1200. Minimum Absolute Difference(Solution || Leetcode easy || Java)

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:

class Solution {
    public List<List<Integer>> minimumAbsDifference(int[] arr) {
        //first we are going to sort our array
        Arrays.sort(arr);
    
        //initialise a variable for minimum difference
        int minDiff = Integer.MAX_VALUE;
        
        //initialise a list of list of integer
        List<List<Integer>> ans = new ArrayList<List<Integer>>();
        
        //search for the minimum difference
        for(int i = 0; i < arr.length -1; i++){
            if(Math.abs(arr[i] - arr[i+1]) < minDiff){
                minDiff = Math.abs(arr[i] - arr[i+1]);
            }
        }
        
        //the pairs between which the minimum difference exist add 
        //that pair as list in our list of list integers
        for(int i = 0; i < arr.length-1; i++){
            if(Math.abs(arr[i] - arr[i+1]) == minDiff){
                ans.add(Arrays.asList(arr[i],arr[i+1]));
            }
        }
        
        return ans;
    }
}

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