Concatenated Words

Published: Sep 14, 2022

Hard Depth-First Search String Dynamic Programming Trie Array


This might be solved by the dynamic programming. Also, the recursion (depth-first search) with prefix and suffix would work. In any case, this is hard to solve.

Problem Description

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.


  • 1 <= words.length <= 10**4
  • 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) <= 10**5


Example 1:
Input: words = ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]
Output: ["catsdogcats","dogcatsdog","ratcatdogcat"]
Example 2:
Input: words = ["cat","dog","catdog"]
Output: ["catdog"]


Divide each word to prefix and suffix. The next deeper level checks previous suffix. In the end, if suffix is in the given word list, the suffix is a part of concatenated word.


class ConcatenatedWords:
    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:
        word_set = set(words)

        def dfs(w):
            n = len(w)
            for i in range(1, n):
                prefix, suffix = w[:i], w[i:]
                if prefix in word_set and (suffix in word_set or dfs(suffix)):
                    return True
            return False
        return [word for word in words if dfs(word)]


  • Time: O()
  • Space: O()