[Codeforces 873B]Balanced Substring

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.

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.

Sample Input

Sample Output

Hint

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.

题解

令所有$0$为$-1$，做一遍前缀和，若$L$,$R$前缀和相同，则$[L+1,R]$是一段可行区间。

 1 //It is made by Awson on 2017.10.15
 2 #include <set>
 3 #include <map>
 4 #include <cmath>
 5 #include <ctime>
 6 #include <cmath>
 7 #include <stack>
 8 #include <queue>
 9 #include <vector>
10 #include <string>
11 #include <cstdio>
12 #include <cstdlib>
13 #include <cstring>
14 #include <iostream>
15 #include <algorithm>
16 #define LL long long
17 #define Min(a, b) ((a) < (b) ? (a) : (b))
18 #define Max(a, b) ((a) > (b) ? (a) : (b))
19 #define sqr(x) ((x)*(x))
20 using namespace std;
21 const int N = 100000;
22 const int INF = ~0u>>1;
23 void read(int &x) {
24     char ch; bool flag = 0;
25     for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == ‘-‘)) || 1); ch = getchar());
26     for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
27     x *= 1-2*flag;
28 }
29 
30 int n, a, zero[N+5], one[N+5];
31 int sum, ans;
32 char ch[N+5];
33 
34 void work() {
35     read(n); scanf("%s", ch+1);
36     for (int i = 1; i <= n; i++) {
37         a = ch[i]-48;
38         if (a == 1) sum++;
39         else sum--;
40         if (sum > 0) {
41             if (one[sum]) ans = Max(ans, i-one[sum]);
42             else one[sum] = i;
43         }else if (sum < 0) {
44             if (zero[sum]) ans = Max(ans, i-zero[sum]);
45             else zero[sum] = i;
46         }
47         else ans = i;
48     }
49     printf("%d\n", ans);
50 }
51 int main() {
52     work();
53     return 0;
54 }