980. Unique Paths III

Problem

You are given an m x n integer array grid where grid[i][j] could be:

• 1 representing the starting square. There is exactly one starting square.

• 2 representing the ending square. There is exactly one ending square.

• 0 representing empty squares we can walk over.

• -1 representing obstacles that we cannot walk over.

Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.

  Example 1:

Input: grid = [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
Output: 2
Explanation: We have the following two paths: 
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)

Example 2:

Input: grid = [[1,0,0,0],[0,0,0,0],[0,0,0,2]]
Output: 4
Explanation: We have the following four paths: 
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3)
3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3)
4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)

Example 3:

Input: grid = [[0,1],[2,0]]
Output: 0
Explanation: There is no path that walks over every empty square exactly once.
Note that the starting and ending square can be anywhere in the grid.

  Constraints:

• m == grid.length

• n == grid[i].length

• 1 <= m, n <= 20

• 1 <= m * n <= 20

• -1 <= grid[i][j] <= 2

• There is exactly one starting cell and one ending cell.

Solution

/**
 * @param {number[][]} grid
 * @return {number}
 */
var uniquePathsIII = function(grid) {
    var start;
    var m = grid.length;
    var n = grid[0].length;
    var emptyNum = 0;
    for (var i = 0; i < m; i++) {
        for (var j = 0; j < n; j++) {
            if (grid[i][j] === 1) start = [i, j];
            else if (grid[i][j] === 0) emptyNum++;
        }
    }
    return getPathNum(start[0], start[1], grid, 0, emptyNum);
};

var getPathNum = function(i, j, grid, visitedNum, emptyNum) {
    var res = 0;
    var directions = [
        [1, 0], // up
        [-1, 0], // down
        [0, -1], // left
        [0, 1], // right
    ];
    for (var k = 0; k < 4; k++) {
        var [diffX, diffY] = directions[k];
        if (grid[i + diffX] && grid[i + diffX][j + diffY] === 0) {
            grid[i + diffX][j + diffY] = -1;
            res += getPathNum(i + diffX, j + diffY, grid, visitedNum + 1, emptyNum);
            grid[i + diffX][j + diffY] = 0;
        } else if (grid[i + diffX] && grid[i + diffX][j + diffY] === 2) {
            res += (visitedNum === emptyNum ? 1 : 0);
        }
    }
    return res;
};

Explain:

nope.

Complexity:

• Time complexity : O(n!).
• Space complexity : O(1).