3001. Minimum Moves to Capture The Queen

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Problem Description

In this problem, we are presented with a chess scenario on a standard 8x8 chess board. There are three pieces on the board: a white rook, a white bishop, and a black queen. Their positions are represented by pairs of integers (a, b for the rook, c, d for the bishop, and e, f for the queen) with coordinates that are 1-indexed.

The goal is to determine the minimum number of moves required to capture the black queen using only the white pieces. To achieve this, we need to consider the unique movement rules for each piece:

  • The rook can move any number of squares either vertically or horizontally, but it cannot jump over other pieces.
  • The bishop can move any number of squares diagonally, but, like the rook, it can't jump over other pieces.

A capture is possible if the rook or bishop can move to the square occupied by the queen. It's worth noting that in this particular problem, the queen does not move at all.


To approach this solution effectively, we need to understand the possible moves for both the rook and bishop and how they can capture the queen. Our main questions are:

  • Can the rook capture the queen? If it's in the same row or column, and there is no bishop blocking its path, then yes.
  • Can the bishop capture the queen? If it's on the same diagonal without the rook in its path, then yes.

Given these conditions, we can summarize the solution approach as follows:

  1. Check if the rook can capture the queen directly. If the rook is in the same row or column as the queen and there's no bishop on the path, this move is possible.
  2. Check if the bishop can capture the queen directly. If the bishop shares a diagonal with the queen and there's no rook in the way, it can capture the queen.
  3. If neither the rook nor the bishop can capture the queen directly, it will always be possible to capture in two moves: one to realign either the rook or the bishop, followed by the capturing move.

The check function constructed in the code represents this logic by iterating in all possible directions a piece can move (specified by dirs) and checking if a collision with the queen happens before any border or the other piece is encountered.

The final return statement uses a ternary operator to succinctly express that if either the rook or the bishop can capture the queen on the first check, then only one move is required (return 1), otherwise, two moves are needed (return 2).

Solution Approach

The solution uses a direct simulation strategy to determine if either the rook or the bishop can reach the queen's position. Let's examine each part of the implementation based on the provided Python function:

  1. Check Function - Linear Simulation: The check function simulates the movement of the chess pieces. It accepts dirs, a tuple containing directions for iteration, sx and sy for starting coordinates of the piece being moved, and bx and by for the position of the other piece used as a blocking reference.

    1def check(dirs, sx, sy, bx, by) -> bool:
    2    for dx, dy in pairwise(dirs):
    3        for k in range(1, 8):
    4            x = sx + dx * k
    5            y = sy + dy * k
    6            if not (1 <= x <= 8 and 1 <= y <= 8) or (x, y) == (bx, by):
    7                break
    8            if (x, y) == (e, f):
    9                return True
    10    return False

    Here, the pairwise utility (assumed to be similar to itertools.pairwise) constructs directional pairs based on the dirs parameter. These pairs dictate the movement of the piece in horizontal, vertical, or diagonal steps. The simulation continues as long as the piece remains within the bounds of the board and does not encounter the position of the other piece. If the piece's position overlaps with the queen's position during this process, the function returns True, indicating a potential capture.

  2. Rook and Bishop Movement Directions: Two direction tuples, dirs1 and dirs2, represent the rook's and bishop's potential movements. The rook's moves are constrained to horizontal and vertical paths while the bishop's moves are diagonal.

    1dirs1 = (-1, 0, 1, 0, -1)  # Directions for the rook: Up, Right, Down, Left
    2dirs2 = (-1, 1, 1, -1, -1)  # Directions for the bishop: NE, SE, SW, NW
  3. Calculate the Minimum Number of Moves: The main function evaluates both the rook's and bishop's ability to capture the queen by calling check with the appropriate movement directions and positions.

    1return 1 if check(dirs1, a, b, c, d) or check(dirs2, c, d, a, b) else 2

    This statement is an elegant expression that boils down the problem to a single line. If either the rook (using dirs1) or the bishop (using dirs2) can capture the queen, the minimum number of moves is 1 (since one piece can immediately move to capture the queen). Otherwise, if neither can capture in one move, it is known that we can rearrange in one move and capture in the next, hence the number of moves is 2.

