Solution Explanation for Minimum Knight Moves

This problem asks for the minimum number of moves a knight needs to reach a target position (x, y) from (0, 0) on an infinite chessboard. We can efficiently solve this using Breadth-First Search (BFS) or Bidirectional BFS.

Approach 1: BFS (Single-Source)

This approach explores the chessboard level by level, starting from (0, 0). It uses a queue to manage the nodes to visit and a set to track visited nodes to avoid cycles.

Algorithm:

1. Initialization: Start with a queue containing the initial position (0, 0), a set of visited nodes (initially empty), and a step counter (initially 0).
2. Iteration: While the queue is not empty:
   • Dequeue a node (current position).
   • If the current position is the target, return the step counter.
   • Otherwise, for each of the eight possible knight moves from the current position:
     • Calculate the new position.
     • If the new position is not in the visited set:
       • Add it to the visited set.
       • Enqueue it.
   • Increment the step counter.
3. No Path: If the queue becomes empty before reaching the target, there's no path (which is not possible in this problem's constraints).

Time Complexity: O(N), where N is the number of nodes visited. In the worst case, this is proportional to the distance from (0,0) to (x,y). Since the search space is restricted, this is practically efficient. Space Complexity: O(N), to store the queue and the visited set.

Approach 2: Bidirectional BFS

This approach simultaneously explores from both the starting point (0, 0) and the target point (x, y). It meets in the middle, significantly reducing the search space for faraway targets.

Algorithm:

1. Initialization: Create two queues, q1 (starting from (0, 0)) and q2 (starting from (x, y)). Two hash maps (m1, m2) store visited nodes and their distances from their respective start points.
2. Iteration: While both queues are not empty:
   • Choose the smaller queue to expand.
   • Extend the queue: For each node in the queue, explore its neighbors.
     • If a neighbor is in the other queue's visited set, the shortest path is found, and the total distance is returned.
     • Otherwise, add the neighbor to the visited set and its queue.
3. No Path: If either queue becomes empty before finding a path, there's no path (which is again impossible given the constraints).

Time Complexity: O(sqrt(N)) on average, where N is the number of nodes in a single-directional BFS. This is a significant improvement over single-directional BFS for distant targets. Space Complexity: O(sqrt(N)), for the two queues and visited sets.

Code Examples (Python3)

Approach 1 (BFS):

from collections import deque
 
class Solution:
    def minKnightMoves(self, x: int, y: int) -> int:
        q = deque([(0, 0, 0)]) # (x, y, steps)
        visited = {(0, 0)}
        while q:
            curr_x, curr_y, steps = q.popleft()
            if curr_x == x and curr_y == y:
                return steps
            for dx, dy in [(2, 1), (2, -1), (-2, 1), (-2, -1), (1, 2), (1, -2), (-1, 2), (-1, -2)]:
                next_x, next_y = curr_x + dx, curr_y + dy
                if (next_x, next_y) not in visited:
                    visited.add((next_x, next_y))
                    q.append((next_x, next_y, steps + 1))

Approach 2 (Bidirectional BFS): (More complex; refer to the provided solution for a complete implementation). The core idea is outlined in the algorithm description above.

The Bidirectional BFS approach generally offers superior performance, especially for larger input values of x and y, because it explores the search space more efficiently. The single-source BFS, while simpler to implement, can become noticeably slower for distant targets.