Substring with Concatenation of All Words

https://leetcode.com/problems/substring-with-concatenation-of-all-words/

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.

For example, given: s: "barfoothefoobarman" words: ["foo", "bar"]

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

Similar problems:

• Longest Substring Which Contains 2 Unique Characters

  http://www.programcreek.com/2013/02/longest-substring-which-contains-2-unique-characters/

• Longest Substring Which Contains at most k Unique Characters

• Longest Substring Without Repeating Characters

• Minimum Window Substring

Solution:

public List<Integer> findSubstring(String s, String[] words) {
    ArrayList<Integer> result = new ArrayList<Integer>();
    if(s==null||s.length()==0||words==null||words.length==0){
        return result;
    } 

    //frequency of words
    HashMap<String, Integer> map = new HashMap<String, Integer>();
    for(String w: words){
        if(map.containsKey(w)){
            map.put(w, map.get(w)+1);
        }else{
            map.put(w, 1);
        }
    }

    int len = words[0].length();

    for(int j=0; j<len; j++){
        HashMap<String, Integer> currentMap = new HashMap<String, Integer>();
        int start = j;//start index of start
        int count = 0;//count totoal qualified words so far

        for(int i=j; i<=s.length()-len; i=i+len){
            String sub = s.substring(i, i+len);
            if(map.containsKey(sub)){
                //set frequency in current map
                if(currentMap.containsKey(sub)){
                    currentMap.put(sub, currentMap.get(sub)+1);
                }else{
                    currentMap.put(sub, 1);
                }

                count++;

                while(currentMap.get(sub)>map.get(sub)){
                    String left = s.substring(start, start+len);
                    currentMap.put(left, currentMap.get(left)-1);

                    count--;
                    start = start + len;
                }


                if(count==words.length){
                    result.add(start); //add to result

                    //shift right and reset currentMap, count & start point         
                    String left = s.substring(start, start+len);
                    currentMap.put(left, currentMap.get(left)-1);
                    count--;
                    start = start + len;
                }
            }else{
                currentMap.clear();
                start = i+len;
                count = 0;
            }
        }
    }

    return result;
}