3001. Minimum Moves to Capture The Queen LeetCode Solution
In this guide, you will get 3001. Minimum Moves to Capture The Queen LeetCode Solution with the best time and space complexity. The solution to Minimum Moves to Capture The Queen problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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.
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
