3001. Minimum Moves to Capture The Queen LeetCode Solution

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Problem Statement
Complexity Analysis
Minimum Moves to Capture The Queen solution in C++
Minimum Moves to Capture The Queen solution in Java
Minimum Moves to Capture The Queen solution in Python
Additional Resources

3001. Minimum Moves to Capture The Queen LeetCode Solution image

Problem Statement of Minimum Moves to Capture The Queen

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.

Complexity Analysis

Time Complexity: O(1)
Space Complexity: O(1)

3001. Minimum Moves to Capture The Queen LeetCode Solution in C++

class Solution {
 public:
  int minMovesToCaptureTheQueen(int a, int b, int c, int d, int e, int f) {
    // The rook is in the same row as the queen.
    if (a == e)
      // The bishop blocks the rook or not.
      return (c == a && (b < d && d < f || b > d && d > f)) ? 2 : 1;
    // The rook is in the same column as the queen.
    if (b == f)
      // The bishop blocks the rook or not.
      return (d == f && (a < c && c < e || a > c && c > e)) ? 2 : 1;
    // The bishop is in the same up-diagonal as the queen.
    if (c + d == e + f)
      // The rook blocks the bishop or not.
      return (a + b == c + d && (c < a && a < e || c > a && a > e)) ? 2 : 1;
    // The bishop is in the same down-diagonal as the queen.
    if (c - d == e - f)
      // The rook blocks the bishop or not.
      return (a - b == c - d && (c < a && a < e || c > a && a > e)) ? 2 : 1;
    // The rook can always get the green in two steps.
    return 2;
  }
};
/* code provided by PROGIEZ */

3001. Minimum Moves to Capture The Queen LeetCode Solution in Java

class Solution {
  public int minMovesToCaptureTheQueen(int a, int b, int c, int d, int e, int f) {
    // The rook is in the same row as the queen.
    if (a == e)
      // The bishop blocks the rook or not.
      return (c == a && (b < d && d < f || b > d && d > f)) ? 2 : 1;
    // The rook is in the same column as the queen.
    if (b == f)
      // The bishop blocks the rook or not.
      return (d == f && (a < c && c < e || a > c && c > e)) ? 2 : 1;
    // The bishop is in the same up-diagonal as the queen.
    if (c + d == e + f)
      // The rook blocks the bishop or not.
      return (a + b == c + d && (c < a && a < e || c > a && a > e)) ? 2 : 1;
    // The bishop is in the same down-diagonal as the queen.
    if (c - d == e - f)
      // The rook blocks the bishop or not.
      return (a - b == c - d && (c < a && a < e || c > a && a > e)) ? 2 : 1;
    // The rook can always get the green in two steps.
    return 2;
  }
}
// code provided by PROGIEZ

3001. Minimum Moves to Capture The Queen LeetCode Solution in Python

class Solution:
  def minMovesToCaptureTheQueen(
      self, a: int, b: int, c: int, d: int, e: int, f: int,
  ) -> int:
    # The rook is in the same row as the queen.
    if a == e:
      # The bishop blocks the rook or not.
      return 2 if c == a and (b < d < f or b > d > f) else 1
    # The rook is in the same column as the queen.
    if b == f:
      # The bishop blocks the rook or not.
      return 2 if d == f and (a < c < e or a > c > e) else 1
    # The bishop is in the same up-diagonal as the queen.
    if c + d == e + f:
      # The rook blocks the bishop or not.
      return 2 if a + b == c + d and (c < a < e or c > a > e) else 1
    # The bishop is in the same down-diagonal as the queen.
    if c - d == e - f:
      # The rook blocks the bishop or not.
      return 2 if a - b == c - d and (c < a < e or c > a > e) else 1
    # The rook can always get the green in two steps.
    return 2
 # code by PROGIEZ