You are given a string S, and a list of words L i.e array/vector of strings (Words in list L are all of the same length). Find the starting indices of the substrings in string S, which contains all the words present in list L.
The order of words of list L appearing inside string S does not matter i.e if string S is “barfooapplefoobar” and list of words (L) is [“foo”, “bar”] then we have to look for substrings “foobar”, “barfoo” in string S.
Note : Words inside the list L can repeat.
Input : S: "barfoothefoobarman" L: ["foo", "bar"] Output : 0 9 Explanation : // at index 0 : barfoo // at index 9 : foobar Input : S: "catbatatecatatebat" L: ["cat", "ate", "bat"] Output : 0 3 9 Explanation : // at index 0 : catbatate // at index 3 : batatecat // at index 9 : catatebat Input : S : "abcdababcd" L : ["ab", "ab", "cd"] Output : 0 2 4 Explanation : // at index 0 : abcdab // at index 2 : cdabab // at index 4 : ababcd Input : S : "abcdababcd" L : ["ab", "ab"] Output : 4
We can use Hashing Technique to solve the above problem. Let’s see the steps :
- Declare a map (hash_map) which stores all words of List L corresponding to their occurrences inside list L.
- Traverse through all possible substrings in string S which are equal to size_L(total number of characters produced if all the words in list L are concatenated).
- Create a temporary map (temp_hash_map) and initialize it with original map(hash_map) for every possible substring.
- Extract the words from the substring and if the word is present in temp_hash_map we decrease it’s corresponding count, if it’s not present in temp_hash_map we simply break.
- After traversing the substring we traverse temp_hash_map and look for any key which has it’s count > 0. If we found no such key it means that all the words in list L were found in substring and store the given starting index of the substring, if we find a key which has it’s count > 0 it means we did not traversed whole substring because we came across a word which was not in temp_hash_map.
Below is the implementation of above approach :
Time Complexity : O(N – K) * K
N : length of string S.
K : total length of list L if all the words are concatenated. If L : [“ab”, “cd”] then K = 4.
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