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| 1 | +package com.fishercoder.solutions; |
| 2 | + |
| 3 | +import java.util.PriorityQueue; |
| 4 | +import java.util.TreeMap; |
| 5 | + |
| 6 | +/** |
| 7 | + * 1057. Campus Bikes |
| 8 | + * |
| 9 | + * On a campus represented as a 2D grid, there are N workers and M bikes, with N <= M. Each worker and bike is a 2D coordinate on this grid. |
| 10 | + * Our goal is to assign a bike to each worker. |
| 11 | + * Among the available bikes and workers, we choose the (worker, bike) pair with the shortest Manhattan distance between each other, |
| 12 | + * and assign the bike to that worker. (If there are multiple (worker, bike) pairs with the same shortest Manhattan distance, |
| 13 | + * we choose the pair with the smallest worker index; if there are multiple ways to do that, |
| 14 | + * we choose the pair with the smallest bike index). We repeat this process until there are no available workers. |
| 15 | + * |
| 16 | + * The Manhattan distance between two points p1 and p2 is Manhattan(p1, p2) = |p1.x - p2.x| + |p1.y - p2.y|. |
| 17 | + * Return a vector ans of length N, where ans[i] is the index (0-indexed) of the bike that the i-th worker is assigned to. |
| 18 | + * |
| 19 | + * Example 1: |
| 20 | + * Input: workers = [[0,0],[2,1]], bikes = [[1,2],[3,3]] |
| 21 | + * Output: [1,0] |
| 22 | + * Explanation: |
| 23 | + * Worker 1 grabs Bike 0 as they are closest (without ties), and Worker 0 is assigned Bike 1. So the output is [1, 0]. |
| 24 | + * |
| 25 | + * Example 2: |
| 26 | + * Input: workers = [[0,0],[1,1],[2,0]], bikes = [[1,0],[2,2],[2,1]] |
| 27 | + * Output: [0,2,1] |
| 28 | + * Explanation: |
| 29 | + * Worker 0 grabs Bike 0 at first. Worker 1 and Worker 2 share the same distance to Bike 2, |
| 30 | + * thus Worker 1 is assigned to Bike 2, and Worker 2 will take Bike 1. So the output is [0,2,1]. |
| 31 | + * |
| 32 | + * Note: |
| 33 | + * 0 <= workers[i][j], bikes[i][j] < 1000 |
| 34 | + * All worker and bike locations are distinct. |
| 35 | + * 1 <= workers.length <= bikes.length <= 1000 |
| 36 | + * */ |
| 37 | +public class _1057 { |
| 38 | + public static class Solution1 { |
| 39 | + public int[] assignBikes(int[][] workers, int[][] bikes) { |
| 40 | + int w = workers.length; |
| 41 | + int b = bikes.length; |
| 42 | + TreeMap<Integer, PriorityQueue<int[]>> treeMap = new TreeMap<>(); |
| 43 | + for (int i = 0; i < w; i++) { |
| 44 | + for (int j = 0; j < b; j++) { |
| 45 | + int distance = Math.abs(workers[i][0] - bikes[j][0]) + Math.abs(workers[i][1] - bikes[j][1]); |
| 46 | + if (!treeMap.containsKey(distance)) { |
| 47 | + treeMap.put(distance, new PriorityQueue<>((x, y) -> x[0] == y[0] ? x[1] - y[1] : x[0] - y[0])); |
| 48 | + } |
| 49 | + treeMap.get(distance).add(new int[]{i, j}); |
| 50 | + } |
| 51 | + } |
| 52 | + int[] ans = new int[w]; |
| 53 | + for (int i = 0; i < w; i++) { |
| 54 | + ans[i] = -1; |
| 55 | + } |
| 56 | + boolean[] assigned = new boolean[b]; |
| 57 | + int workersHaveBikes = 0; |
| 58 | + for (int dist : treeMap.keySet()) { |
| 59 | + PriorityQueue<int[]> workerBikePairs = treeMap.get(dist); |
| 60 | + while (!workerBikePairs.isEmpty()) { |
| 61 | + int[] workerBikePair = workerBikePairs.poll(); |
| 62 | + if (ans[workerBikePair[0]] == -1) { |
| 63 | + if (!assigned[workerBikePair[1]]) { |
| 64 | + assigned[workerBikePair[1]] = true; |
| 65 | + ans[workerBikePair[0]] = workerBikePair[1]; |
| 66 | + workersHaveBikes++; |
| 67 | + } |
| 68 | + } |
| 69 | + } |
| 70 | + if (workersHaveBikes == w) { |
| 71 | + return ans; |
| 72 | + } |
| 73 | + } |
| 74 | + return ans; |
| 75 | + } |
| 76 | + } |
| 77 | +} |
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