双指针

• 寻找超过3个x的个数并求和，减2即应该删去的x的数量

时间复杂度 $O(n)$

C++ 代码

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

int main()
{
    int n;
    string s;
    int res = 0;
    cin >> n >> s;
    for(int i = 0; i < n; i++ )
    {
        if(s[i] != 'x') continue;
        int j = i + 1; //error
        while(j < n && s[j] == 'x') j++ ;
        res += max(0, j - i - 2); //error
        i = j - 1;

    }
    cout << res;
    return 0;
}