The first approach involves sorting the array and then iterating through it to find the pairs with the minimum absolute difference. By sorting, the smallest differences can only occur between consecutive elements. Hence, we sort the array and compute differences between adjacent elements to find the minimum difference.
Steps:
Time Complexity: O(n log n), due to sorting the array. Space Complexity: O(1), not considering the output space.
1using System;
2using System.Collections.Generic;
3
4public class Solution {
5 public IList<IList<int>> MinimumAbsDifference(int[] arr) {
6 Array.Sort(arr);
7 int minDiff = int.MaxValue;
8 List<IList<int>> result = new List<IList<int>>();
9 for (int i = 1; i < arr.Length; i++) {
10 int diff = arr[i] - arr[i - 1];
11 if (diff < minDiff) {
12 minDiff = diff;
13 result.Clear();
14 result.Add(new List<int> { arr[i - 1], arr[i] });
15 } else if (diff == minDiff) {
16 result.Add(new List<int> { arr[i - 1], arr[i] });
17 }
18 }
19 return result;
20 }
21}
22The C# solution similarly sorts the array and examines consecutive element differences. It maintains the minimum difference and populates the resultant list with pairs having that difference.
The second approach compares all pairs of elements to determine their absolute differences. This method ensures that all possible pairs are considered, and the pairs with the minimum difference are identified. Although not as efficient as the first approach, it explicitly checks every possible pair.
Steps:
Time Complexity: O(n^2), considering all pair combinations. Space Complexity: O(1), excluding result space.
1import java.util.ArrayList;
2import java.util.List;
3
4class Solution {
5 public List<List<Integer>> minimumAbsDifference(int[] arr) {
6 int minDiff = Integer.MAX_VALUE;
7 List<List<Integer>> result = new ArrayList<>();
8 for (int i = 0; i < arr.length; i++) {
9 for (int j = i + 1; j < arr.length; j++) {
10 int diff = Math.abs(arr[i] - arr[j]);
11 if (diff < minDiff) {
12 minDiff = diff;
13 result.clear();
14 result.add(List.of(arr[i], arr[j]));
15 } else if (diff == minDiff) {
16 result.add(List.of(arr[i], arr[j]));
17 }
18 }
19 }
20 return result;
21 }
22}
23The Java approach explicitly investigates each element pair, calculating their differences, updating the minimum difference, and storing viable pairs in a list that is returned on completion.