LeetCode //C - 688. Knight Probability in Chessboard

688. 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
  

From: LeetCode
Link: 688. Knight Probability in Chessboard

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Solution:

Ideas:

• Directions: The 8 possible knight moves are defined in the directions array.
  
• DP Array: We maintain a DP array dp[k][j] to store the probability of being at position (i, j) after k moves.
  
• Base Case: Initially, the knight is at (row, column) with probability 1.0.
  
• Transition: For each move k, we update the probability at each cell (i, j) by considering all the possible previous positions from which the knight could have arrived.
  
• Final Result: After k moves, the final result is the sum of the probabilities for all valid positions (i, j) on the board.
  

Code:

1. double knightProbability(int n, int k, int row, int column) {
2.     // Define the 8 possible moves of a knight
3.     int directions[8][2] = {
4.         {2, 1}, {2, -1}, {-2, 1}, {-2, -1},
5.         {1, 2}, {1, -2}, {-1, 2}, {-1, -2}
6.     };
7.    
8.     // dp[k][i][j] represents the probability of the knight being at (i, j) after k moves
9.     double dp[101][25][25] = {0}; // Maximum k = 100, n = 25
10.    
11.     // Initial position
12.     dp[0][row][column] = 1.0;
13.    
14.     // Perform k moves
15.     for (int move = 1; move <= k; ++move) {
16.         for (int i = 0; i < n; ++i) {
17.             for (int j = 0; j < n; ++j) {
18.                 // If knight is at position (i, j), check all previous 8 moves
19.                 for (int d = 0; d < 8; ++d) {
20.                     int prev_i = i + directions[d][0];
21.                     int prev_j = j + directions[d][1];
22.                     // Check if the previous position is valid
23.                     if (prev_i >= 0 && prev_i < n && prev_j >= 0 && prev_j < n) {
24.                         dp[move][i][j] += dp[move - 1][prev_i][prev_j] / 8.0;
25.                     }
26.                 }
27.             }
28.         }
29.     }
30.    
31.     // Calculate the probability that the knight remains on the board after k moves
32.     double result = 0.0;
33.     for (int i = 0; i < n; ++i) {
34.         for (int j = 0; j < n; ++j) {
35.             result += dp[k][i][j];
36.         }
37.     }
38.    
39.     return result;
40. }

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