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Problem solution · Java

Maximum Number of Moves to Kill All Pawns

Maximum Number of Moves to Kill All Pawns: a Java solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Breadth-first search
Source
walkccc LeetCode Solutions
Length
87 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Breadth-first search

For Maximum Number of Moves to Kill All Pawns, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 87 lines of Java from the credited upstream file 3283.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup, work queue, cached states.
  • 11 loop blocks detected.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximum Number of Moves to Kill All Pawns · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int maxMoves(int kx, int ky, int[][] positions) {    final int n = positions.length;    List<int[]> positionsList = new ArrayList<>(List.of(positions));    positionsList.add(new int[] {kx, ky});    Map<Integer, Integer> hashedPositionToIndex = new HashMap<>();    // dist[i][j] := the minimum distance from positions[i] to positions[j]    int[][] dist = new int[n + 1][n + 1];     for (int i = 0; i < positionsList.size(); ++i) {      final int x = positionsList.get(i)[0];      final int y = positionsList.get(i)[1];      hashedPositionToIndex.put(hash(x, y), i);    }     for (int sourceIndex = 0; sourceIndex < n + 1; ++sourceIndex)      bfs(positionsList, sourceIndex, hashedPositionToIndex, dist);     int MAX_MASK = 1 << (n + 1);    // dp[i][mask][turn] := the maximum (Alice) or the minimum (Bob) cost to    // kill all pawns, where i is the current pawn, mask is the set of pawns    // that have been killed, and turn is the current player's turn (0 for Alice    // and 1 for Bob)    int[][][] dp = new int[n + 1][1 << (n + 1)][2];     for (int i = 0; i < n + 1; ++i)      for (int mask = 0; mask < MAX_MASK - 1; ++mask)        dp[i][mask] = new int[] {-MAX, MAX};     for (int mask = MAX_MASK - 2; mask >= 0; --mask)      for (int i = 0; i < n + 1; ++i)        for (int turn = 0; turn < 2; ++turn)          for (int j = 0; j < n; ++j) {            if ((mask >> j & 1) == 1)              continue;            final int moves = dist[i][j] + dp[j][mask | 1 << j][1 - turn];            dp[i][mask][turn] = turn == 0 ? Math.max(dp[i][mask][turn], moves) //                                          : Math.min(dp[i][mask][turn], moves);          }     // Returns the maximum cost to kill all pawns, i.e., the original positions    // array without the knight (kx, ky).    return dp[n][1 << n][0];  }   private static final int SIZE = 50;  private static final int MAX = 1_000_000;  private static final int[][] DIRS = {{1, 2},   {2, 1},   {2, -1}, {1, -2},                                       {-1, -2}, {-2, -1}, {-2, 1}, {-1, 2}};   private int hash(int x, int y) {    return x * SIZE + y;  }   // Computes the distance between positions[sourceIndex] and other positions.  private void bfs(List<int[]> positions, int sourceIndex,                   Map<Integer, Integer> hashedPositionToIndex, int[][] dist) {    final int sx = positions.get(sourceIndex)[0];    final int sy = positions.get(sourceIndex)[1];    Queue<Pair<Integer, Integer>> q = new ArrayDeque<>(List.of(new Pair<>(sx, sy)));    boolean[][] seen = new boolean[SIZE][SIZE];    seen[sx][sy] = true;    int seenPositions = 0;     for (int step = 0; !q.isEmpty() && seenPositions < positions.size(); ++step)      for (int sz = q.size(); sz > 0; --sz) {        final int i = q.peek().getKey();        final int j = q.poll().getValue();        final int hashedPosition = hash(i, j);        if (hashedPositionToIndex.containsKey(hashedPosition)) {          dist[sourceIndex][hashedPositionToIndex.get(hashedPosition)] = step;          ++seenPositions;        }        for (int[] dir : DIRS) {          final int x = i + dir[0];          final int y = j + dir[1];          if (x < 0 || x >= SIZE || y < 0 || y >= SIZE)            continue;          if (seen[x][y])            continue;          q.offer(new Pair<>(x, y));          seen[x][y] = true;        }      }  }}

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