LeetCode-in-Java

1200. Minimum Absolute Difference

Easy

Given an array of distinct integers arr, find all pairs of elements with the minimum absolute difference of any two elements.

Return a list of pairs in ascending order(with respect to pairs), each pair [a, b] follows

• a, b are from arr
• a < b
• b - a equals to the minimum absolute difference of any two elements in arr

Example 1:

Input: arr = [4,2,1,3]

Output: [[1,2],[2,3],[3,4]]

Explanation: The minimum absolute difference is 1. List all pairs with difference equal to 1 in ascending order.

Example 2:

Input: arr = [1,3,6,10,15]

Output: [[1,3]]

Example 3:

Input: arr = [3,8,-10,23,19,-4,-14,27]

Output: [[-14,-10],[19,23],[23,27]]

Constraints:

• 2 <= arr.length <= 105
• -106 <= arr[i] <= 106

Solution

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class Solution {
    public List<List<Integer>> minimumAbsDifference(int[] arr) {
        List<List<Integer>> result = new ArrayList<>();
        int min = 10000000;
        Arrays.sort(arr);
        for (int i = 0; i + 1 < arr.length; i++) {
            int diff = arr[i + 1] - arr[i];
            if (diff <= min) {
                if (diff < min) {
                    min = diff;
                    result.clear();
                }
                result.add(Arrays.asList(arr[i], arr[i + 1]));
            }
        }
        return result;
    }
}

This site is open source. Improve this page.