leetcode Question 106: Substring with Concatenation of All Words

Substring with Concatenation of All Words

You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.
For example, given:
L["foo", "bar"]
You should return the indices: [0,9].
(order does not matter).


Try to think this problem straightforward:
Say in L there are m strings with length n. 
What string is required to match in S?     A length of m*n string start with each position in S.
What is a match?  In the m*n long string, every string in L appear only once.

So the algorithm is:
Scan every m*n long string start from each position in S, see if all the strings in L have been appeared only once using Map data structure. If so, store the starting position.

Yes, do not consider any DP or DFS solutions, just using the hash map and loop.


class Solution {
    vector<int> findSubstring(string S, vector<string> &L) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> res;
        int num = L.size();
        int len = L[0].size();
        if (num==0){return res;}
        map<string,int> mp;  
        for (int i=0;i<num;i++){mp[L[i]]++;}      
        int i=0;
        while ((i+num*len-1)<S.size()){
            map<string,int> mp2;
            int j=0;
            while (j<num){
                string subs = S.substr(i+j*len,len);
                if (mp.find(subs)==mp.end()){
                    if (mp2[subs]>mp[subs]){
            if (j==num){res.push_back(i);}
        return res;


class Solution:
    # @param S, a string
    # @param L, a list of string
    # @return a list of integer
    def findSubstring(self, S, L):
        res = []            # result list
        num = len(L)        # length of the str list 
        ls = len(S)
        if num == 0:
            return []
        str_len = len(L[0]) # length of each str
        #create the map: count the occurrance of each string
        #Note that set(L) is used to reduce the time, otherwise will not pass the large test
        map_str = dict((x,L.count(x)) for x in set(L))
        i = 0
        while i + num * str_len - 1 < ls:
            map_str2 = {}
            j = 0
            while j < num:
                subs = S[i + j * str_len:i + j * str_len + str_len ]
                if not subs in map_str:
                    # Note that dict.get(key, default_val) is used to handel the case that key NOT exist
                    map_str2[subs] = map_str2.get(subs, 0) + 1
                    if map_str2[subs]>map_str[subs]:
                    j = j + 1
            if j == num:
            i = i + 1
        return res