LeetCode 1200. Minimum absolute difference

【LetMeFly】1200. Minimum absolute difference

Force button topic link ：https://leetcode.cn/problems/minimum-absolute-difference/

Here's an array of integers  arr, Each of these elements is inequality .

Please find all the elements with the least absolute difference , And return in ascending order .

 

Example 1：

 Input ：arr = [4,2,1,3]
 Output ：[[1,2],[2,3],[3,4]]

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]]

 

Tips ：

• 2 <= arr.length <= 10^5
• -10^6 <= arr[i] <= 10^6

Method 1 ： Sort

The data range of this question is $10^5$, So it's impossible to $O(n^2)$ violence

It's not hard , because After ordering “ The element pair with the smallest absolute difference ” It must be next to

So we can sort directly , Then traverse twice

For the first time, find the minimum value of only difference in absolute value , For the second time, the difference between the absolute values is equal to that of the minimum value pair Just put it in the answer .

• Time complexity $O(n\log n)$, among $n$ It's an array $\text{arr}$ The length of
• Spatial complexity $O(\log n)$, All are ordered

AC Code

C++

class Solution {
    
public:
    vector<vector<int>> minimumAbsDifference(vector<int>& arr) {
    
        sort(arr.begin(), arr.end());
        int m = arr[1] - arr[0];
        for (int i = 1; i < arr.size(); i++) {
    
            m = min(m, arr[i] - arr[i - 1]);
        }
        vector<vector<int>> ans;
        for (int i = 1; i < arr.size(); i++) {
    
            if (arr[i] - arr[i - 1] == m) {
    
                ans.push_back({
    arr[i - 1], arr[i]});
            }
        }
        return ans;
    }
};

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