Word Search

Given a 2D board and a word, find if the word exists in the grid.

• Time: O(N)
• Space: O(1)

public boolean exist(char[][] board, String word) {
    if (board == null || board.length == 0 
    || board[0].length == 0 || word == null) return false;
    char[] target = word.toCharArray();
    boolean[][] visited = new boolean[board.length][board[0].length];
    for (int i = 0; i < board.length; i++) {
        for (int j = 0; j < board[0].length; j++) {
            if (board[i][j] == target[0]) {
                if (dfs(board, target, 0, i, j, visited)) return true;
            }
        }
    }
    return false;
}

public boolean dfs(char[][] board, char[] target, 
            int start, int i, int j, boolean[][] visited) {
    if (board[i][j] == target[start]) {
        visited[i][j] = true;
        if (start == target.length - 1) return true;
        if (i - 1 >= 0 && !visited[i - 1][j]) {
            if (dfs(board, target, start + 1, i - 1, j, visited)) 
              return true;
        }
        if (i + 1 < board.length && !visited[i + 1][j]) {
            if (dfs(board, target, start + 1, i + 1, j, visited)) 
              return true;
        }
        if (j - 1 >= 0 && !visited[i][j - 1]) {
            if (dfs(board, target, start + 1, i, j - 1, visited)) 
              return true;
        }
        if (j + 1 < board[0].length && !visited[i][j + 1]) {
            if (dfs(board, target, start + 1, i, j + 1, visited)) 
              return true;
        }
    }
    visited[i][j] = false;
    return false;
}