How do you solve Knight Probability in Chessboard on LeetCode?

Knight Probability in Chessboard (LeetCode #688) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n^2 * k) time complexity and O(n^2) space complexity. Key insight: Dynamic programming can significantly reduce redundant calculations.

What is the brute force solution for Knight Probability in Chessboard?

The brute force approach for Knight Probability in Chessboard is Brute Force, with O(8^k) time complexity and O(k) space complexity. The brute-force approach involves simulating all possible moves the knight can make for exactly k moves. We will count how many of these moves keep the knight on the board and calculate the probability based on that count. The optimal Optimal Solution approach improves this to O(n^2 * k).

What algorithm or data structure is used to solve Knight Probability in Chessboard?

Knight Probability in Chessboard uses the following concepts: Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Recursion.

What are the interview tips for Knight Probability in Chessboard?

Explain your thought process clearly, especially when transitioning from brute force to optimal. Be prepared to discuss the time and space complexities of your solutions. Practice writing out the DP table and its transitions on paper.

What are common mistakes when solving Knight Probability in Chessboard?

Not considering edge cases where k = 0. Forgetting to normalize probabilities by dividing by total possible moves.

#688

Knight Probability in Chessboard

Medium

Dynamic ProgrammingDynamic ProgrammingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(8^k)	O(n^2 * k)
Space	O(k)	O(n^2)

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Intuition

Time O(n^2 * k)Space O(n^2)

The optimal solution uses dynamic programming to store the probabilities of the knight being on the board after each move. This avoids redundant calculations and significantly reduces the time complexity.

⚙️

Algorithm

3 steps

1Step 1: Create a 3D DP array where dp[moves][r][c] represents the probability of the knight being at (r, c) after 'moves' moves.
2Step 2: Initialize dp[0][row][column] = 1, as the knight starts at the given position with 100% probability.
3Step 3: Iterate through each move from 1 to k, updating the DP table based on the possible knight moves.

solution.py12 lines

1def knightProbability(n, k, row, column):
2    moves = [(2, 1), (2, -1), (-2, 1), (-2, -1), (1, 2), (1, -2), (-1, 2), (-1, -2)]
3    dp = [[[0] * n for _ in range(n)] for _ in range(k + 1)]
4    dp[0][row][column] = 1
5    for moves_left in range(1, k + 1):
6        for r in range(n):
7            for c in range(n):
8                for dr, dc in moves:
9                    nr, nc = r + dr, c + dc
10                    if 0 <= nr < n and 0 <= nc < n:
11                        dp[moves_left][nr][nc] += dp[moves_left - 1][r][c] / 8
12    return sum(dp[k][r][c] for r in range(n) for c in range(n))

ℹ

Complexity note: The time complexity is O(n^2 * k) because we iterate through each cell on the board for each of the k moves. The space complexity is O(n^2) due to the DP table storing probabilities for each cell.

1Dynamic programming can significantly reduce redundant calculations.
2Understanding the knight's movement is crucial for setting up the problem.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.