On a campus represented on the X-Y plane, there are
n workers and
m bikes, with
n <= m.
You are given an array
workers of length
workers[i] = [xi, yi] is the position of the
ith worker. You are also given an array
bikes of length
bikes[j] = [xj, yj] is the position of the
jth bike. All the given positions are unique.
Assign a bike to each worker. Among the available bikes and workers, we choose the
(workeri, bikej) pair with the shortest Manhattan distance between each other and assign the bike to that worker.
If there are multiple
(workeri, bikej) pairs with the same shortest Manhattan distance, we choose the pair with the smallest worker index. If there are multiple ways to do that, we choose the pair with the smallest bike index. Repeat this process until there are no available workers.
Return an array
answer of length
answer[i] is the index (0-indexed) of the bike that the
ith worker is assigned to.
The Manhattan distance between two points
Manhattan(p1, p2) = |p1.x - p2.x| + |p1.y - p2.y|.
Input: workers = [[0,0],[2,1]], bikes = [[1,2],[3,3]] Output: [1,0] Explanation: Worker 1 grabs Bike 0 as they are closest (without ties), and Worker 0 is assigned Bike 1. So the output is [1, 0].
Input: workers = [[0,0],[1,1],[2,0]], bikes = [[1,0],[2,2],[2,1]] Output: [0,2,1] Explanation: Worker 0 grabs Bike 0 at first. Worker 1 and Worker 2 share the same distance to Bike 2, thus Worker 1 is assigned to Bike 2, and Worker 2 will take Bike 1. So the output is [0,2,1].
n == workers.length
m == bikes.length
1 <= n <= m <= 1000
workers[i].length == bikes[j].length == 2
0 <= xi, yi < 1000
0 <= xj, yj < 1000
- All worker and bike locations are unique.
Build a distance map so that we can assign bike worker pairs in ascending order. Use hashsets to track assigned bikes and workers.
class Solution: def assignBikes(self, workers: List[List[int]], bikes: List[List[int]]) -> List[int]: results = [None for i in range(len(workers))] distances = defaultdict(list) for worker_index, worker in enumerate(workers): for bike_index, bike in enumerate(bikes): distance = self.get_distance(worker, bike) distances[distance].append([worker_index, bike_index]) assigned_workers = set() assigned_bikes = set() for key in sorted(distances.keys()): for pair in distances[key]: if pair not in assigned_workers and pair not in assigned_bikes: results[pair] = pair assigned_workers.add(pair) assigned_bikes.add(pair) return results def get_distance(self, worker, bike): return abs(worker - bike) + abs(worker - bike)
- Time Complexity: O(MN).
- Space Complexity: O(N).