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Minimum Knight Moves

Updated on 18 June, 2025
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Problem Statement

In the realm of a vast, infinite chessboard that stretches in all directions from negative to positive infinity, you control a knight stationed initially at the coordinate [0, 0]. In chess, a knight moves in an "L" shape: it can move two squares in one of the four cardinal directions (north, south, east, west) and then one square perpendicular to that direction, resulting in eight possible moves from any position.

The task is to determine the minimal number of moves the knight requires to reach a specified square [x, y]. This calculation must account for the knight’s unique movement capabilities and the infinite nature of the board. Considering the board's infiniteness ensures that any target position can indeed be reached from the starting point.

Examples

Example 1

Input:

x = 2, y = 1

Output:

1

Explanation:

[0, 0] → [2, 1]

Example 2

Input:

x = 5, y = 5

Output:

4

Explanation:

[0, 0] → [2, 1] → [4, 2] → [3, 4] → [5, 5]

Constraints

  • -300 <= x, y <= 300
  • 0 <= |x| + |y| <= 300

Approach and Intuition

The challenge fundamentally revolves around the breadth-first search (BFS) algorithm, commonly used in grid-based pathfinding problems like this one involving the knight’s movement on an infinite chessboard. The idea is to explore all possible moves from a given position in systematic "waves" outwards until the target is reached. Here’s a breakdown based on the provided examples and constraints:

  • Initial Thoughts:

    • A knight can potentially move to 8 different positions from any given point.
    • Given the condition of the board’s infinite size and symmetry, the movement strategy remains consistent no matter where on the board the knight is positioned.

  • Utilizing BFS:

    1. Start from the initial position [0, 0] and consider it as the root node in a BFS traversal.
    2. From each position, calculate all valid moves that can be made by the knight.
    3. Each of these moves represents a potential node or vertex in the search space.
    4. Continue expanding outwards, visiting each position by moving from each node to its neighbors according to the knight’s movement rules.
    5. Use a queue to manage the sequence of positions to explore and a set or map to keep track of already visited positions (avoiding cycles and redundant work).
    6. The search halts when the target [x, y] is reached.

  • Efficiency Considerations:

    • Because the board is symmetric relative to the knight’s movements, calculations often only need to consider one quadrant and mirror the results.
    • Use the constraints 0 <= |x| + |y| <= 300 effectively to manage and anticipate the resources needed for the computation—knowing the maximum distance helps in optimizing the BFS traversal.

By applying this method, it's possible to efficiently determine the minimum steps required for the knight to reach the target square on the chessboard, leaning on BFS to systematically explore the infinite set of moves while adhering to the game's rules.

Solutions

  • Java
  • Python
java 
class Solution {
    private Map<String, Integer> cache = new HashMap<>();

    private int search(int x, int y) {
        String id = x + "," + y;
        if (cache.containsKey(id)) {
            return cache.get(id);
        }

        if (x + y == 0) {
            return 0;
        } else if (x + y == 2) {
            return 2;
        } else {
            Integer result = Math.min(search(Math.abs(x - 1), Math.abs(y - 2)),
                    search(Math.abs(x - 2), Math.abs(y - 1))) + 1;
            cache.put(id, result);
            return result;
        }
    }

    public int minKnightMoves(int x, int y) {
        return search(Math.abs(x), Math.abs(y));
    }
}

The Java solution provided for calculating the minimum moves a knight needs to reach a specific position on an infinite chessboard efficiently utilizes dynamic programming with memoization.

Key components of the solution are:

  • Memoization Strategy: A HashMap named cache stores calculated results for positions (x, y) to avoid redundant computations. Positions are stored as strings in "x,y" format.

  • Recursive Function search: This function computes the minimum number of moves for the knight to reach position (x, y). It implements the following logic:

    1. Checks if the result for (x, y) is already computed by querying cache.
    2. Base cases handle simple scenarios:
      • (x + y == 0): Directly at the target (0, 0) requires no moves.
      • (x + y == 2): Positions like (1,1), (2,0), or (0,2) from (0,0) require exactly two moves.
    3. It recursively calculates the minimum moves for the two permissible knight jumps in reverse direction that move closer to (0, 0), then adds one to account for the current step.

  • Positive Coordinate Enforcement: By using Math.abs() for x and y in the recursive function, the solution avoids unnecessary calculations for negative coordinates, leveraging the symmetric property of the chess board.

  • Main Function minKnightMoves: Starts the recursive computation process with absolute coordinates, ensuring correct results irrespective of the quadrant of the target position.

This solution uses recursion combined with memoization (caching intermediate results) which efficiently reduces the computational complexity, leveraging both time-saving and space-optimization via storing minimum steps required for specific coordinates.

python 
class Solution:
    def minimumKnightMoves(self, x: int, y: int) -> int:

        @lru_cache(maxsize=None)
        def find_moves(dx, dy):
            if dx + dy == 0:
                return 0
            elif dx + dy == 2:
                return 2
            else:
                return min(find_moves(abs(dx - 1), abs(dy - 2)), find_moves(abs(dx - 2), abs(dy - 1))) + 1

        return find_moves(abs(x), abs(y))

The provided code solves the problem of determining the minimum number of moves a knight in chess needs to reach a specific point (x, y) from the origin (0, 0). It uses Python with a recursive approach, enhanced by memoization through the lru_cache decorator to optimize performance.

  • First, the minimumKnightMoves function is defined within a class named Solution. This function takes two integers x and y, which represent the target coordinates on the chessboard.
  • The find_moves function, defined inside minimumKnightMoves, uses recursion to calculate the minimum moves. It is decorated with lru_cache to store the results of computations and avoid redundant calculations in future calls.
  • In the recursion:
    • If dx + dy == 0, the knight is already at the target, so it returns 0.
    • If dx + dy == 2, it returns 2, handling a base case in chess movement.
    • Otherwise, it recursively calculates the minimum moves needed from the two possible knight moves that bring it closest to the origin (since knights move in an "L" shape) and adds 1 for the current move.
  • The calls to the recursive function use the absolute values of differences dx and dy between the current position and the target, ensuring the calculations are based on the minimal moves in a non-negative coordinate space.

This approach effectively uses the properties of knight's movement and recursion, backed by caching, to efficiently solve for the minimum moves in a chess-based scenario. By always reducing the problem into smaller sub-problems aligned with the knight's movement rules and caching these results, the solution is both scalable and performance-optimized.

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