The solution's effectiveness lies in its simplicity. There is no need for complex algorithms or data structures—just a straightforward simulation of movement and a conditional return based on whether a capture is possible in one move. It encapsulates the chess rules accurately within programmed control flows and loops.

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Example Walkthrough

Let's consider a small example to illustrate the solution approach. Assume we have the following positions for the rook, bishop, and queen:

  • Rook at (3, 3)
  • Bishop at (4, 5)
  • Queen at (3, 7)

According to the rules of chess, the rook moves horizontally or vertically, and the bishop moves diagonally. Let's analyze how many moves it would require to capture the queen with the given positions.

  1. Check Rook's Capture Capability:

    • The rook is at (3, 3). It can move to any square in the third row or third column unless blocked.
    • Check if rook can capture the queen directly. The queen is on (3, 7), which is in the same row as the rook.
    • There is no bishop between the rook and the queen; the bishop is located at (4, 5), which does not intersect the path from the rook to the queen.
    • Therefore, the rook can move to (3, 7) and capture the queen in one move.
  2. Check Bishop's Capture Capability (Even though not needed in this case):

    • The bishop is at (4, 5). It moves diagonally.
    • Check if bishop can capture the queen directly. The queen at (3, 7) is not on the same diagonal as the bishop, thus, the bishop cannot capture the queen in one move.
  3. Calculate Minimum Number of Moves:

    • Since the rook can directly capture the queen, there's no need to move the bishop. The minimum number of moves required is 1.

The implementation would process this example as follows:

  • It uses the check function with dirs1 to simulate the rook's movement and dirs2 for the bishop's movement.
  • For the rook with dirs1, the positions (3, 4), (3, 5), and (3, 6) are checked sequentially until (3, 7) is reached, which matches the queen's position.
  • Since the condition is met (rook can capture the queen), the check function would return True.
  • The final return statement in the main function would return 1 as either the rook or bishop can capture the queen directly.

Our simulation accurately represents the process and quickly determines the minimum moves for this scenario.

