1200. Minimum Absolute Difference
1200. 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]]
Constraints:
2 <= arr.length <= 10^5
-10^6 <= arr[i] <= 10^6
- Sorting
public List<List<Integer>> minimumAbsDifference(int[] arr) {
List<List<Integer>> result = new ArrayList<>();
Arrays.sort(arr);
int min = Integer.MAX_VALUE;
for(int i = 1; i < arr.length; i++) {
int diff = Math.abs(arr[i]-arr[i-1]);
if(diff <= min) {
if(diff < min) {
min = diff;
result.clear();
}
result.add(Arrays.asList(arr[i-1], arr[i]));
}
}
return result;
}