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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
class Solution {
public List<List<Integer>> minimumAbsDifference(int[] arr) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
Arrays.sort(arr);
int min = arr[1] - arr[0];
int len = arr.length;
for(int i =0 ; i <len - 1 ; i++){
if( arr[i+1] -arr[i] < min) min = arr[i+1] -arr[i];
}
for(int i =0 ; i <len - 1 ; i++){
List<Integer> list = new ArrayList<>();
if( arr[i+1] -arr[i] == min) {
list.add(arr[i]);
list.add(arr[i+1]);
}
if(list.size() == 2){
res.add(list);
list = new ArrayList<>();
}
if(list.size() == 2){
res.add(list);
list = new ArrayList<>();
}
}
return res;
}
}
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