Solution Implementation

1class Solution:
2    def minMovesToCaptureTheQueen(self, rook_x: int, rook_y: int, bishop_x: int, bishop_y: int, queen_x: int, queen_y: int) -> int:
4        def check(directions, start_x, start_y, blocker_x, blocker_y) -> bool:
5            """
6            Check if the piece can move to capture the queen, considering the blocker.
8            :param directions: Tuple of directions in which the piece will move.
9            :param start_x: The starting x-coordinate of the piece.
10            :param start_y: The starting y-coordinate of the piece.
11            :param blocker_x: The x-coordinate of the blocking piece.
12            :param blocker_y: The y-coordinate of the blocking piece.
13            :return: True if the piece can capture the queen, False otherwise.
14            """
15            # Iterate over the directions pairwise (e.g., up, right, down, left for a rook)
16            for dx, dy in zip(directions, directions[1:]):
17                # Try moving the piece in the current direction
18                for k in range(1, 8):  # The chessboard is 8x8
19                    x = start_x + dx * k
20                    y = start_y + dy * k
21                    # Check board boundaries and blocker position
22                    if not (1 <= x <= 8 and 1 <= y <= 8) or (x, y) == (blocker_x, blocker_y):
23                        break
24                    # Check if we've reached the queen's position
25                    if (x, y) == (queen_x, queen_y):
26                        return True
27            return False
29        # Directions that the rook can move: up, right, down, left
30        rook_directions = (-1, 0, 1, 0, -1)
31        # Directions that the bishop can move: up-right, down-right, down-left, up-left
32        bishop_directions = (-1, 1, 1, -1, -1)
34        # Check if the rook can capture the queen considering the bishop as a blocker, or
35        # if the bishop can capture the queen considering the rook as a blocker
36        # If either can capture in one move, return 1; otherwise, it will take at least 2 moves
37        return 1 if check(rook_directions, rook_x, rook_y, bishop_x, bishop_y) or check(bishop_directions, bishop_x, bishop_y, rook_x, rook_y) else 2
1class Solution {
3    // Directions for a chess piece to move in a straight line (up, down, left, right)
4    private final int[] straightDirections = {-1, 0, 1, 0, -1};
6    // Directions for a chess piece to move diagonally
7    private final int[] diagonalDirections = {-1, 1, 1, -1, -1};
9    // Coordinates for the targeted queen
10    private int queenX, queenY;
12    // Method to find the minimum moves needed to capture the queen
13    // a, b are coordinates for the bishop
14    // c, d are coordinates for the knight
15    // e, f are coordinates for the queen
16    public int minMovesToCaptureTheQueen(int bishopX, int bishopY, int knightX, int knightY, int queenX, int queenY) {
17        this.queenX = queenX;
18        this.queenY = queenY;
19        boolean straightMove = check(straightDirections, bishopX, bishopY, knightX, knightY);
20        boolean diagonalMove = check(diagonalDirections, knightX, knightY, bishopX, bishopY);
21        // If either a straight line or a diagonal move can capture the queen, return 1
22        return straightMove || diagonalMove ? 1 : 2;
23    }
25    // Helper method to check whether the queen can be captured
26    private boolean check(int[] directions, int startX, int startY, int blockX, int blockY) {
27        // Try all four directions
28        for (int d = 0; d < 4; ++d) {
29            // Check each position in the direction up to 7 squares away (size of the board)
30            for (int k = 1; k < 8; ++k) {
31                int x = startX + directions[d] * k; // X-coordinate after moving
32                int y = startY + directions[d + 1] * k; // Y-coordinate after moving
34                // Check if the move is out of bounds or blocked by the other piece
35                if (x < 1 || x > 8 || y < 1 || y > 8 || (x == blockX && y == blockY)) {
36                    break; // Invalid move, so break out of the loop
37                }
39                // Check if the current move captures the queen
40                if (x == queenX && y == queenY) {
41                    return true; // Queen can be captured
42                }
43            }
44        }
45        // The queen is not capturable within the moves checked
46        return false;
47    }
1class Solution {
3    int minMovesToCaptureTheQueen(int knightX, int knightY, int bishopX, int bishopY, int queenX, int queenY) {
4        // Define directions for knight (L-shape) and bishop (diagonals).
5        // For the knight, the directions are combined L-shapes, i.e., 2 along x-axis and 1 along y-axis or vice versa.
6        // For the bishop, the directions are the four main diagonals.
7        int directions[2][5] = {{-1, 0, 1, 0, -1}, {-1, 1, 1, -1, -1}};
9        // Lambda function to check if a move captures the queen.
10        auto canCaptureQueen = [&](int pieceIndex, int startX, int startY, int blockX, int blockY) {
11            for (int d = 0; d < 4; ++d) { // Loop through each direction
12                for (int k = 1; k < 8; ++k) { // Loop through possible steps in a direction
13                    // Calculate the next position based on direction and step
14                    int x = startX + directions[pieceIndex][d] * k;
15                    int y = startY + directions[pieceIndex][d + 1] * k;
17                    // Check if the move is out of bounds or blocked by the other piece
18                    if (x < 1 || x > 8 || y < 1 || y > 8 || (x == blockX && y == blockY)) {
19                        break; // Move is not valid, break out of the loop
20                    }
22                    // If the calculated position captures the queen, return true
23                    if (x == queenX && y == queenY) {
24                        return true;
25                    }
26                }
27            }
28            return false; // No move captures the queen
29        };
31        // Check if either the knight or the bishop can capture the queen in one move.
32        // If either of them can, return 1. Otherwise, return 2 for multiple moves required.
33        return canCaptureQueen(0, knightX, knightY, bishopX, bishopY) || 
34               canCaptureQueen(1, bishopX, bishopY, knightX, knightY) ? 1 : 2;
35    }
1// Function to calculate the minimum number of moves to capture the queen on a chess board
2// (a, b) represents the starting position of the bishop
3// (c, d) represents the position of the blocker (a piece that can block the bishop)
4// (e, f) represents the position of the queen
5function minMovesToCaptureTheQueen(
6    bishopX: number,
7    bishopY: number,
8    blockerX: number,
9    blockerY: number,
10    queenX: number,
11    queenY: number,
12): number {
13    // Directions array for the bishop 4 diagonals: top-left, top-right, bottom-right, bottom-left
14    const directions: number[][] = [
15        [-1, 0, 1, 0, -1], // x-direction deltas
16        [-1, 1, 1, -1, -1], // y-direction deltas
17    ];
19    // Check if the queen can be captured by moving the bishop in one of the directions
20    const canCaptureQueen = (
21        directionIndex: number, 
22        startX: number, 
23        startY: number, 
24        blockX: number, 
25        blockY: number
26    ): boolean => {
27        // Loop through the 4 possible directions
28        for (let dir = 0; dir < 4; ++dir) {
29            // Move the bishop in the current direction ('k' represents distance)
30            for (let distance = 1; distance < 8; ++distance) {
31                const x = startX + directions[directionIndex][dir] * distance;
32                const y = startY + directions[directionIndex][dir + 1] * distance;
34                // If the next position is outside the chess board, stop checking this direction
35                if (x < 1 || x > 8 || y < 1 || y > 8) {
36                    break;
37                }
38                // If we encounter the blocker, stop checking this direction
39                if (x === blockX && y === blockY) {
40                    break;
41                }
42                // If we reach the position of the queen, we can capture her
43                if (x === queenX && y === queenY) {
44                    return true;
45                }
46            }
47        }
48        // If we cannot reach the queen in any direction, return false
49        return false;
50    };
52    // Check if the queen can be captured with 1 move via the bishop or the blocker
53    // If any of them can capture the queen in 1 move, return 1, otherwise return 2
54    return (
55        canCaptureQueen(0, bishopX, bishopY, blockerX, blockerY) || 
56        canCaptureQueen(1, blockerX, blockerY, bishopX, bishopY)
57    ) ? 1 : 2;

