[Leetcode 30, Hard] Substring with Concatenation of All Words

Problem:

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 wordsexactly 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).

Analysis:

Solutions:

C++:

    vector<int> findSubstring(string s, vector<string>& words) 
    {
        map<string, int> candidates;
        for(int i = 0; i < words.size(); ++i) {
            if(candidates.find(words[i]) == candidates.end())
                candidates[words[i]] = 1;
            else
                ++candidates[words[i]];
        }
            
        int num_test_words = words.size();
        int word_len = words[0].size();
        vector<int> results;
        for(int start = 0; start < (int)s.size() - num_test_words * word_len + 1; ++start) {
            bool is_valid = true;
            map<string, int> local_count;
            int end = start + num_test_words * words[0].size() - 1;
            for(int j = start; j <= end;) {
                if(candidates.find(s.substr(j, word_len)) == candidates.end()) {
                    is_valid = false;
                    break;
                } else if(local_count.find(s.substr(j, word_len)) != local_count.end()
                        && candidates[s.substr(j, word_len)] == local_count[s.substr(j, word_len)]) {
                    is_valid = false;
                    break;
                } else {
                    if(local_count.find(s.substr(j, word_len)) == local_count.end())
                        local_count[s.substr(j, word_len)] = 1;
                    else
                        ++local_count[s.substr(j, word_len)];
                    j += word_len;
                }
            }
            if(is_valid)
                results.push_back(start);
        }
        
        return results;
    }

Java:

Python: