4

inputCopy

3

111

outputCopy

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.

题目就是要你找出最长的0的数量和1的数量相等的长度是多少

唉，一开始觉得2分能做，还做了半天，后面发现有问题有找不出答案，后面一看结果太简单了

把0当成-1，建一个结构体，存前缀和，并保持当前的位置的序号，然后sort，前缀和相同的位置序号相减，保留最大值就好了，第0个为0,0，否则有问题

#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<stdio.h>
#include<float.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define sf scanf
#define pf printf
#define fi first
#define se second
#define mp make_pair
#define pii pair<int,int>
#define scf(x) scanf("%d",&x)
#define prf(x) printf("%d\n",x)
#define scff(x,y) scanf("%d%d",&x,&y)
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
#define mm(x,b) memset((x),(b),sizeof(x))
#define scfff(x,y,z) scanf("%d%d%d",&x,&y,&z)
typedef long long ll;
const ll mod=1e9+7;
using namespace std;

const double eps=1e-8;
const int inf=0x3f3f3f3f;
const double pi=acos(-1.0);
const int N=2e5+10;
char a[N];
struct node{
	int num;
	int pos;
}q[N];
bool cmp(node x,node y)
{
	if(x.num == y.num ) return x.pos < y.pos ;
	return x.num <y.num ;
 }
int main()
{
	int n;scf(n);
	sf("%s",a+1);

	rep(i,0,n+1)
	{
		if(i==0)
		{
			q[i].num = 0;
			q[i].pos= 0;
		}else
		{
			if(a[i]=='1')
			{
				q[i].num = q[i-1].num +1;
			}else
			{
				q[i].num = q[i-1].num -1;
			}
			q[i].pos=i;
		}
	}
	sort(q,q+n+1,cmp);
	int ans=0;
	rep(i,1,n+1)
	{
		if(q[i].num == q[i-1].num )
		{
			int j=i;
			while(q[j].num ==q[i-1].num&&j<=n) j++;
			j--;
			ans=max(ans,q[j].pos -q[i-1].pos);
			i=j;
		}
	}
	prf(ans);
	return 0;
}