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425. Word Squares πŸ”’

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 <= 4
  • All words[i] have the same length.
  • words[i] consists of only lowercase English letters.
  • All words[i] are unique.

Solutions

Solution 1

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class Trie:
    def __init__(self):
        self.children = [None] * 26
        self.v = []

    def insert(self, w, i):
        node = self
        for c in w:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                node.children[idx] = Trie()
            node = node.children[idx]
            node.v.append(i)

    def search(self, w):
        node = self
        for c in w:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                return []
            node = node.children[idx]
        return node.v


class Solution:
    def wordSquares(self, words: List[str]) -> List[List[str]]:
        def dfs(t):
            if len(t) == len(words[0]):
                ans.append(t[:])
                return
            idx = len(t)
            pref = [v[idx] for v in t]
            indexes = trie.search(''.join(pref))
            for i in indexes:
                t.append(words[i])
                dfs(t)
                t.pop()

        trie = Trie()
        ans = []
        for i, w in enumerate(words):
            trie.insert(w, i)
        for w in words:
            dfs([w])
        return ans

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class Trie {
    Trie[] children = new Trie[26];
    List<Integer> v = new ArrayList<>();

    void insert(String word, int i) {
        Trie node = this;
        for (char c : word.toCharArray()) {
            c -= 'a';
            if (node.children[c] == null) {
                node.children[c] = new Trie();
            }
            node = node.children[c];
            node.v.add(i);
        }
    }

    List<Integer> search(String pref) {
        Trie node = this;
        for (char c : pref.toCharArray()) {
            c -= 'a';
            if (node.children[c] == null) {
                return Collections.emptyList();
            }
            node = node.children[c];
        }
        return node.v;
    }
}

class Solution {
    private Trie trie = new Trie();
    private String[] words;
    private List<List<String>> ans = new ArrayList<>();

    public List<List<String>> wordSquares(String[] words) {
        this.words = words;
        for (int i = 0; i < words.length; ++i) {
            trie.insert(words[i], i);