Educational Codeforces Round 30 B.Balanced Substring

B. Balanced Substring

Problem Statement

    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

Example 1
    Input
        8
        11010111
    Output
        4
Example 2
    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.

题意

    给你一个长度为n的01子串，求最长的子串使得该子串内部0和1的个数相同。

思路

    对于串内的每个0，我们把它的贡献记为-1，对于每个1，我们把它的贡献记为1，之后我们对原串求一遍前缀和，将所有前缀和相等的位置记录在一起，因为如果两个位置前缀和相同，那么这两个位置中间这一部分就是满足题目要求01个数相同的子串。
    P.S. 刚开始我写完后忘记判整个串正好是满足01个数相同的情况了，就wa掉了

Code

#pragma GCC optimize(3)
#include
using namespace std;
typedef long long ll;
inline void readInt(int &x) {
    x=0;int f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    x*=f;
}
inline void readLong(ll &x) {
    x=0;int f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    x*=f;
}
/*================Header Template==============*/
string s;
int sum[100010];
map<int,vector<int> >pos;
map<int,bool> vis;
int main() {
    int n;
    readInt(n);
    cin>>s;
    sum[0]=0;
    pos[0].push_back(0);
    for(int i=0;iif(s[i]=='0')
            sum[i+1]=sum[i]-1;
        else
            sum[i+1]=sum[i]+1;
        pos[sum[i+1]].push_back(i+1);
    }
    int mxlen=0;
    for(int i=1;i<=s.length();i++) {
        if(vis[sum[i]])
            continue;
        else {
            vis[sum[i]]=1;
            if(pos[sum[i]].size()>=2)
                mxlen=max(mxlen,pos[sum[i]][pos[sum[i]].size()-1]-pos[sum[i]][0]);
        }
    }
    printf("%d\n",mxlen);
    return 0;
}