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.

Example 1:

Input:
  s = "barfoothefoobarman",
  words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.

Example 2:

Input:
  s = "wordgoodstudentgoodword",
  words = ["word","student"]
Output: []

 

AC code:

class Solution {
public:
    vector<int> findSubstring(string s, vector<string>& words) {
        unordered_map<string, int> mario;
        vector<int> res={};
        int len1 = s.length(), num = words.size();
        if (len1 == 0 || num == 0) return res;
        int len2 = words[0].length();
        for (int i = 0; i < words.size(); ++i) {
            mario[words[i]]++;
        }
        for (int i = 0; i < len1-num*len2+1; ++i) {
            unordered_map<string, int> seen;
            int j = 0;
            for (; j < num; ++j) {
                string word = s.substr(i+j*len2, len2);
                if (mario.find(word) != mario.end()) {
                    seen[word]++;
                    if (seen[word] > mario[word])
                        break;
                } else 
                    break;
            }
            if (j == num)
                res.push_back(i);
        }
        return res;
    }
};

Runtime: 132 ms, faster than 49.73% of C++ online submissions for Substring with Concatenation of All Words.