#3283

Hard

LeetCode Problem Workspace

Maximum Number of Moves to Kill All Pawns

Calculate the maximum number of moves to eliminate all pawns using BFS, bitmasking, and precise array position math efficiently.

array math bit manipulation breadth first search game theory

Problem Statement

You are given a 50x50 chessboard with one knight located at coordinates (kx, ky) and a list of pawn positions positions[i] = [xi, yi]. The knight can move in the standard L-shaped pattern, and each pawn must be captured. Alice and Bob play a turn-based game with Alice going first, aiming to maximize the total number of moves, while Bob tries to minimize them.

Determine the maximum total number of moves required to capture all pawns on the board assuming both players play optimally. The input includes integers kx, ky for the knight's start position and a 2D array positions for the pawns' positions, all unique and not overlapping with the knight's position.

Examples

Example 1

Input: kx = 1, ky = 1, positions = [[0,0]]

Output: 4

The knight takes 4 moves to reach the pawn at (0, 0) .

Example 2

Input: kx = 0, ky = 2, positions = [[1,1],[2,2],[3,3]]

Output: 8

Example 3

Input: kx = 0, ky = 0, positions = [[1,2],[2,4]]

Output: 3

Constraints

0 <= kx, ky <= 49
1 <= positions.length <= 15
positions[i].length == 2
0 <= positions[i][0], positions[i][1] <= 49
All positions[i] are unique.
The input is generated such that positions[i] != [kx, ky] for all 0 <= i < positions.length.

Solution Approach

Preprocess moves with BFS

Use BFS from each pawn and the knight's starting position to calculate minimum moves between any two positions. Store results in a distance matrix to avoid recalculating moves repeatedly, which is crucial given the L-shaped knight movement.

Use bitmask DP for pawn capture states

Represent each combination of captured pawns as a bitmask and recursively compute the maximum total moves for each state. Use memoization to store intermediate results, leveraging the array plus math pattern to efficiently track remaining pawns.

Combine array math and game strategy

Iterate over all possible next pawns to capture using the distance matrix and current bitmask state. Apply optimal play logic: Alice maximizes and Bob minimizes moves. This approach ensures correct total move count while adhering to array indexing and bit manipulation constraints.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on the number of pawns (up to 15), producing O(2^n * n^2) due to bitmask states and pairwise distance checks. Space complexity is O(2^n * n + board_size^2) for memoization and BFS distance storage.

What Interviewers Usually Probe

The problem tests combining BFS for knight moves with bitmask dynamic programming.
Expect discussion of array indexing errors and optimizing state transitions.
Watch for understanding how Alice and Bob's optimal strategies affect total move count.

Common Pitfalls or Variants

Common pitfalls

Not correctly preprocessing knight moves with BFS, leading to repeated calculations.
Mismanaging bitmask states or forgetting memoization, causing timeouts.
Failing to handle optimal game strategy, giving incorrect maximum total moves.

Follow-up variants

Maximize moves with a bishop or rook instead of a knight, changing movement constraints.
Use larger chessboards or more pawns to test scalability of bitmask DP approach.
Minimize total moves instead of maximizing, reversing Alice and Bob strategies.

FAQ

What pattern does Maximum Number of Moves to Kill All Pawns use?

It uses an array plus math pattern combined with BFS preprocessing and bitmask dynamic programming to handle all pawn capture states.

How do I preprocess knight moves efficiently?

Run BFS from the knight and each pawn to build a distance matrix storing minimum moves between positions.

Why is bitmask DP needed in this problem?

Bitmask DP efficiently represents which pawns are captured, allowing recursion with memoization to compute maximum total moves.

Can I apply this approach to larger boards?

Yes, but BFS distance storage and bitmask DP scale exponentially with the number of pawns, so practical limits are around 15 pawns.

What common mistakes should I avoid?

Avoid recomputing distances repeatedly, mismanaging bitmask states, and ignoring Alice/Bob optimal play, all of which lead to incorrect move totals.

terminal

Solution

Solution 1: BFS + State Compression + Memoization

First, we preprocess the shortest distance for each pawn to any position on the chessboard and record it in the array $\textit{dist}$, where $\textit{dist}[i][x][y]$ represents the shortest distance for the $i$-th pawn to the position $(x, y)$.

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class Solution:
    def maxMoves(self, kx: int, ky: int, positions: List[List[int]]) -> int:
        @cache
        def dfs(last: int, state: int, k: int) -> int:
            if state == 0:
                return 0
            if k:
                res = 0
                for i, (x, y) in enumerate(positions):
                    if state >> i & 1:
                        t = dfs(i, state ^ (1 << i), k ^ 1) + dist[last][x][y]
                        if res < t:
                            res = t
                return res
            else:
                res = inf
                for i, (x, y) in enumerate(positions):
                    if state >> i & 1:
                        t = dfs(i, state ^ (1 << i), k ^ 1) + dist[last][x][y]
                        if res > t:
                            res = t
                return res

        n = len(positions)
        m = 50
        dist = [[[-1] * m for _ in range(m)] for _ in range(n + 1)]
        dx = [1, 1, 2, 2, -1, -1, -2, -2]
        dy = [2, -2, 1, -1, 2, -2, 1, -1]
        positions.append([kx, ky])
        for i, (x, y) in enumerate(positions):
            dist[i][x][y] = 0
            q = deque([(x, y)])
            step = 0
            while q:
                step += 1
                for _ in range(len(q)):
                    x1, y1 = q.popleft()
                    for j in range(8):
                        x2, y2 = x1 + dx[j], y1 + dy[j]
                        if 0 <= x2 < m and 0 <= y2 < m and dist[i][x2][y2] == -1:
                            dist[i][x2][y2] = step
                            q.append((x2, y2))

        ans = dfs(n, (1 << n) - 1, 1)
        dfs.cache_clear()
        return ans

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