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 map = new HashMap(); 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 currentMap = new HashMap(); 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; }