[LeetCode] 76. Minimum Window Substring

When require[ch] is positive or zero, letter ch in window exists enough to construct t. Else, -value is the required number of 'ch' to construct t.

class Solution {
public:
    string minWindow(string s, string t) {
        int ank, hok;
        int s_len = s.length();

        int require[128] = { 0, };

        int chk = 0;
        int res_len, res_idx;

        ank = 0; hok = 0;
        for(char ch: t) { 
            if (!require[ch]) { chk--; }
            require[ch]--;
        }
        res_len = 0; res_idx = -1;
        for(; hok < s_len; hok++) {
            char ch = s[hok];
            require[ch]++;
            if (require[ch] == 0) { chk++; }
            while (require[s[ank]] > 0) {
                require[s[ank]]--; ank++;
            }
            if (chk < 0) { continue; }
            if (res_idx == -1 || hok - ank + 1 < res_len) {
                res_len = hok - ank + 1;
                res_idx = ank;
            }
        }
        if (res_idx == -1) { return ""; }
        else { return s.substr(res_idx, res_len); }
    }
};