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 fromarr
-
a < b
-
b - a
equals 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 <= 105
-
-106 <= arr[i] <= 106
SOLUTION:
class Solution:
def minimumAbsDifference(self, arr: List[int]) -> List[List[int]]:
n = len(arr)
mindiff = float('inf')
mindiffindexes = []
arr.sort()
for i in range(n - 1):
currdiff = arr[i + 1] - arr[i]
if currdiff < mindiff:
mindiff = currdiff
mindiffindexes = []
if currdiff == mindiff:
mindiffindexes.append(i)
return [[arr[i], arr[i + 1]] for i in mindiffindexes]
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