79. Word Search

Given an m x n grid of characters board and a string word, return true if word exists in the grid.

The word can 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.

Example 1:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
Output: true

Example 2:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: true

Example 3:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
Output: false

Solution:

class Solution {
    public boolean exist(char[][] board, String word) {
        // base case
        if (board == null || board.length == 0 || board[0].length == 0){
            return false;
        }

        int row = board.length;
        int col = board[0].length;

        boolean[][] visited = new boolean[row][col];

        char[] wordArray = word.toCharArray();

        for (int i = 0; i < row; i++){ // row
            for (int j = 0; j < col; j++){ // col
                int index = 0;
                if (dfs(board, wordArray, index, i, j, visited) == true){
                    return true;
                }
            }
        } 
        return false;
    }

    private boolean dfs(char[][] board, char[] wordArray, int index, int x, int y, boolean[][] visited){
        if (index == wordArray.length){
            return true;
        }

        int row = board.length;
        int col = board[0].length;

        if (x >= row || x < 0 || y >= col || y < 0 || visited[x][y] == true || wordArray[index] != board[x][y]){
            return false;
        }

        visited[x][y] = true;

        if (dfs(board, wordArray, index+1, x+1, y, visited) ||
        dfs(board, wordArray, index + 1, x-1, y, visited)||
        dfs(board, wordArray, index +1, x, y-1, visited)||
        dfs(board, wordArray, index + 1, x, y+1, visited)){
            return true;
        }

        visited[x][y] = false;
        return false;

    }
}


// TC: O(row * col * (dfs)) = O(row * col * (branch ^ level)) = O(row * col * 4 ^ length)
// SC: O(Math.max(row * col, length)) 

class Solution {
    public boolean exist(char[][] board, String word) {
        // base case 
        if (board == null || board.length == 0 || board[0] == null || board[0].length == 0){
            return false;
        }

        int row = board.length;
        int col = board[0].length; 

        boolean[][] visited = new boolean[row][col];
        char[] array = word.toCharArray();

        for (int i = 0; i < row; i++){ /// where to start
            for (int j = 0; j < col; j++){
                int index = 0;
                if (dfs(board, index, array, i, j, visited)){
                    return true;
                }
            }
        }
        return false;

    }

    private boolean dfs(char[][] board, int index, char[] array, int x, int y, boolean[][] visited){
        int row = board.length;
        int col = board[0].length;

        if (index == array.length){
            return true;
        }

        // base case check
        if (x >= row || x < 0 || y >= col || y < 0 || visited[x][y] || array[index] != board[x][y]){
            return false;
        }

        visited[x][y] = true;

        // go int next level four directions
        if (dfs(board, index + 1, array, x + 1, y, visited) ||
        dfs(board, index+1, array, x -1 , y , visited) || 
        dfs(board, index+1, array, x, y - 1, visited) || 
        dfs(board, index + 1, array, x, y + 1, visited)){
            return true;
        }
        visited[x][y] = false;
        return false;
    }
}

// TC: O(row * col * 4 ^ length)
// SC: O(row * col)