human
Find the starting indices of the substrings in string ( S ) which is made by concatenating all words from a list ( L ) | Returns an integer vector consisting of starting indices of substrings present inside the string S.
Number of a characters of a word in list L ..
Number of words present inside list L ..
Total characters present in list L ..
Resultant vector which stores indices ..
If the total number of characters in list L is more than length of string S itself ..
Map stores the words present in list L against it 's occurrences inside list L.
Traverse the substring.
Extract the word.
If word not found or if frequency of current word is more than required simply break ..
Else decrement the count of word from hash_map.
Store the starting index of that substring when all the words in the list are in substring.
Driver Code"
Solution in Python:

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What is a code to find the starting indices of the substrings in string ( S ) which is made by concatenating all words from a list ( L ) and the other steps required?
assistant
#Here is code to find the starting indices of the substrings in string ( S ) which is made by concatenating all words from a list ( L ) and the other steps required.
 def findSubStringIndices ( s , L ) :
	size_word = len ( L [ 0 ] )
	word_count = len ( L )
	size_L = size_word * word_count
	res = [ ]
	if size_L > len ( s ) :
		return res
	hash_map = dict ( )
	for i in range ( word_count ) :
		if L [ i ] in hash_map :
			hash_map [ L [ i ] ] += 1
		else :
			hash_map [ L [ i ] ] = 1
	for i in range ( 0 , len ( s ) - size_L + 1 , 1 ) :
		temp_hash_map = hash_map . copy ( )
		j = i
		count = word_count
		while j < i + size_L :
			word = s [ j : j + size_word ]
			if ( word not in hash_map or temp_hash_map [ word ] == 0 ) :
				break
			else :
				temp_hash_map [ word ] -= 1
				count -= 1
			j += size_word
		if count == 0 :
			res . append ( i )
	return res
if __name__ == " _ _ main _ _ " :
	s = " barfoothefoobarman "
	L = [ " foo " , " bar " ]
	indices = findSubStringIndices ( s , L )
	print ( * indices )