# An interesting problem leetcode 910. Smallest Range II

2 min readDec 16, 2019

An nice explanation is from the following link. As it is too clear, I will not reinvent the wheel. Credits go to the original author.

All the following is from above link:

- “Sort the vector.
- Assuming there is a
`point`

, on the left of the`point`

, all elements add K, on the right of the`point`

, all elements subtract K, check each possible point. (a`point`

is between two numbers). - Cause there are two segments
`(A[0]+K, A[1]+K, ..., A[i]+K, A[i+1]-K, ..., A[n]-K)`

, the first one is on the left of the current`point`

(`A[i] + K`

is the last element of the first segment), the second one is on the right of the current`point`

(`A[i + 1] - K`

is the first element of the second segment). - For the first segment, the smallest element is
`left`

, the largest is`A[i] + K`

; For the second segment,`A[i + 1] - K`

is the smallest element, the largest is`right`

. Thus, for each point, the largest element should be either`A[i] + K`

or`right`

, the smallest element should be either`left`

or`A[i + 1] - K`

.”

Here is my contribution by provding a Python code:

`class Solution:`

def smallestRangeII(self, A: List[int], K: int) -> int:

A.sort()

if len(A) == 1:return 0

res = A[-1] - A[0]

for i in range(len(A)-1):

this_max = max(A[i]+K, A[-1]-K)

this_min = min(A[0]+K, A[i+1]-K)

res = min(res, this_max - this_min)

return res

Complexity is the sort part which is O(nlogn)