0688. Knight Probability in Chessboard

https://leetcode.com/problems/knight-probability-in-chessboard

Description

On an n x n chessboard, a knight starts at the cell (row, column) and attempts to make exactly k moves. The rows and columns are 0-indexed, so the top-left cell is (0, 0), and the bottom-right cell is (n - 1, n - 1).

A chess knight has eight possible moves it can make, as illustrated below. Each move is two cells in a cardinal direction, then one cell in an orthogonal direction.

The knight continues moving until it has made exactly k moves or has moved off the chessboard.

Return the probability that the knight remains on the board after it has stopped moving.

Example 1:

**Input:** n = 3, k = 2, row = 0, column = 0
**Output:** 0.06250
**Explanation:** There are two moves (to (1,2), (2,1)) that will keep the knight on the board.
From each of those positions, there are also two moves that will keep the knight on the board.
The total probability the knight stays on the board is 0.0625.

Example 2:

**Input:** n = 1, k = 0, row = 0, column = 0
**Output:** 1.00000

Constraints:

  • 1 <= n <= 25

  • 0 <= k <= 100

  • 0 <= row, column <= n

ac

class Solution {
    int[][] dirs = {{-2, 1}, {-1, 2}, {1, 2}, {2, 1}, {2, -1}, {1, -2}, {-1, -2}, {-2, -1}};
    public double knightProbability(int N, int K, int r, int c) {
        double[][][] memo = new double[N][N][K+1];

        return dfs(N, K, r, c, memo);
    }

    public double dfs(int n, int k, int r, int c, double[][][] memo) {
        // exit
        if (k == 0)  return 1;
        if (memo[r][c][k] > 0) return memo[r][c][k];

        for (int[] d : dirs) {
            int r2 = r + d[0];
            int c2 = c + d[1];
            if (r2 < 0 || r2 >= n || c2 < 0 || c2 >= n) continue; // out of boundary
            memo[r][c][k] += dfs(n, k-1, r2, c2, memo) / 8;
        }

        return memo[r][c][k];
    }
}

/*
DFS + memorization, memo[r][c][k]: start at r-c, after k moves, the possibility stay in board
*/

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