Leetcode 76: Minimum Window Substring

https://leetcode.com/problems/minimum-window-substring/description/

Notes:

The basic idea is counting and comparing with a sliding window or two-pointer method.

One detail is how to compare? If we compare the counting arrays or hash table directly, it will be costly. One better way is we just need a counting variable to count the total number of valid letters.
See the code below:

class Solution {
public:
    string minWindow(string s, string t) {
        int len1 = s.size(), len2 = t.size();
        if(len1 < len2) return "";
        int start = 0, len = 0;
        vector v1(256, 0), v2(256, 0);
        for(auto &a : t) ++v2[a];
        int i = 0, j = 0, ct = 0;
        while(i < len1 && j < len1) {
            ++v1[s[j]];//move right pointer
            if(v1[s[j]] <= v2[s[j]]) ++ct;//only count when the letter is valid;
            if(ct == len2) {
                while(i < j && v2[s[i]] < v1[s[i]]) --v1[s[i++]];//move left pointer
                if(len == 0 || len > j - i + 1) {
                    len = j - i + 1;
                    start  = i;
                }
            }
            ++j;
        }
        return s.substr(start, len);
    }
};