How do you solve Maximum Number of Moves to Kill All Pawns on LeetCode?

Maximum Number of Moves to Kill All Pawns (LeetCode #3283) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n² * m) time complexity and O(n²) space complexity. Key insight: Understanding BFS is crucial for calculating minimum moves in grid-based problems.

What is the brute force solution for Maximum Number of Moves to Kill All Pawns?

The brute force approach for Maximum Number of Moves to Kill All Pawns is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calculating the number of moves for the knight to capture each pawn one by one. This is done by simulating all possible moves until the knight reaches the pawn, which can be inefficient but straightforward. The optimal Optimal Solution approach improves this to O(n² * m).

What are the interview tips for Maximum Number of Moves to Kill All Pawns?

Explain your thought process clearly while coding; it shows your understanding. Practice BFS and dynamic programming problems to build intuition. Always consider edge cases, such as no pawns or pawns in unreachable positions.

What are common mistakes when solving Maximum Number of Moves to Kill All Pawns?

Not considering the knight's unique movement pattern when calculating distances. Failing to account for both players' optimal strategies in turn-based games.

#3283

Maximum Number of Moves to Kill All Pawns

Hard

ArrayMathBit ManipulationBreadth-First SearchGame TheoryBitmaskBreadth-First SearchDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n² * m)
Space	O(1)	O(n²)

💡

Intuition

Time O(n² * m)Space O(n²)

The optimal approach involves preprocessing the minimum moves between all pawns and the knight using BFS, then using dynamic programming to simulate the game. This allows us to efficiently calculate the maximum moves Alice can achieve while considering Bob's optimal play.

⚙️

Algorithm

3 steps

1Step 1: Use BFS to calculate the minimum moves between the knight and each pawn and between each pair of pawns.
2Step 2: Store these distances in a matrix.
3Step 3: Use dynamic programming to simulate the game, where Alice maximizes the total moves and Bob minimizes them.

solution.py39 lines

1# Full working Python code
2from collections import deque
3
4def knight_moves(kx, ky, positions):
5    directions = [(2, 1), (2, -1), (-2, 1), (-2, -1), (1, 2), (1, -2), (-1, 2), (-1, -2)]
6    n = len(positions)
7    dist = [[0] * (n + 1) for _ in range(n + 1)]
8
9    def bfs(start):
10        queue = deque([start])
11        visited = set([start])
12        moves = 0
13        while queue:
14            for _ in range(len(queue)):
15                x, y = queue.popleft()
16                for dx, dy in directions:
17                    nx, ny = x + dx, y + dy
18                    if 0 <= nx < 50 and 0 <= ny < 50 and (nx, ny) not in visited:
19                        visited.add((nx, ny))
20                        queue.append((nx, ny))
21                        if (nx, ny) in positions:
22                            dist[positions.index((nx, ny))][n] = moves + 1
23            moves += 1
24
25    bfs((kx, ky))
26    for i in range(n):
27        bfs(positions[i])
28
29    dp = [[0] * (1 << n) for _ in range(n + 1)]
30    for mask in range(1 << n):
31        for i in range(n):
32            if mask & (1 << i):
33                for j in range(n):
34                    if not (mask & (1 << j)):
35                        dp[i][mask] = max(dp[i][mask], dist[i][j] + dp[j][mask | (1 << j)])
36    return max(dp[i][(1 << n) - 1] for i in range(n))
37
38# Example usage
39print(knight_moves(1, 1, [(0, 0)]))  # Output: 4

ℹ

Complexity note: The time complexity is O(n² * m) where n is the number of pawns and m is the maximum number of moves required to reach a pawn. The space complexity is O(n²) due to the distance matrix storing the moves between pawns.

1Understanding BFS is crucial for calculating minimum moves in grid-based problems.
2Dynamic programming can optimize turn-based games by considering both players' strategies.

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