|
4 | 4 | importjava.util.Arrays; |
5 | 5 | importjava.util.List; |
6 | 6 |
|
7 | | -/** |
8 | | - * 1200. Minimum Absolute Difference |
9 | | - * |
10 | | - * Given an array of distinct integers arr, find all pairs of elements with the minimum absolute difference of any two elements. |
11 | | - * Return a list of pairs in ascending order(with respect to pairs), each pair [a, b] follows |
12 | | - * a, b are from arr |
13 | | - * a < b |
14 | | - * b - a equals to the minimum absolute difference of any two elements in arr |
15 | | - * |
16 | | - * Example 1: |
17 | | - * Input: arr = [4,2,1,3] |
18 | | - * Output: [[1,2],[2,3],[3,4]] |
19 | | - * Explanation: The minimum absolute difference is 1. List all pairs with difference equal to 1 in ascending order. |
20 | | - * |
21 | | - * Example 2: |
22 | | - * Input: arr = [1,3,6,10,15] |
23 | | - * Output: [[1,3]] |
24 | | - * |
25 | | - * Example 3: |
26 | | - * Input: arr = [3,8,-10,23,19,-4,-14,27] |
27 | | - * Output: [[-14,-10],[19,23],[23,27]] |
28 | | - * |
29 | | - * Constraints: |
30 | | - * 2 <= arr.length <= 10^5 |
31 | | - * -10^6 <= arr[i] <= 10^6 |
32 | | - * */ |
33 | 7 | publicclass_1200 { |
34 | 8 | publicstaticclassSolution1 { |
35 | 9 | /** |
36 | 10 | * Time: O(nlogn) due to sorting |
37 | 11 | * Space: O(k) where k is the distinct number of differences between two numbers in the given array |
38 | | - * */ |
| 12 | + */ |
39 | 13 | publicList<List<Integer>>minimumAbsDifference(int[]arr) { |
40 | 14 | Arrays.sort(arr); |
41 | 15 | intminimumDiff =arr[1] -arr[0]; |
|