0425. Word Squares

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

Description

Given an array of unique strings words, return all the word squares 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

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
*/