leetcode@ [30/76] Substring with Concatenation of All Words & Minimum Window Substring (Hashtable, Two Pointers)

https://leetcode.com/problems/substring-with-concatenation-of-all-words/

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.

For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

class Solution {
public:
    bool func(string s, vector<string>& words, map<string, int>& h) {
        int n = words.size();
        int each = words[].length();

        map<string, int> mh; mh.clear();
        int i = ;
        for(; i<=s.length()-each; i+=each) {
            string sub = s.substr(i, each);

            if(h.find(sub) != h.end()) {
                ++mh[sub];
                if(mh[sub] > h[sub]) return false;
            }
            else return false;
        }

        return true;
    }

    vector<int> findSubstring(string s, vector<string>& words) {
        vector<int> res;
        int n = words.size();
        if(n == )  return res;

        map<string, int> h;
        for(int i=; i<n; ++i) {
            if(h.find(words[i]) == h.end()) {
                h.insert(make_pair(words[i], ));
            }
            else {
                h[words[i]]++;
            }
        }
        int each = words[].length();
        int tot = each * n;
        if(s.length() < tot) return res;

        for(int i = ; i <= s.length() - tot; ++i) {
            string sub = s.substr(i, tot);
            //cout << sub << endl;
            if(func(sub, words, h))  res.push_back(i);
        }

        return res;
    }
};

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

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

For example,
S = "ADOBECODEBANC"
T = "ABC"

Minimum window is "BANC".

Note:
If there is no such window in S that covers all characters in T, return the empty string "".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

class Solution {
public:
    string minWindow(string s, string t) {
        if(s.length() ==  || s.length() < t.length())  return "";
        if(s.length() == t.length()) {
            if(s == t)  return s;
        }

        map<char, int> h; h.clear();
        map<char, bool> mh; mh.clear();
        for(int i=; i<t.length(); ++i) {
            h[t[i]]++;
            mh[t[i]] = true;
        }

        int cnt = t.length(), l = , minRange = INT_MAX, mini, minj;
        for(int r=; r<s.length(); ++r) {
            if(mh[s[r]]) {
                --h[s[r]];
                if(h[s[r]] >= )  --cnt;
            }

            if(cnt == ) {
                while(!mh[s[l]] || h[s[l]] < ) {
                    ++h[s[l]];
                    ++l;
                }

                if(minRange > r - l + ) {
                    minRange = r - l + ;
                    mini = l;
                }
            }
        }

        if(minRange == INT_MAX)  return "";

        return s.substr(mini, minRange);
    }
};