concatenated words leetcode 472 python

certainly! the problem of "concatenated words" (leetcode 472) is a classic problem that involves finding words in a list that can be formed by concatenating other words from the same list.

problem statement

you are given an array of strings `words`. a word is called a concatenated word if it is made up of at least two shorter words from the same list. for example, if the list contains the words `"cat"`, `"cats"`, and `"dog"`, then the word `"cats"` is a concatenated word because it can be formed by concatenating the word `"cat"` and the word `"s"` (which isn't in the list, so `"cats"` should not be included).

example

**input:**

**output:**

approach

1. **set for fast lookup:** use a set to store the words for o(1) average-time complexity when checking if a substring is in the list.

2. **dfs with memoization:** use depth first search (dfs) to check if a word can be formed by concatenating other words. we can use memoization to store results of previously computed words to avoid redundant calculations.

3. **iterate through the list:** for each word in the list, check if it can be formed by concatenating other words.

implementation

here's a python implementation of the above approach:

explanation of the code

1. **data structures:**
- `word_set`: a set containing all words for quick lookup.
- `memo`: a dictionary for memoization to store whether a word can be formed.

2. **function `can_form`:**
- this function checks if a given word can be formed by concatenating other words.
- if the word is already computed (exists in `memo`), return the stored result.
- if the word exists in the `word_set`, it is a valid concatenated word (can be formed from itself).
- iterate over the word to split it into a prefix and suffix. if the prefix exists in the set and the suffix can be formed, mark it as a valid concatenated word.

3. **main loop:**
- for each word, temporarily remove it from the set to avoid considering it as part of itself ...

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concatenated words
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