Minimum Absolute Difference in Python
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Minimum Absolute Difference in Python

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

Solution:

class Solution:
    def minimumAbsDifference(arr):
        if not arr:
            return []
        
        arr.sort()
        min_diff = arr[1] - arr[0]
        for i in range(2, len(arr)):
            min_diff = min(min_diff, (arr[i] - arr[i - 1]))
        
        result = []
        for i in range(1, len(arr)):
            if arr[i] - arr[i - 1] == min_diff:
                result.append([arr[i - 1], arr[i]])
        return result


arr = [4,2,1,3]
print(Solution.minimumAbsDifference(arr))

arr = [1,3,6,10,15]
print(Solution.minimumAbsDifference(arr))

arr = [3,8,-10,23,19,-4,-14,27]
print(Solution.minimumAbsDifference(arr))