LeetCode 1057. Campus Bikes

Description

https://leetcode.com/problems/campus-bikes/

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 n where workers[i] = [xi, yi] is the position of the ith worker. You are also given an array bikes of length m where 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 n, where answer[i] is the index (0-indexed) of the bike that the ith worker is assigned to.

The Manhattan distance between two points p1 and p2 is Manhattan(p1, p2) = |p1.x - p2.x| + |p1.y - p2.y|.

Example 1:

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].

Example 2:

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].

Constraints:

  • 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.

Python Solution

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[0] not in assigned_workers and pair[1] not in assigned_bikes:
                    results[pair[0]] = pair[1]

                    assigned_workers.add(pair[0])
                    assigned_bikes.add(pair[1])

                
        return results
        
        
    def get_distance(self, worker, bike):
        return abs(worker[0] - bike[0]) + abs(worker[1] - bike[1])
  • Time Complexity: O(MN).
  • Space Complexity: O(N).

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