CPMagical CP Library

Problem solution · Java

Number of Valid Move Combinations On Chessboard

Number of Valid Move Combinations On Chessboard: a Java solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
92 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

Depth-first search

For Number of Valid Move Combinations On Chessboard, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 92 lines of Java from the credited upstream file 2056.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 7 loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 codeNumber of Valid Move Combinations On Chessboard · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int countCombinations(String[] pieces, int[][] positions) {    final int n = pieces.length;    Set<Long> hashedBoards = new HashSet<>();    // Stores all possible move combinations for `pieces`. Each element is a    // vector of moves, one for each piece in the input order. e.g., if pieces =    // ["rook", "bishop"], one element might be [[1,0], [1,1]], representing a    // rook moving right and a bishop moving diagonally up-right.    List<List<int[]>> combMoves = new ArrayList<>();     getCombMoves(pieces, 0, new ArrayList<>(), combMoves);     for (List<int[]> combMove : combMoves)      dfs(positions, n, combMove, (1 << n) - 1, hashedBoards);     return hashedBoards.size();  }   private static Map<String, int[][]> MOVES = new HashMap<>();   static {    MOVES.put("rook", new int[][] {{1, 0}, {-1, 0}, {0, 1}, {0, -1}});    MOVES.put("bishop", new int[][] {{1, 1}, {1, -1}, {-1, 1}, {-1, -1}});    MOVES.put("queen",              new int[][] {{1, 0}, {-1, 0}, {0, 1}, {0, -1}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}});  }   // Generates all possible combinations of moves for the given pieces.  private void getCombMoves(String[] pieces, int ithPiece, List<int[]> path,                            List<List<int[]>> combMoves) {    if (ithPiece == pieces.length) {      combMoves.add(new ArrayList<>(path));      return;    }     for (int[] move : MOVES.get(pieces[ithPiece])) {      path.add(move);      getCombMoves(pieces, ithPiece + 1, path, combMoves);      path.remove(path.size() - 1);    }  }   // Performs a depth-first search to explore all possible board states.  private void dfs(int[][] board, int n, List<int[]> combMove, int activeMask,                   Set<Long> hashedBoards) {    if (activeMask == 0)      return;    hashedBoards.add(getHash(board));     for (int nextActiveMask = 1; nextActiveMask < 1 << n; ++nextActiveMask) {      if ((activeMask & nextActiveMask) != nextActiveMask)        continue;       // Copy the board.      int[][] nextBoard = new int[n][];      for (int i = 0; i < n; ++i)        nextBoard[i] = board[i].clone();       // Move the pieces that are active in this turn.      for (int i = 0; i < n; ++i)        if ((nextActiveMask >> i & 1) == 1) {          nextBoard[i][0] += combMove.get(i)[0];          nextBoard[i][1] += combMove.get(i)[1];        }       // No two or more pieces occupy the same square.      if (getUniqueSize(nextBoard) < n)        continue;       // Every piece needs to be in the boundary.      if (Arrays.stream(nextBoard).allMatch(p -> 1 <= p[0] && p[0] <= 8 && 1 <= p[1] && p[1] <= 8))        dfs(nextBoard, n, combMove, nextActiveMask, hashedBoards);    }  }   // Generates a unique hash for the given board state.  private long getHash(int[][] board) {    long hash = 0;    for (int[] pos : board)      hash = (hash * 64) + (pos[0] - 1 << 3) + (pos[1] - 1);    return hash;  }   // Counts the number of unique positions on the board occupied by the pieces.  private int getUniqueSize(int[][] board) {    Set<Integer> unique = new HashSet<>();    for (int[] pos : board)      unique.add(pos[0] * 8 + pos[1]);    return unique.size();  }}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