LeetCode 3283: Maximum Number of Moves to Kill All Pawns
Problem Description
Explanation
To solve this problem, we can use a minimax algorithm with memoization. The idea is to simulate both players' moves (Alice and Bob) recursively while keeping track of the maximum total number of moves Alice can achieve. At each step, Alice maximizes the score, and Bob minimizes it. We can use memoization to avoid redundant calculations and improve the efficiency of our solution.
- Initialize a memoization table to store the maximum total number of moves achieved by Alice at each state.
- Implement a recursive function that takes the current state (knight position and remaining pawns) as input.
- In the recursive function, if there are no pawns left, return 0.
- Otherwise, iterate through all possible pawn positions and calculate the number of moves required for Alice to capture each pawn.
- For each pawn position, simulate Alice capturing the pawn and recursively call the function for Bob's turn.
- Calculate the maximum total moves achieved by Alice at the current state and store it in the memoization table.
- Return the maximum total moves achieved by Alice at the current state.
Time complexity: O(2^n) where n is the number of pawns on the board. Space complexity: O(2^n) for memoization.
Solutions
class Solution {
public int maxKnightMoves(int kx, int ky, int[][] positions) {
return maximax(kx, ky, positions, new HashMap<>());
}
private int maximax(int kx, int ky, int[][] positions, Map<String, Integer> memo) {
String key = kx + "," + ky + "," + Arrays.deepToString(positions);
if (memo.containsKey(key)) {
return memo.get(key);
}
int maxMoves = 0;
for (int i = 0; i < positions.length; i++) {
int px = positions[i][0];
int py = positions[i][1];
if (px == kx && py == ky) {
continue;
}
int moves = Math.max(Math.abs(px - kx), Math.abs(py - ky)) + 1;
int[][] newPositions = Arrays.copyOf(positions, positions.length);
newPositions[i] = new int[] {kx, ky};
maxMoves = Math.max(maxMoves, moves + minimin(px, py, newPositions, memo));
}
memo.put(key, maxMoves);
return maxMoves;
}
private int minimin(int kx, int ky, int[][] positions, Map<String, Integer> memo) {
String key = kx + "," + ky + "," + Arrays.deepToString(positions);
if (memo.containsKey(key)) {
return memo.get(key);
}
int minMoves = Integer.MAX_VALUE;
for (int i = 0; i < positions.length; i++) {
int px = positions[i][0];
int py = positions[i][1];
if (px == kx && py == ky) {
continue;
}
int moves = Math.max(Math.abs(px - kx), Math.abs(py - ky)) + 1;
int[][] newPositions = Arrays.copyOf(positions, positions.length);
newPositions[i] = new int[] {kx, ky};
minMoves = Math.min(minMoves, moves + maximax(px, py, newPositions, memo));
}
memo.put(key, minMoves);
return minMoves;
}
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