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Harsh Rajpal
Harsh Rajpal

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472. Concatenated Words

Problem Statement:

Given an array of strings words (without duplicates), return all the concatenated words in the given list of words.

A concatenated word is defined as a string that is comprised entirely of at least two shorter words in the given array.

Example 1:

Input: words = ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]

Output: ["catsdogcats","dogcatsdog","ratcatdogcat"]
Explanation: "catsdogcats" can be concatenated by "cats", "dog" and "cats";
"dogcatsdog" can be concatenated by "dog", "cats" and "dog";
"ratcatdogcat" can be concatenated by "rat", "cat", "dog" and "cat".

Example 2:

Input: words = ["cat","dog","catdog"]
Output: ["catdog"]


  • 1 <= words.length <= 104
  • 1 <= words[i].length <= 30
  • words[i] consists of only lowercase English letters.
  • All the strings of words are unique.
  • 1 <= sum(words[i].length) <= 105



class Solution {
  public List<String> findAllConcatenatedWordsInADict(String[] words) {
    List<String> ans = new ArrayList<>();
    Set<String> wordSet = new HashSet<>(Arrays.asList(words));
    Map<String, Boolean> memo = new HashMap<>();

    for (final String word : words)
      if (wordBreak(word, wordSet, memo))

    return ans;

  private boolean wordBreak(final String word, Set<String> wordSet, Map<String, Boolean> memo) {
    if (memo.containsKey(word))
      return memo.get(word);

    for (int i = 1; i < word.length(); ++i) {
      final String prefix = word.substring(0, i);
      final String suffix = word.substring(i);
      if (wordSet.contains(prefix) &&
          (wordSet.contains(suffix) || wordBreak(suffix, wordSet, memo))) {
        memo.put(word, true);
        return true;

    memo.put(word, false);
    return false;
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