1200. Minimum Absolute Difference

Description

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 <= 1⁰⁵
• -106 <= arr[i] <= 1⁰⁶

💡 Hint 1:
Find the minimum absolute difference between two elements in the array.

💡 Hint 2:
The minimum absolute difference must be a difference between two consecutive elements in the sorted array.

Submitted Code

class Solution:
    def minimumAbsDifference(self, arr: List[int]) -> List[List[int]]:
        arr.sort()
        min_diff = float('inf')
        result = []

        for a, b in zip(arr, arr[1:]):
            min_diff = min(min_diff, (b - a))

        for a, b in zip(arr, arr[1:]):
            if b - a == min_diff:
                result.append([a, b])
        
        return result

Runtime: 50 ms | Beats 66.07%
Memory: 31.41 MB | Beats 12.42%

Other Solutions

1st

class Solution:
    def minimumAbsDifference(self, arr: List[int]) -> List[List[int]]:
        minDiff = math.inf
        dic = collections.defaultdict(list)
        arr.sort()                               #O(n log n) time
        
        for i in range(len(arr)-1):              #O(n) time
          diff = arr[i+1] - arr[i]
          dic[diff].append([arr[i], arr[i+1]])   #O(n) space if all the pairs have the same minimum difference
          minDiff = min(minDiff, diff)
        return dic[minDiff]

time complexity: 𝑂(𝑛log𝑛)
space complexity: 𝑂(𝑛)

연속한 두 원소의 차이를 키로 하는 딕셔너리를 이용할 수도 있다. collections의 defaultdict는 키가 없을 때 KeyError가 나지 않고 자동으로 빈 리스트를 만들어주기 때문에 편리하다.