Time and Space Complexity

The given code defines a class Solution with a method minMovesToCaptureTheQueen that takes the positions of a bishop, a rook, and a queen on a chessboard and returns the minimum number of moves required for either the bishop or rook to capture the queen. The chessboard is assumed to be an 8x8 grid, and the positions are given by their respective coordinates.

Time Complexity

The time complexity of the minMovesToCaptureTheQueen method is derived from the helper function check which is performing a search in certain directions until it hits the boundaries of the chessboard (the conditions 1 <= x <= 8 and 1 <= y <= 8), or encounters the blocking piece's position ((x, y) == (bx, by)), or finds the queen's position ((x, y) == (e, f)).

This function loops over a list of directions, given by dirs, and for each direction, it potentially loops a maximum of 7 times (as the furthest distance on a chessboard in any direction is 7 squares away from any given square).

  • Since there are 4 directions to check from any starting point for straight moves (dirs1) and 4 directions for diagonal moves (dirs2), we have a fixed number of directions to iterate over.
  • For each direction, the loop can execute a maximum of 7 times.

So, the maximum number of iterations for a single check function is 4 (directions) * 7 (steps), and since we are calling the check function twice, once for straight moves and once for diagonal moves, the total maximum iteration count will be 2 * 4 * 7.

Hence, the time complexity is O(1). Since there is a fixed upper limit to the number of iterations (56), it does not scale with the size of the input.

Space Complexity

The space complexity of the method is also O(1). There are no data structures used that scale with the size of the input. The variables used in the function, including the tuples generated by pairwise(dirs), are of constant size, and the function does not have any recursive calls that would increase the stack space.


Both the time and the space complexities of the method minMovesToCaptureTheQueen are O(1) since the number of operations and the amount of space used do not depend on the input size but are determined by the fixed size of the chessboard.

Learn more about how to find time and space complexity quickly using problem constraints.

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