# 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

• 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){

//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;
}
``````