Substring with Concatenation of All Words

/* 
You are given a string s and an array of strings words. All the strings of words are of the same length.

A concatenated substring in s is a substring that contains all the strings of any permutation of words concatenated.

For example, if words = ["ab","cd","ef"], then "abcdef", "abefcd", "cdabef", "cdefab", "efabcd", and "efcdab" are all concatenated strings. "acdbef" is not a concatenated substring because it is not the concatenation of any permutation of words.
Return the starting indices of all the concatenated substrings in s. You can return the answer in any order.

 

Example 1:

Input: s = "barfoothefoobarman", words = ["foo","bar"]
Output: [0,9]
Explanation: Since words.length == 2 and words[i].length == 3, the concatenated substring has to be of length 6.
The substring starting at 0 is "barfoo". It is the concatenation of ["bar","foo"] which is a permutation of words.
The substring starting at 9 is "foobar". It is the concatenation of ["foo","bar"] which is a permutation of words.
The output order does not matter. Returning [9,0] is fine too.
Example 2:

Input: s = "wordgoodgoodgoodbestword", words = ["word","good","best","word"]
Output: []
Explanation: Since words.length == 4 and words[i].length == 4, the concatenated substring has to be of length 16.
There is no substring of length 16 in s that is equal to the concatenation of any permutation of words.
We return an empty array.
Example 3:

Input: s = "barfoofoobarthefoobarman", words = ["bar","foo","the"]
Output: [6,9,12]
Explanation: Since words.length == 3 and words[i].length == 3, the concatenated substring has to be of length 9.
The substring starting at 6 is "foobarthe". It is the concatenation of ["foo","bar","the"] which is a permutation of words.
The substring starting at 9 is "barthefoo". It is the concatenation of ["bar","the","foo"] which is a permutation of words.
The substring starting at 12 is "thefoobar". It is the concatenation of ["the","foo","bar"] which is a permutation of words.
 

Constraints:

1 <= s.length <= 104
1 <= words.length <= 5000
1 <= words[i].length <= 30
s and words[i] consist of lowercase English letters.
*/

Approach: 
    1. brute: store all combined combination of concatenated substrings.
       get all the indexes of matched combinations in the given full input string.

Approach 2:
    If traversing on the given input string, if the substring of word length is existing in given dictionary , then proceed for further word, or slide.
    hence keep count of the words of the dictionary in the word hash with the count of each word. will be use ful for repeated words.
    and slide till you are getting the substring matched with any of the words present in the dictionary hash. else start from other index.
    unordered_map & sliding.


class Solution {
public:
    bool checkCombinedSubString( string & s, unordered_map< string, int > wordHash, int i, int wl, int totalWL ) {
        for( ; totalWL > 0 ; i += wl, totalWL -= wl ) {
            string subS = s.substr( i, wl ) ;
            if( wordHash.find( subS ) == wordHash.end() ) return false ;
            if( -- wordHash[ subS ] < 0 ) return false ;
        }
        return true ;
    }
    
    vector< int > findSubstring(string s, vector<string>& words) {
        unordered_map< string, int > wordHash;
        int totalWL = 0;
        for(string & word : words) {
            ++ wordHash [ word ] ;
            totalWL += word.length();
        }
        int i=0, wl = words[0].length();
        vector< int > ans;
        while( i < s.length() - totalWL + 1 ) {
            if( checkCombinedSubString(s, wordHash, i, wl, totalWL) ) {
                ans.push_back( i );
            }
            ++i;
        }
        return ans;
    }
};