LeetCode - Arrays - Minimum Absolute Difference

 

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]]
class Solution {
    public List<List<Integer>> minimumAbsDifference(int[] arr) {
        List<List<Integer>> resultList = new ArrayList<List<Integer>>();
        Arrays.sort(arr);
        //System.out.println(Arrays.toString(arr));
        //System.out.println("Min Diff >> "+(arr[1]-arr[0]));
        int min_diff = arr[1]-arr[0];
        int counter = 1;
        while(counter<arr.length){
            if(arr[counter]-arr[counter-1] < min_diff){
                min_diff = arr[counter]-arr[counter-1];
                resultList.clear();
            }
            if(arr[counter]-arr[counter-1] == min_diff){
                List<Integer> temp = new ArrayList<Integer>();
                temp.add(arr[counter-1]);
                temp.add(arr[counter]);
                resultList.add(temp);
            }
            counter++;
        }
        return resultList;
    }
}

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