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, bare fromarra < bb - aequals to the minimum absolute difference of any two elements inarr
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).