LeetCode #3001 — MEDIUM

Minimum Moves to Capture The Queen

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

There is a 1-indexed 8 x 8 chessboard containing 3 pieces.

You are given 6 integers a, b, c, d, e, and f where:

(a, b) denotes the position of the white rook.
(c, d) denotes the position of the white bishop.
(e, f) denotes the position of the black queen.

Given that you can only move the white pieces, return the minimum number of moves required to capture the black queen.

Note that:

Rooks can move any number of squares either vertically or horizontally, but cannot jump over other pieces.
Bishops can move any number of squares diagonally, but cannot jump over other pieces.
A rook or a bishop can capture the queen if it is located in a square that they can move to.
The queen does not move.

Example 1:

Input: a = 1, b = 1, c = 8, d = 8, e = 2, f = 3
Output: 2
Explanation: We can capture the black queen in two moves by moving the white rook to (1, 3) then to (2, 3).
It is impossible to capture the black queen in less than two moves since it is not being attacked by any of the pieces at the beginning.

Example 2:

Input: a = 5, b = 3, c = 3, d = 4, e = 5, f = 2
Output: 1
Explanation: We can capture the black queen in a single move by doing one of the following: 
- Move the white rook to (5, 2).
- Move the white bishop to (5, 2).

Constraints:

1 <= a, b, c, d, e, f <= 8
No two pieces are on the same square.

Step 01

Brute Force Baseline

Problem summary: There is a 1-indexed 8 x 8 chessboard containing 3 pieces. You are given 6 integers a, b, c, d, e, and f where: (a, b) denotes the position of the white rook. (c, d) denotes the position of the white bishop. (e, f) denotes the position of the black queen. Given that you can only move the white pieces, return the minimum number of moves required to capture the black queen. Note that: Rooks can move any number of squares either vertically or horizontally, but cannot jump over other pieces. Bishops can move any number of squares diagonally, but cannot jump over other pieces. A rook or a bishop can capture the queen if it is located in a square that they can move to. The queen does not move.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

Example 2

Core Insight

What unlocks the optimal approach

The minimum number of moves can be either <code>1</code> or <code>2</code>.
The answer will be <code>1</code> if the queen is on the path of the rook or bishop and none of them is in between.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3001: Minimum Moves to Capture The Queen
class Solution {
    public int minMovesToCaptureTheQueen(int a, int b, int c, int d, int e, int f) {
        if (a == e && (c != a || (d - b) * (d - f) > 0)) {
            return 1;
        }
        if (b == f && (d != b || (c - a) * (c - e) > 0)) {
            return 1;
        }
        if (c - e == d - f && (a - e != b - f || (a - c) * (a - e) > 0)) {
            return 1;
        }
        if (c - e == f - d && (a - e != f - b || (a - c) * (a - e) > 0)) {
            return 1;
        }
        return 2;
    }
}

// Accepted solution for LeetCode #3001: Minimum Moves to Capture The Queen
func minMovesToCaptureTheQueen(a int, b int, c int, d int, e int, f int) int {
	if a == e && (c != a || (d-b)*(d-f) > 0) {
		return 1
	}
	if b == f && (d != b || (c-a)*(c-e) > 0) {
		return 1
	}
	if c-e == d-f && (a-e != b-f || (a-c)*(a-e) > 0) {
		return 1
	}
	if c-e == f-d && (a-e != f-b || (a-c)*(a-e) > 0) {
		return 1
	}
	return 2
}

# Accepted solution for LeetCode #3001: Minimum Moves to Capture The Queen
class Solution:
    def minMovesToCaptureTheQueen(
        self, a: int, b: int, c: int, d: int, e: int, f: int
    ) -> int:
        if a == e and (c != a or (d - b) * (d - f) > 0):
            return 1
        if b == f and (d != b or (c - a) * (c - e) > 0):
            return 1
        if c - e == d - f and (a - e != b - f or (a - c) * (a - e) > 0):
            return 1
        if c - e == f - d and (a - e != f - b or (a - c) * (a - e) > 0):
            return 1
        return 2

// Accepted solution for LeetCode #3001: Minimum Moves to Capture The Queen
impl Solution {
    pub fn min_moves_to_capture_the_queen(a: i32, b: i32, c: i32, d: i32, e: i32, f: i32) -> i32 {
        if a == e && (c != a || (d - b) * (d - f) > 0) {
            return 1;
        }
        if b == f && (d != b || (c - a) * (c - e) > 0) {
            return 1;
        }
        if c - e == d - f && (a - e != b - f || (a - c) * (a - e) > 0) {
            return 1;
        }
        if c - e == f - d && (a - e != f - b || (a - c) * (a - e) > 0) {
            return 1;
        }
        return 2;
    }
}

// Accepted solution for LeetCode #3001: Minimum Moves to Capture The Queen
function minMovesToCaptureTheQueen(
    a: number,
    b: number,
    c: number,
    d: number,
    e: number,
    f: number,
): number {
    if (a === e && (c !== a || (d - b) * (d - f) > 0)) {
        return 1;
    }
    if (b === f && (d !== b || (c - a) * (c - e) > 0)) {
        return 1;
    }
    if (c - e === d - f && (a - e !== b - f || (a - c) * (a - e) > 0)) {
        return 1;
    }
    if (c - e === f - d && (a - e !== f - b || (a - c) * (a - e) > 0)) {
        return 1;
    }
    return 2;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(1)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.