# leetcode\_1200

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

## Solutions

1. **sort with two pointers**

```cpp
class Solution {
public:
    vector<vector<int>> minimumAbsDifference(vector<int>& arr) {
        sort(arr.begin(), arr.end());
        int mindiff = INT_MAX;
        for (int i = 1; i < arr.size(); i++)
            mindiff = min(mindiff, arr[i] - arr[i - 1]);

        vector<vector<int>> res;
        int i = 0, j = 1;
        while (j < arr.size()) {
            while (arr[j] - arr[i] > mindiff) i++;
            while (arr[j] - arr[i] == mindiff)
                res.push_back({arr[i++], arr[j]});
            j++;
        }

        return res;
    }
};
```