Minimum Knight Moves

 

1. Clarify the problem: Before diving into solving the problem, let's clarify the requirements:

  • We are given a starting position and a target position on an infinite chessboard.
  • We need to find the minimum number of moves required for a knight to move from the starting position to the target position.

2. Analyze the problem: To solve this problem, we can use a breadth-first search (BFS) approach to explore all possible moves of the knight until we reach the target position.

3. Design an algorithm: Here is the algorithm to solve the problem:

  1. Define a set visited to keep track of visited positions on the chessboard.
  2. Define two lists queue and nextQueue to perform the BFS.
  3. Enqueue the starting position into queue.
  4. Initialize a variable moves to 0.
  5. While queue is not empty:
    • Dequeue a position from queue.
    • If the dequeued position is equal to the target position, return moves.
    • Generate all possible valid moves from the current position and enqueue them into nextQueue if they have not been visited before.
    • If queue is empty, swap queue with nextQueue and increment moves by 1.
  6. If the target position is not reachable, return -1.

4. Explain your approach: The approach involves using a breadth-first search to explore all possible moves of the knight on the chessboard. We start from the starting position and enqueue it. Then, we dequeue positions one by one, generate all possible valid moves, and enqueue them if they have not been visited before. We continue this process until we reach the target position or exhaust all possible moves. If the target position is reachable, we return the number of moves taken; otherwise, we return -1.

5. Write clean and readable code:

python
def minKnightMoves(x: int, y: int) -> int:
    # Helper function to generate possible knight moves
    def generateMoves(x: int, y: int):
        moves = [(x + 2, y + 1), (x + 1, y + 2), (x - 1, y + 2), (x - 2, y + 1),
                 (x - 2, y - 1), (x - 1, y - 2), (x + 1, y - 2), (x + 2, y - 1)]
        return moves

    # Normalize the coordinates to the positive quadrant
    x, y = abs(x), abs(y)

    queue = [(0, 0)]
    visited = set([(0, 0)])
    moves = 0

    while queue:
        for _ in range(len(queue)):
            currX, currY = queue.pop(0)
            if currX == x and currY == y:
                return moves

            nextMoves = generateMoves(currX, currY)

            for nextX, nextY in nextMoves:
                if (nextX, nextY) not in visited and nextX >= -2 and nextY >= -2:
                    visited.add((nextX, nextY))
                    queue.append((nextX, nextY))

        moves += 1

    return -1

# Test cases
print(minKnightMoves(2, 1))  # Expected output: 1
print(minKnightMoves(5, 5))  # Expected output: 4

6. Test your code: Let's test the code with the given test cases:

  • Test case 1: minKnightMoves(2, 1).
    • The expected output is 1 because the knight can move from (0, 0) to (2, 1) in one move.
  • Test case 2: minKnightMoves(5, 5).
    • The expected output is 4 because the knight can move from (0, 0) to (5, 5) in four moves: (0, 0) -> (2, 1) -> (3, 3) -> (4, 5) -> (5, 5).

7. Optimize if necessary: The current solution has a time complexity of O(N), where N is the maximum of the absolute values of x and y. This is because we explore all possible moves on the chessboard until we reach the target position.

8. Handle error cases: The code assumes that the input values of x and y are integers. If the input is not valid (e.g., None, float, or string), the code will raise a TypeError.

9. Discuss complexity analysis:

  • The time complexity of the solution is O(N), where N is the maximum of the absolute values of x and y. This is because we explore all possible moves on the chessboard until we reach the target position.
  • The space complexity is O(N) as well since we use a queue and a set to store the visited positions on the chessboard.
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