# Solving Leetcode Interviews in Seconds with AI: Concatenated Words


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "472" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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 (not necessarily distinct) 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"]    Constraints:  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  

	# Explanation
	Here's a breakdown of the approach and the Python code:

*   **High-Level Approach:**
    *   Sort the words by length in ascending order. This allows us to efficiently check if a longer word can be formed by concatenating shorter words that have already been processed.
    *   Use dynamic programming to determine if a word can be formed by concatenating shorter words. `dp[i]` will be true if the substring `word[:i]` can be segmented into smaller words from the input.
    *   Maintain a set of valid words encountered so far.

*   **Complexity:**
    *   Runtime Complexity: O(n * l^2), where n is the number of words and l is the maximum length of a word. Sorting takes O(n log n), but it is dominated by the DP part.
    *   Storage Complexity: O(n + l), where n is the number of words (for the result and word set) and l is the maximum length of a word (for the DP array).

	
	# Code
	```python
	def findAllConcatenatedWordsInADict(words):
    """
    Finds all concatenated words in a given list of words.

    Args:
        words: A list of strings.

    Returns:
        A list of strings, representing the concatenated words.
    """
    word_set = set(words)
    result = []

    def can_form(word):
        if not word:
            return True
        
        n = len(word)
        dp = [False] * (n + 1)
        dp[0] = True

        for i in range(1, n + 1):
            for j in range(i):
                if dp[j] and word[j:i] in word_set:
                    dp[i] = True
                    break
        return dp[n]

    words.sort(key=len)

    for word in words:
        word_set.remove(word) # Temporarily remove the word itself
        if can_form(word):
            result.append(word)
        word_set.add(word)  # Add it back for future checks

    return result
	```