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 fromarr
a < b
b - a
equals 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: Sorting
The min abs difference could only happen between consecutive numbers in sorted form.
Time complexity: O(nlogn)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: vector<vector<int>> minimumAbsDifference(vector<int>& arr) { sort(begin(arr), end(arr)); vector<vector<int>> ans; int best = INT_MAX; for (int i = 1; i < arr.size(); ++i) { int d = abs(arr[i] - arr[i - 1]); if (d < best) { ans.clear(); best = d; } if (d == best) ans.push_back({arr[i - 1], arr[i]}); } return ans; } }; |
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