LeetCode 1200. Minimum Absolute Difference

Description

https://leetcode.com/problems/minimum-absolute-difference/

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

Explanation

Calculate pair distances and find those pairs with the minimum distance.

Python Solution

class Solution:
    def minimumAbsDifference(self, arr: List[int]) -> List[List[int]]:

        
        arr = sorted(arr)

        pair_distances = []
        
        for i in range(0, len(arr)):
            num1 = arr[i]
            
            local_min_distance = sys.maxsize
            for j in range(i + 1, len(arr)):                
                num2 = arr[j]
                distance = abs(num2 - num1)
                
                if distance > local_min_distance:
                    break
                else:
                    local_min_distance = distance
                    pair_distances.append([num1, num2, distance])
        
        distance_map = defaultdict(list)
        
        min_distance = sys.maxsize
        for pair_distance in pair_distances:
            distance_map[pair_distance[2]].append([pair_distance[0], pair_distance[1]])  
            min_distance = min(min_distance, pair_distance[2])
        
        
        return distance_map[min_distance]
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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