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, bare fromarra < bb - aequals to the minimum absolute difference of any two elements inarr
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;
}
}