LeetCode Problem 3001

Difficulty: 🟡 Medium
Topics: Math, Enumeration

Solution

LeetCode 3001 - Minimum Moves to Capture The Queen

Problem Understanding

This problem gives us the positions of three chess pieces on a standard 8 x 8 chessboard:

• A white rook at (a, b)
• A white bishop at (c, d)
• A black queen at (e, f)

We are allowed to move only the white pieces. The queen remains stationary.

Our goal is to determine the minimum number of moves required for either the rook or the bishop to capture the queen.

A rook moves horizontally or vertically any number of squares, but it cannot jump over another piece. A bishop moves diagonally any number of squares, and it also cannot jump over another piece.

The key observation is that the answer can only be 1 or 2.

If either the rook or the bishop can capture the queen immediately, then the answer is 1.

Otherwise, the answer is always 2. This is because a rook can always reposition itself in one move to align with the queen and capture it on the next move.

The constraints are extremely small. Every coordinate lies between 1 and 8, and there are only three pieces on the board. This means we do not need any sophisticated search algorithm. Instead, we can directly analyze the geometric relationships between the pieces.

Important edge cases include situations where:

• The rook and queen are on the same row, but the bishop blocks the path.
• The rook and queen are on the same column, but the bishop blocks the path.
• The bishop and queen lie on the same diagonal, but the rook blocks the path.
• Both rook and bishop can capture in one move.
• Neither piece can capture immediately.

The problem guarantees that no two pieces occupy the same square.

Approaches

Brute Force Approach

A brute force solution could explicitly simulate legal moves on the chessboard.

We could generate every square the rook can move to, every square the bishop can move to, and perform a breadth-first search over board states until a capture occurs. Since the board is only 8 x 8, this would eventually find the correct answer.

This approach is correct because BFS explores states in increasing order of move count, guaranteeing that the first capture found uses the minimum number of moves.

However, it is unnecessarily complicated. The board contains only three pieces, and the answer depends solely on whether a direct attack exists. Performing state exploration is excessive for such a small geometric problem.

Optimal Approach

The key observation is that the answer is always either 1 or 2.

To determine whether the answer is 1, we check:

1. Can the rook attack the queen directly?

• Same row, with the bishop not blocking.
• Same column, with the bishop not blocking.

2. Can the bishop attack the queen directly?

• Same diagonal, with the rook not blocking.

If any of these conditions hold, we return 1.

Otherwise, we return 2.

The blocking condition is important because neither piece may jump over another piece. When two pieces lie on the same attack line, we must verify that the third piece does not lie strictly between them.

Approach Comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(64²)	O(64²)	Explore board states using BFS
Optimal	O(1)	O(1)	Direct geometric analysis of attack lines

Algorithm Walkthrough

1. Check whether the rook and queen are on the same row.

If a == e, the rook could potentially attack horizontally. We then determine whether the bishop lies on that same row and whether its column lies strictly between the rook and queen. If not blocked, return 1. 2. Check whether the rook and queen are on the same column.

If b == f, the rook could potentially attack vertically. We then determine whether the bishop lies on that same column and whether its row lies strictly between the rook and queen. If not blocked, return 1. 3. Check whether the bishop and queen are on the same diagonal.

Two squares lie on the same diagonal if:

abs(c - e) == abs(d - f)

If this condition holds, determine whether the rook lies on the same diagonal and is strictly between the bishop and queen. If not blocked, return 1. 4. If none of the previous checks succeeded, return 2.

Why it works

A rook captures in one move exactly when the queen lies on the same row or column and no piece blocks the path. A bishop captures in one move exactly when the queen lies on the same diagonal and no piece blocks the path.

These are the only ways to capture in a single move. If none are possible, one of the white pieces can always reposition in one move and capture on the next, making the answer 2. Therefore the algorithm correctly returns 1 whenever an immediate capture exists and 2 otherwise.

Python Solution

class Solution:
    def minMovesToCaptureTheQueen(
        self,
        a: int,
        b: int,
        c: int,
        d: int,
        e: int,
        f: int
    ) -> int:
        
        # Rook attacks along a row
        if a == e:
            if not (c == a and min(b, f) < d < max(b, f)):
                return 1

        # Rook attacks along a column
        if b == f:
            if not (d == b and min(a, e) < c < max(a, e)):
                return 1

        # Bishop attacks along a diagonal
        if abs(c - e) == abs(d - f):
            if not (
                abs(a - e) == abs(b - f) and
                min(c, e) < a < max(c, e) and
                min(d, f) < b < max(d, f)
            ):
                return 1

        return 2

The implementation directly follows the geometric observations.

The first section checks whether the rook and queen share a row. If they do, the only possible blocker is the bishop. The bishop blocks exactly when it lies on the same row and its column falls strictly between the rook and queen.

The second section performs the analogous check for a shared column.

The third section checks whether the bishop and queen share a diagonal. If they do, the rook may potentially block that diagonal attack. We verify whether the rook lies on the same diagonal and between the bishop and queen.

If none of the three attack scenarios succeeds, immediate capture is impossible, so the answer is 2.

