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| 1 | +package com.fishercoder.solutions; |
| 2 | + |
| 3 | +/** |
| 4 | + * 1066. Campus Bikes II |
| 5 | + * |
| 6 | + * 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. |
| 7 | + * We assign one unique bike to each worker so that the sum of the Manhattan distances between each worker and their assigned bike is minimized. |
| 8 | + * The Manhattan distance between two points p1 and p2 is Manhattan(p1, p2) = |p1.x - p2.x| + |p1.y - p2.y|. |
| 9 | + * Return the minimum possible sum of Manhattan distances between each worker and their assigned bike. |
| 10 | + * |
| 11 | + * Example 1: |
| 12 | + * Input: workers = [[0,0],[2,1]], bikes = [[1,2],[3,3]] |
| 13 | + * Output: 6 |
| 14 | + * Explanation: |
| 15 | + * We assign bike 0 to worker 0, bike 1 to worker 1. The Manhattan distance of both assignments is 3, so the output is 6. |
| 16 | + * |
| 17 | + * Example 2: |
| 18 | + * Input: workers = [[0,0],[1,1],[2,0]], bikes = [[1,0],[2,2],[2,1]] |
| 19 | + * Output: 4 |
| 20 | + * Explanation: |
| 21 | + * We first assign bike 0 to worker 0, then assign bike 1 to worker 1 or worker 2, bike 2 to worker 2 or worker 1. |
| 22 | + * Both assignments lead to sum of the Manhattan distances as 4. |
| 23 | + * |
| 24 | + * Note: |
| 25 | + * 0 <= workers[i][0], workers[i][1], bikes[i][0], bikes[i][1] < 1000 |
| 26 | + * All worker and bike locations are distinct. |
| 27 | + * 1 <= workers.length <= bikes.length <= 10 |
| 28 | + * */ |
| 29 | +public class _1066 { |
| 30 | + public static class Solution1 { |
| 31 | + int minSum = Integer.MAX_VALUE; |
| 32 | + |
| 33 | + public int assignBikes(int[][] workers, int[][] bikes) { |
| 34 | + backtracking(workers, bikes, 0, new boolean[bikes.length], 0); |
| 35 | + return minSum; |
| 36 | + } |
| 37 | + |
| 38 | + private void backtracking(int[][] workers, int[][] bikes, int workersIndex, boolean[] bikesAssigned, int currentSum) { |
| 39 | + if (workersIndex >= workers.length) { |
| 40 | + minSum = Math.min(minSum, currentSum); |
| 41 | + return; |
| 42 | + } |
| 43 | + if (currentSum > minSum) { |
| 44 | + return; |
| 45 | + } |
| 46 | + for (int j = 0; j < bikes.length; j++) { |
| 47 | + if (!bikesAssigned[j]) { |
| 48 | + bikesAssigned[j] = true; |
| 49 | + backtracking(workers, bikes, workersIndex + 1, bikesAssigned, currentSum + dist(workers[workersIndex], bikes[j])); |
| 50 | + bikesAssigned[j] = false; |
| 51 | + } |
| 52 | + } |
| 53 | + } |
| 54 | + |
| 55 | + private int dist(int[] worker, int[] bike) { |
| 56 | + return Math.abs(worker[0] - bike[0]) + Math.abs(worker[1] - bike[1]); |
| 57 | + } |
| 58 | + } |
| 59 | +} |
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