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 arra < bb - 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] <= 10^6class Solution {
public:
vector<vector<int>> minimumAbsDifference(vector<int>& arr) {
vector<vector<int> > res;
int diff = INT_MAX;
sort(arr.begin(), arr.end());
for(int i = 1; i < arr.size(); i++){
int temp = arr[i]-arr[i-1];
cout << "$" << temp << endl;
if (temp>diff)
continue;
if (temp<diff){
diff = temp;
res.clear();
}
if (temp<=diff)
res.push_back({arr[i-1],arr[i]});
}
return res;
}
};