CODE 103: 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:
S: "barfoothefoobarman"
L: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

public ArrayList<Integer> findSubstring(String S, String[] L) {// IMPORTANT: Please reset any member data you declared, as// the same Solution instance will be reused for each test case.Map<String, Integer> wordMap = new HashMap<String, Integer>();ArrayList<Integer> starts = new ArrayList<Integer>();int wordLength = L[0].length();int wordNumber = L.length;int allLength = wordLength * wordNumber;for (String l : L) {if (wordMap.containsKey(l)) {int num = wordMap.get(l);wordMap.put(l, num + 1);} else {wordMap.put(l, 1);}}for (int i = 0; i <= S.length() - allLength; i++) {Map<String, Integer> tmpWordMap = new HashMap<String, Integer>();boolean flag = true;for (int k = i, curWordNum = 0; curWordNum < wordNumber; k += wordLength, curWordNum++) {String tmpStr = S.substring(k, k + wordLength);if (tmpWordMap.containsKey(tmpStr)) {int number = tmpWordMap.get(tmpStr);tmpWordMap.put(tmpStr, number + 1);} else {tmpWordMap.put(tmpStr, 1);}if (!wordMap.containsKey(tmpStr)|| tmpWordMap.get(tmpStr) > wordMap.get(tmpStr)) {flag = false;break;}}if (flag) {starts.add(i);}}return starts;}