# LeetCode – Substring with Concatenation of All Words (Java)

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"], return [0,9].

Analysis

This problem is similar (almost the same) to Longest Substring Which Contains 2 Unique Characters.

Since each word in the dictionary has the same length, each of them can be treated as a single character.

Java 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; }```
Category >> Algorithms >> Interview
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```<pre><code>
String foo = "bar";
</code></pre>
```

I solved by a better solution which happens to be more clear and has a better performance using the indexOf method and comparing the indices as follows:

public static List searchIndecies(String s, String arr[]) {

if (s == null || s.length() == 0 || arr == null || arr.length == 0) {
return null;
}
int wordLength = arr[0].length();
List returnedIndecies = new ArrayList();
List indecies = new ArrayList();
for (int i = 0; i < s.length(); i++) {
indecies.clear();
for (int m = 0; m < arr.length; m++) {
int index = s.indexOf(arr[m], i);
if (index != -1) {
}
}
Collections.sort(indecies);
boolean validSeq = false;
for (int n = 1; n < indecies.size() && indecies.size() == arr.length; n++) {
if (indecies.get(n – 1) != indecies.get(n) – wordLength) {//Comparing indices based on words same length
i = i + wordLength – 1;
break;
} else {
i = indecies.get(n);
validSeq = true;
}
}
if (validSeq) {
}

}
return returnedIndecies;

}

• Larry Okeke

A simple to understand soluton.

Just get every possible concatenation of words in words array, and check whether each occurs in the main string.
if the result is not -1, we have an index of substring!

``` import java.util.*; public class duplicate_substring{ public static ArrayList allPossibleArrangements = new ArrayList(); public static void main(String[] args){ String[] arr = new String[] {"chinelo", "arukwe", "larry"}; ArrayList words = new ArrayList(Arrays.asList(arr)); String s = "chineloarukwelarryokekearukwechinelolarrytestinglarryarukwechinelo"; permutate(words, new Stack(), arr.length); List solution = solution(allPossibleArrangements.toArray(new String[0]), s); System.out.println(solution.toString()); } public static List solution(String[] words, String s){ //use combinatorial algorithm to get every possible concatenation of words //find the index of each derived word in s. StringBuilder build = new StringBuilder(); ArrayList result = new ArrayList(); for(int i = 0; i < words.length; i++){ String currentConcatenation = words[i]; if(s.indexOf(currentConcatenation)!=-1){ result.add(s.indexOf(currentConcatenation)); } build.setLength(0); } return result; } public static void permutate(List items, Stack permutation, int size) { /* permutation stack has become equal to size that we require */ if(permutation.size() == size) { /* print the permutation */ //System.out.println(Arrays.toString(permutation.toArray(new Integer[0]))); String s = ""; for(String str: permutation){ s+=str; } allPossibleArrangements.add(s); } /* items available for permutation */ String[] availableItems = items.toArray(new String[0]); for(String i : availableItems) { /* add current item */ permutation.push(i); /* remove item from available item set */ items.remove(i); /* pass it on for next permutation */ permutate(items, permutation, size); /* pop and put the removed item back */ items.add(permutation.pop()); } } } ```