1200

1200. Minimum Absolute Difference (Easy)

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

Companies:
Salesforce, Twitter

Related Topics:
Array, Sorting

Solution 1.

// OJ: https://leetcode.com/problems/minimum-absolute-difference/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1) extra space
class Solution {
public:
    vector<vector<int>> minimumAbsDifference(vector<int>& A) {
        sort(begin(A), end(A));
        int N = A.size(), minDiff = INT_MAX;
        vector<vector<int>> ans;
        for (int i = 1; i < N; ++i) minDiff = min(minDiff, A[i] - A[i - 1]);
        for (int i = 1; i < N; ++i) {
            if (A[i] - A[i - 1] == minDiff) ans.push_back({ A[i - 1], A[i] });
        }
        return ans;
    }
};