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]]
Constraints:
2 <= arr.length <= 10^5-10^6 <= arr[i] <= 10^6
Solution:
Easy.
class Solution {
public:
vector<vector<int>> minimumAbsDifference(vector<int>& arr) {
sort(arr.begin(),arr.end());
int mina = INT_MAX;
for(int i =0 ;i <arr.size()-1;i++){
if(arr[i+1]-arr[i] < mina) mina = arr[i+1]-arr[i];
}
vector<vector<int>> ans;
for(int i =0 ;i <arr.size()-1;i++){
if(arr[i+1]-arr[i] == mina) ans.push_back(vector<int>{arr[i],arr[i+1]});
}
return ans;
}
};