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 in python:
class Solution:
def minimumAbsDifference(self, arr: List[int]) -> List[List[int]]:
min_dif = 2*10**6
arr.sort()
for i in range(len(arr)-1):
dif = abs(arr[i]-arr[i+1])
if dif < min_dif:
min_dif = dif
result = []
for i in range(len(arr)-1):
dif = abs(arr[i]-arr[i+1])
if dif == min_dif:
result.append([arr[i], arr[i+1]])
return result
留言