3001. Minimum Moves to Capture The Queen

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.

Intuition

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.

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

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.

Conclusion

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.