[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: