[LeetCode] 30. 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 {
    void helper(string s, int lenword, int totallen, unordered_map<string, int>& ht, unordered_map<string, int>& ht_bk, vector<int>& allsol){

        for (int i = 0; i<= s.size() - totallen; i++){
            int j;
            for (j = i; j <= i + totallen - lenword; j += lenword){
                string tmp = s.substr(j, lenword);
                //if (ht.count(tmp) && ht[tmp] > 0 ){
                if (ht[tmp] > 0 ){
                    ht[tmp]--;
                }else{
                    break;
                }
            }
            // !! to clear the ht
            ht = ht_bk;

            if (j == i + totallen){
                allsol.push_back(i);
            }
        }
        
        return;
    }
public:
    vector<int> findSubstring(string s, vector<string>& words) {
        vector<int> allsol;
        int sol = 0;
        int start;
        
        if (words.empty() || s.empty()) return allsol;
        
        unordered_map<string, int> ht;
        unordered_map<string, int> ht_bk;
        int n = s.size();
        int lenword = words[0].size();
        int ntotalwords = words.size();
        int totallen = ntotalwords * lenword;
        if (n < totallen) return allsol;
        
        for (int i = 0; i < words.size(); i++){
            ht[words[i]]++;
        }
        ht_bk = ht;
        
        helper(s, lenword, totallen, ht, ht_bk,allsol);
        
        return allsol;
    }
};