| comments | difficulty | edit_url | tags | |
|---|---|---|---|---|
true |
Medium |
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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.
Each time the knight is to move, it chooses one of eight possible moves uniformly at random (even if the piece would go off the chessboard) and moves there.
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 <= 250 <= k <= 1000 <= row, column <= n - 1
Let
When
When
where
The final answer is
The time complexity is
class Solution:
def knightProbability(self, n: int, k: int, row: int, column: int) -> float:
f = [[[0] * n for _ in range(n)] for _ in range(k + 1)]
for i in range(n):
for j in range(n):
f[0][i][j] = 1
for h in range(1, k + 1):
for i in range(n):
for j in range(n):
for a, b in pairwise((-2, -1, 2, 1, -2, 1, 2, -1, -2)):
x, y = i + a, j + b
if 0 <= x < n and 0 <= y < n:
f[h][i][j] += f[h - 1][x][y] / 8
return f[k][row][column]class Solution {
public double knightProbability(int n, int k, int row, int column) {
double[][][] f = new double[k + 1][n][n];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
f[0][i][j] = 1;
}
}
int[] dirs = {-2, -1, 2, 1, -2, 1, 2, -1, -2};
for (int h = 1; h <= k; ++h) {
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
for (int p = 0; p < 8; ++p) {
int x = i + dirs[p], y = j + dirs[p + 1];
if (x >= 0 && x < n && y >= 0 && y < n) {
f[h][i][j] += f[h - 1][x][y] / 8;
}
}
}
}
}
return f[k][row][column];
}
}class Solution {
public:
double knightProbability(int n, int k, int row, int column) {
double f[k + 1][n][n];
memset(f, 0, sizeof(f));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
f[0][i][j] = 1;
}
}
int dirs[9] = {-2, -1, 2, 1, -2, 1, 2, -1, -2};
for (int h = 1; h <= k; ++h) {
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
for (int p = 0; p < 8; ++p) {
int x = i + dirs[p], y = j + dirs[p + 1];
if (x >= 0 && x < n && y >= 0 && y < n) {
f[h][i][j] += f[h - 1][x][y] / 8;
}
}
}
}
}
return f[k][row][column];
}
};func knightProbability(n int, k int, row int, column int) float64 {
f := make([][][]float64, k+1)
for h := range f {
f[h] = make([][]float64, n)
for i := range f[h] {
f[h][i] = make([]float64, n)
for j := range f[h][i] {
f[0][i][j] = 1
}
}
}
dirs := [9]int{-2, -1, 2, 1, -2, 1, 2, -1, -2}
for h := 1; h <= k; h++ {
for i := 0; i < n; i++ {
for j := 0; j < n; j++ {
for p := 0; p < 8; p++ {
x, y := i+dirs[p], j+dirs[p+1]
if x >= 0 && x < n && y >= 0 && y < n {
f[h][i][j] += f[h-1][x][y] / 8
}
}
}
}
}
return f[k][row][column]
}function knightProbability(n: number, k: number, row: number, column: number): number {
const f = new Array(k + 1)
.fill(0)
.map(() => new Array(n).fill(0).map(() => new Array(n).fill(0)));
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
f[0][i][j] = 1;
}
}
const dirs = [-2, -1, 2, 1, -2, 1, 2, -1, -2];
for (let h = 1; h <= k; ++h) {
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
for (let p = 0; p < 8; ++p) {
const x = i + dirs[p];
const y = j + dirs[p + 1];
if (x >= 0 && x < n && y >= 0 && y < n) {
f[h][i][j] += f[h - 1][x][y] / 8;
}
}
}
}
}
return f[k][row][column];
}const DIR: [(i32, i32); 8] = [
(-2, -1),
(2, -1),
(-1, -2),
(1, -2),
(2, 1),
(-2, 1),
(1, 2),
(-1, 2),
];
const P: f64 = 1.0 / 8.0;
impl Solution {
#[allow(dead_code)]
pub fn knight_probability(n: i32, k: i32, row: i32, column: i32) -> f64 {
// Here dp[i][j][k] represents through `i` steps, the probability that the knight stays on the board
// Starts from row: `j`, column: `k`
let mut dp: Vec<Vec<Vec<f64>>> =
vec![vec![vec![0 as f64; n as usize]; n as usize]; k as usize + 1];
// Initialize the dp vector, since dp[0][j][k] should be 1
for j in 0..n as usize {
for k in 0..n as usize {
dp[0][j][k] = 1.0;
}
}
// Begin the actual dp process
for i in 1..=k {
for j in 0..n {
for k in 0..n {
for (dx, dy) in DIR {
let x = j + dx;
let y = k + dy;
if Self::check_bounds(x, y, n, n) {
dp[i as usize][j as usize][k as usize] +=
P * dp[(i as usize) - 1][x as usize][y as usize];
}
}
}
}
}
dp[k as usize][row as usize][column as usize]
}
#[allow(dead_code)]
fn check_bounds(i: i32, j: i32, n: i32, m: i32) -> bool {
i >= 0 && i < n && j >= 0 && j < m
}
}