src/main/java/com/fishercoder/solutions/_688.java

package com.fishercoder.solutions;

import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;

public class _688 {

    public static class Solution1 {
        /**
         * This BFS solution results in TLE on Leetcode.
         */
        public double knightProbability(int N, int K, int r, int c) {
            int[][] directions = {{-2, 1}, {-1, 2}, {1, 2}, {2, 1}, {2, -1}, {1, -2}, {-1, -2}, {-2, -1}};
            Queue<int[]> queue = new LinkedList<>();
            queue.offer(new int[]{r, c});
            int level = K;
            while (level-- > 0) {
                int size = queue.size();
                for (int i = 0; i < size; i++) {
                    int[] curr = queue.poll();
                    for (int[] direction : directions) {
                        int x = curr[0] + direction[0];
                        int y = curr[1] + direction[1];
                        if (x >= 0 && x < N && y >= 0 && y < N) {
                            queue.offer(new int[]{x, y});
                        }
                    }
                }
            }
            double prob = queue.size();
            for (int i = 0; i < K; i++) {
                prob /= 8;
            }
            return prob;
        }
    }

    public static class Solution2 {
        /**
         * Let f[r][c][k] mean the probability that the knight is still on board after k steps,
         * we can deduce a recursion from its k-1 steps
         * In addition, instead of using a 3-d array, we can only keep the most recent two layers,
         * i.e. using only two 2-d arrays.
         */
        public double knightProbability(int N, int K, int r, int c) {
            int[][] directions = {{-2, 1}, {-1, 2}, {1, 2}, {2, 1}, {2, -1}, {1, -2}, {-1, -2}, {-2, -1}};
            double[][] dp0 = new double[N][N];
            for (double[] row : dp0) {
                Arrays.fill(row, 1);
            }
            for (int k = 0; k < K; k++) {
                double[][] dp1 = new double[N][N];
                for (int i = 0; i < N; i++) {
                    for (int j = 0; j < N; j++) {
                        for (int l = 0; l < directions.length; l++) {
                            int[] direction = directions[l];
                            int x = i + direction[0];
                            int y = j + direction[1];
                            if (x >= 0 && y >= 0 && x < N && y < N) {
                                dp1[i][j] += dp0[x][y];
                            }
                        }
                    }
                }
                dp0 = dp1;
            }
            return dp0[r][c] / Math.pow(8, K);
        }
    }

    public static void main(String... args) {
        System.out.println((double) 2 / 8);
    }
}