688. Knight Probability in Chessboard LeetCode Solution

Last updated: February 19, 2025

In this guide, you will get 688. Knight Probability in Chessboard LeetCode Solution with the best time and space complexity. The solution to Knight Probability in Chessboard problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Knight Probability in Chessboard solution in C++
Knight Probability in Chessboard solution in Java
Knight Probability in Chessboard solution in Python
Additional Resources

688. Knight Probability in Chessboard LeetCode Solution image

Problem Statement of Knight Probability in Chessboard

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 <= 25
0 <= k <= 100
0 <= row, column <= n – 1

Complexity Analysis

Time Complexity: O(kn^2)
Space Complexity: O(n^2)

688. Knight Probability in Chessboard LeetCode Solution in C++

class Solution {
 public:
  double knightProbability(int n, int k, int row, int column) {
    constexpr int dirs[8][2] = {{1, 2},   {2, 1},   {2, -1}, {1, -2},
                                {-1, -2}, {-2, -1}, {-2, 1}, {-1, 2}};
    constexpr double kProb = 0.125;
    // dp[i][j] := the probability to stand on (i, j)
    vector<vector<double>> dp(n, vector<double>(n));
    dp[row][column] = 1.0;

    for (int move = 0; move < k; ++move) {
      vector<vector<double>> newDp(n, vector<double>(n));
      for (int i = 0; i < n; ++i)
        for (int j = 0; j < n; ++j)
          if (dp[i][j] > 0.0) {
            for (const auto& [dx, dy] : dirs) {
              const int x = i + dx;
              const int y = j + dy;
              if (x < 0 || x >= n || y < 0 || y >= n)
                continue;
              newDp[x][y] += dp[i][j] * kProb;
            }
          }
      dp = std::move(newDp);
    }

    return accumulate(dp.begin(), dp.end(), 0.0,
                      [](double subtotal, const vector<double>& row) {
      return subtotal + accumulate(row.begin(), row.end(), 0.0);
    });
  }
};
/* code provided by PROGIEZ */

688. Knight Probability in Chessboard LeetCode Solution in Java

class Solution {
  public double knightProbability(int n, int k, int row, int column) {
    final int[][] dirs = {{1, 2}, {2, 1}, {2, -1}, {1, -2}, {-1, -2}, {-2, -1}, {-2, 1}, {-1, 2}};
    final double kProb = 0.125;
    // dp[i][j] := the probability to stand on (i, j)
    double[][] dp = new double[n][n];
    dp[row][column] = 1.0;

    for (int move = 0; move < k; ++move) {
      double[][] newDp = new double[n][n];
      for (int i = 0; i < n; ++i)
        for (int j = 0; j < n; ++j)
          if (dp[i][j] > 0.0) {
            for (int[] dir : dirs) {
              final int x = i + dir[0];
              final int y = j + dir[1];
              if (x < 0 || x >= n || y < 0 || y >= n)
                continue;
              newDp[x][y] += dp[i][j] * kProb;
            }
          }
      dp = newDp;
    }

    return Arrays.stream(dp).flatMapToDouble(Arrays::stream).sum();
  }
}
// code provided by PROGIEZ

688. Knight Probability in Chessboard LeetCode Solution in Python

class Solution:
  def knightProbability(self, n: int, k: int, row: int, column: int) -> float:
    dirs = ((1, 2), (2, 1), (2, -1), (1, -2),
            (-1, -2), (-2, -1), (-2, 1), (-1, 2))
    # dp[i][j] := the probability to stand on (i, j)
    dp = [[0] * n for _ in range(n)]
    dp[row][column] = 1.0

    for _ in range(k):
      newDp = [[0] * n for _ in range(n)]
      for i in range(n):
        for j in range(n):
          for dx, dy in dirs:
            x = i + dx
            y = j + dy
            if 0 <= x < n and 0 <= y < n:
              newDp[i][j] += dp[x][y]
      dp = newDp

    return sum(map(sum, dp)) / 8**k
 # code by PROGIEZ

Additional Resources

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