[LeetCode] Problem 980 - Unique Paths III

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On a 2-dimensional grid, there are 4 types of squares:

  • 1 represents the starting square. There is exactly one starting square.
  • 2 represents the ending square. There is exactly one ending square.
  • 0 represents empty squares we can walk over.
  • -1 represents 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

No.1

Input: [[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)

No.2

Input: [[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)

No.3

Input: [[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.

Note

  1. 1 <= grid.length * grid[0].length <= 20

Code

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public int uniquePathsIII(int[][] grid) {
    int m = grid.length;
    int n = grid[0].length;
    int[][][] dp = new int[m][n][1 << (m * n)];
    int state = 0;
    int startX = 0;
    int startY = 0;
    
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (grid[i][j] == 0 || grid[i][j] == 2)
                state |= getKey(i, j, n);
            else if (grid[i][j] == 1) {
                startX = i;
                startY = j;
            }
        }
    }

    return dfs(grid, dp, state, m, n, startX, startY);
}

private int dfs(int[][] grid, int[][][] dp, int state, int m, int n, int x, int y) {
    if (dp[x][y][state] != 0)
        return dp[x][y][state];
    else if (grid[x][y] == 2)
        return state == 0 ? 1 : 0;
    
    int[][] dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    int result = 0;

    for (int[] dir : dirs){
        int newX = x + dir[0];
        int newY = y + dir[1];

        if (newX < 0 || newY < 0 || newX >= m || newY >= n || grid[newX][newY] == -1)
            continue;

        if ((state & getKey(newX, newY, n)) == 0)
            continue;
        
        result += dfs(grid, dp, state ^ getKey(newX, newY, n), m, n, newX, newY);
    }

    dp[x][y][state] = result;
    return dp[x][y][state];
}

private int getKey(int x, int y, int n) {
    return 1 << (x * n + y);
}