Substring with Concatenation of All Words (Java)

Brute Force Solution, looking forward to better one

	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:
	S: "barfoothefoobarman"
	L: ["foo", "bar"]
	You should return the indices: [0,9].
	(order does not matter).
	Time complexity: O((M - N) * N)
	public class Solution {
	public ArrayList<Integer> findSubstring(String S, String[] L) {
	ArrayList<Integer> result=new ArrayList<Integer> ();
	if (S==null||L==null||S.length()==0 ||L.length==0){
	return result;
	}
	int wordLen=L[0].length();
	int totalLen=wordLen*L.length;
	if (S.length()<totalLen){
	return result;
	}
	HashMap<String, Integer> checker=new HashMap<String, Integer>();
	for (String temp: L){
	if (checker.containsKey(temp)){
	checker.put(temp, checker.get(temp)+1);
	}
	else{
	checker.put(temp,1);
	}
	}
	for(int i=0; i<=S.length()-totalLen; i++){
	if (isValid(checker, S, i,wordLen, totalLen)){
	result.add(i);
	}
	}
	return result;
	}
	private boolean isValid(HashMap<String, Integer> checker,String S, int i, int wordLen, int totalLen){
	HashMap<String, Integer> tmpChecker=new HashMap<String, Integer>(checker);
	int start=i;
	int end=i+wordLen;
	while(end-i<=totalLen){
	String tmpString=S.substring(start, end);
	if (tmpChecker.containsKey(tmpString)){
	if (tmpChecker.get(tmpString)==0){
	return false;
	}
	tmpChecker.put(tmpString, tmpChecker.get(tmpString)-1);
	start=end;
	end=start+wordLen;
	}
	else{
	return false;
	}
	}
	return true;
	}
	}

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