Find List Words Formed by Joining Two or More Other Words From the List

Given a list of distinct lowercase words, return every word in the list that can be formed by joining two or more words from the same list end to end. ### Function Signature ```python def find_compound_words(words: list[str]) -> list[str]: ``` ### Rules - A word `w` in the list is compound when `w` equals the concatenation `p1 + p2 + ... + pm` for some `m >= 2`, where every piece `pi` is a word in the list. - The same word may be used as a piece more than once. Because there are at least two nonempty pieces, every piece is shorter than `w`, so a word never counts as a piece of itself. - Return the compound words in the order in which they appear in `words`. Return an empty list if there are none. ### Constraints - `1 <= len(words) <= 10^4` - `1 <= len(words[i]) <= 30` - The total length of all the words is at most `10^5`. - Every word consists of lowercase English letters, and all words are distinct. ### Examples **Example 1** ```text Input: words = ["sun", "flower", "sunflower", "flowers", "s", "sunflowers", "pot"] Output: ["sunflower", "flowers", "sunflowers"] ``` `"sunflower"` is `"sun" + "flower"`, `"flowers"` is `"flower" + "s"`, and `"sunflowers"` is `"sun" + "flower" + "s"` (or `"sun" + "flowers"`). None of the other words can be split into words from the list. **Example 2** ```text Input: words = ["ab", "abab", "aba", "b"] Output: ["abab"] ``` `"abab"` is `"ab" + "ab"`, using the same word twice. `"aba"` would need the remaining piece `"a"`, which is not in the list, and `"ab"` would need `"a" + "b"`. **Example 3** ```text Input: words = ["x", "y"] Output: [] ```