> For the complete documentation index, see [llms.txt](https://jaywin.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://jaywin.gitbook.io/leetcode/solutions/0425-word-squares.md).

# 0425. Word Squares

<https://leetcode.com/problems/word-squares>

## Description

Given an array of **unique** strings `words`, return *all the* [**word squares**](https://en.wikipedia.org/wiki/Word_square) *you can build from* `words`. The same word from `words` can be used **multiple times**. You can return the answer in **any order**.

A sequence of strings forms a valid **word square** if the `kth` row and column read the same string, where `0 <= k < max(numRows, numColumns)`.

* For example, the word sequence `["ball","area","lead","lady"]` forms a word square because each word reads the same both horizontally and vertically.

**Example 1:**

```
**Input:** words = ["area","lead","wall","lady","ball"]
**Output:** [["ball","area","lead","lady"],["wall","area","lead","lady"]]
**Explanation:**
The output consists of two word squares. The order of output does not matter (just the order of words in each word square matters).
```

**Example 2:**

```
**Input:** words = ["abat","baba","atan","atal"]
**Output:** [["baba","abat","baba","atal"],["baba","abat","baba","atan"]]
**Explanation:**
The output consists of two word squares. The order of output does not matter (just the order of words in each word square matters).
```

**Constraints:**

* `1 <= words.length <= 1000`
* `1 <= words[i].length <= 5`
* All `words[i]` have the same length.
* `words[i]` consists of only lowercase English letters.
* All `words[i]` are **unique**.

## ac

```java
class Solution {
    public List<List<String>> wordSquares(String[] words) {
        List<List<String>> res = new ArrayList<>();
        // edge cases
        if (words == null || words.length == 0) return res;

        // build trie
        TrieNode root = new TrieNode();
        for (String word : words) {
            build(root, word);
            root.candidates.add(word);
        }

        // backtracking
        search(root, new ArrayList<String>(), res);

        return res;
    }
    private void search(TrieNode root, List<String> note, List<List<String>> res) {
        // exit
        if (note.size() > 0 && note.size() == note.get(0).length()) {
            res.add(new ArrayList<>(note));
            return;
        }

        String prefix = buildPrefix(note);
        List<String> candidates = getCandidates(root, prefix);
        for (String candi : candidates) {
            note.add(candi);
            search(root, note, res);
            note.remove(note.size() - 1);
        }
    }
    private List<String> getCandidates(TrieNode root, String prefix) {
        TrieNode r = root;
        for (int i = 0; i < prefix.length(); i++) {
            int pos = prefix.charAt(i) - 'a';
            if (r.next[pos] == null) return new ArrayList<>();
            r = r.next[pos];
        }
        return r.candidates;
    }
    private String buildPrefix(List<String> note) {
        StringBuilder sb = new StringBuilder();
        int col = note.size();
        for (String s : note) {
            if(col < s.length()) sb.append(s.charAt(col));
        }
        return sb.toString();
    }
    private void build(TrieNode root, String word) {
        TrieNode r = root;
        for (int i = 0; i < word.length(); i++) {
            int pos = word.charAt(i) - 'a';
            if (r.next[pos] == null) r.next[pos] = new TrieNode();
            r = r.next[pos];
            r.candidates.add(word);
        }
    }
}

class TrieNode {
    TrieNode[] next = new TrieNode[26];
    List<String> candidates = new ArrayList<>();
}

/*
trie + backtracking
*/
```


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