Minimum Moves for Two Knights to Meet

Two knights start at coordinates first = [x1, y1] and second = [x2, y2] on an infinite chessboard. They take turns, with the first knight moving first. On a turn, the chosen knight must make one standard knight move: two squares along one axis and one square along the other.
Return the minimum total number of moves until the two knights occupy the same coordinate. If they already share a coordinate, return 0.

Function
minimumKnightMeetingMoves(first: int[], second: int[]) → int

Examples
Example 1
first = [0,0]
second = [1,2]
return = 1
The first knight can reach the second knight in one move.
Example 2
first = [0,0]
second = [1,0]
return = 3
The adjacent displacement is the exceptional three-move case.
Example 3
first = [7,-4]
second = [7,-4]
return = 0
The knights already meet.

Constraints
first.length == second.length == 2
-10^9 <= first[i], second[i] <= 10^9

Alternating turns doesn't change the answer. Two knights closing a gap in total moves is the same as one knight traveling the displacement (dx, dy). Take absolute values, swap so dx >= dy, and you're solving the classic infinite-board knight distance. Special cases first: (0,0) is 0, (1,0) is 3, and (2,2) is 4. Otherwise compute delta = dx - dy. If dy > delta, the answer is delta - 2*floor((delta - dy) / 3). If not, it's delta - 2*floor((delta - dy) / 4). Hmm, that's the sign-flipped version people mix up, so test it against your examples. The pitfall is skipping the exceptions, or running BFS with a bounded grid and timing out. Example 2 is the (1,0) case, so it's a hint the grader checks it. With StealthCoder as your hedge in the live OA, you'd still want to verify these three exceptions yourself.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.