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Word Search II Backtracking with Trie

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Given an m x n board of characters and a list of strings words, return all words on the board.

Each word must be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

public List<String> findWords(char[][] board, String[] words) {
    Trie trie = buildTrie(words);
    Set<String> res = new HashSet<>();
    for (int i = 0; i < board.length; i++) {
        for (int j = 0; j < board[0].length; j++) {
            dfs(board, trie, res, i, j);
        }
    }
    return new ArrayList<>(res);
}

public void dfs(char[][] board, Trie node, Set<String> res, int i, int j) {
    if (i < 0 || i >= board.length || j < 0 || j >= board[0].length || 
        board[i][j] == '#' || node.next[board[i][j] - 'a'] == null) {
            return;
    }
    if (node.next[board[i][j] - 'a'].word != null) {
        res.add(node.next[board[i][j] - 'a'].word);
    }

    // Go to next char
    node = node.next[board[i][j] - 'a']; 
    char c = board[i][j];
    board[i][j] = '#';
    dfs(board, node, res, i - 1, j);
    dfs(board, node, res, i + 1, j);
    dfs(board, node, res, i, j - 1);
    dfs(board, node, res, i, j + 1);
    board[i][j] = c;
}   

public Trie buildTrie(String[] words) {
    Trie root = new Trie();
    for (String w : words) {
        Trie p = root;
        for (char c : w.toCharArray()) {
            if (p.next[c - 'a'] == null) {
                p.next[c - 'a'] = new Trie();
            }
            p = p.next[c - 'a'];  // will point to curr char
        }
        p.word = w;
    }
    return root;
}

private class Trie {
    Trie[] next = new Trie[26];
    String word = null;
}

Posted by Jamie Meyer 9 months ago

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