Go Solution

func minMovesToCaptureTheQueen(a int, b int, c int, d int, e int, f int) int {
	// Rook attacks along a row
	if a == e {
		if !(c == a && min(b, f) < d && d < max(b, f)) {
			return 1
		}
	}

	// Rook attacks along a column
	if b == f {
		if !(d == b && min(a, e) < c && c < max(a, e)) {
			return 1
		}
	}

	// Bishop attacks along a diagonal
	if abs(c-e) == abs(d-f) {
		if !(abs(a-e) == abs(b-f) &&
			min(c, e) < a && a < max(c, e) &&
			min(d, f) < b && b < max(d, f)) {
			return 1
		}
	}

	return 2
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

func min(a int, b int) int {
	if a < b {
		return a
	}
	return b
}

func max(a int, b int) int {
	if a > b {
		return a
	}
	return b
}

The Go implementation mirrors the Python solution exactly.

There are no concerns about integer overflow because all coordinates are between 1 and 8. Helper functions for abs, min, and max are provided because Go does not include them for integers in the standard language primitives.

Worked Examples

Example 1

a = 1, b = 1
c = 8, d = 8
e = 2, f = 3

Step-by-step evaluation

Check	Result
Same row? a == e	1 == 2 → No
Same column? b == f	1 == 3 → No
Same diagonal? abs(8-2) == abs(8-3)	6 == 5 → No

No immediate attack exists.

Therefore:

Answer = 2

Example 2

a = 5, b = 3
c = 3, d = 4
e = 5, f = 2

Step-by-step evaluation

Check	Result
Same row? a == e	5 == 5 → Yes
Bishop blocks? c == a	3 == 5 → No
Rook can capture immediately	Yes

Since the rook already attacks the queen directly:

Answer = 1

The bishop also attacks the queen diagonally:

abs(3 - 5) = 2
abs(4 - 2) = 2

but the algorithm already found a one-move capture.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only a fixed number of arithmetic comparisons are performed
Space	O(1)	No additional data structures are used

The board size is fixed at 8 x 8, and the algorithm performs only a handful of coordinate comparisons. The amount of work never depends on input size, so both time and space complexity are constant.

Test Cases

sol = Solution()

assert sol.minMovesToCaptureTheQueen(1, 1, 8, 8, 2, 3) == 2  # example 1
assert sol.minMovesToCaptureTheQueen(5, 3, 3, 4, 5, 2) == 1  # example 2

assert sol.minMovesToCaptureTheQueen(1, 1, 4, 3, 1, 8) == 1  # rook captures along row
assert sol.minMovesToCaptureTheQueen(1, 1, 3, 1, 8, 1) == 2  # bishop blocks rook column attack

assert sol.minMovesToCaptureTheQueen(2, 5, 4, 4, 8, 8) == 1  # bishop diagonal attack
assert sol.minMovesToCaptureTheQueen(5, 5, 1, 1, 8, 8) == 2  # rook blocks bishop attack

assert sol.minMovesToCaptureTheQueen(1, 8, 8, 1, 4, 5) == 2  # neither attacks
assert sol.minMovesToCaptureTheQueen(8, 8, 1, 2, 8, 1) == 1  # rook row attack
assert sol.minMovesToCaptureTheQueen(2, 2, 6, 6, 7, 7) == 1  # bishop diagonal attack

assert sol.minMovesToCaptureTheQueen(1, 4, 1, 6, 1, 8) == 2  # bishop blocks row attack
assert sol.minMovesToCaptureTheQueen(4, 1, 6, 1, 8, 1) == 2  # bishop blocks column attack

assert sol.minMovesToCaptureTheQueen(3, 3, 1, 1, 8, 8) == 2  # rook blocks diagonal attack

Test Summary

Test	Why
Example 1	No immediate attack exists
Example 2	Immediate capture exists
Rook same row	Basic rook attack
Rook blocked vertically	Blocking detection
Bishop diagonal attack	Basic bishop attack
Bishop blocked by rook	Diagonal blocking logic
Neither piece attacks	Requires two moves
Edge row attack	Board boundary handling
Long diagonal attack	Large diagonal distance
Bishop blocks row attack	Horizontal blocking
Bishop blocks column attack	Vertical blocking
Rook blocks diagonal attack	Diagonal obstruction

Edge Cases

Rook and queen share a row, but the bishop blocks the path

A common mistake is checking only whether the rook and queen are on the same row. The bishop may lie directly between them, making the capture illegal because rooks cannot jump over pieces.

For example:

Rook   = (1,1)
Bishop = (1,4)
Queen  = (1,8)

The implementation explicitly checks whether the bishop lies on the same row and strictly between the rook and queen before declaring a one-move capture.

Rook and queen share a column, but the bishop blocks the path

The vertical version of the previous case is equally important.

For example:

Rook   = (1,1)
Bishop = (4,1)
Queen  = (8,1)

Although the rook and queen share a column, the bishop obstructs the path. The implementation correctly identifies this and returns 2.

Bishop and queen share a diagonal, but the rook blocks the path

A bishop attack is only legal if no piece lies between the bishop and queen on that diagonal.

For example:

Bishop = (1,1)
Rook   = (3,3)
Queen  = (8,8)

All three pieces lie on the same diagonal. The rook is between the bishop and queen, preventing the bishop from capturing immediately. The implementation checks both diagonal alignment and positional ordering to detect this obstruction correctly.

Both pieces can capture in one move

Sometimes both the rook and bishop attack the queen simultaneously.

For example:

Rook   = (5,3)
Bishop = (3,4)
Queen  = (5,2)

In this situation the minimum number of moves is still 1. The implementation returns immediately when it finds any valid one-move capture, ensuring the correct minimum is reported.