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Problem

Minimum Knight Moves

LeetCode #1197Medium

BFS = shortest path on the move-graph

On an infinite chessboard with a knight starting at the origin, return the minimum number of knight moves needed to reach a given target square.

Asked atAmazonGoogle

BFS over knight moves

time O(V)space O(V)

step 1 / 47

6 × 6 board · knight from (0,0) to (4,4)

queue · cells to expand

front →

(empty)

← back

BFS over knight moves


▸
1queue = [start]; dist(start) = 0
2while queue not empty:
3    cell = queue.dequeue()
4    if cell == target: return dist(cell)
5    for each of the 8 knight moves:
6        if neighbour in bounds and unvisited:
7            dist = dist(cell) + 1; enqueue it
8return -1   // unreachable

state

start(0,0)
target(4,4)

line 1Minimum Knight Moves: a knight starts at (0,0) and we want the fewest L-shaped hops to reach (4,4). The key insight: treat each square as a graph node whose neighbours are the (up to) 8 squares one knight-move away. On an UNWEIGHTED graph, BFS finds shortest paths — so the BFS layer at which (4,4) first appears is exactly the minimum number of moves.