Substring with concatenation of all Words

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

Example 1:

Input:  s = "Barfoothefoobarman",  [0,9] explanation:substrings starting at index 0 and 9 is "Barfoor" and "Foobar" respectively. The output order does not matter, returning [9,0] is fine too.

Example 2:

Input:  s = "Wordgoodstudentgoodword",  []

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: faster than 49.73% of C + + online submissions for Substring with concatenation of all Words.

Substring with concatenation of all Words