# 908. Smallest Range I

Easy
You are given an integer array `nums` and an integer `k`.
In one operation, you can choose any index `i` where `0 <= i < nums.length` and change `nums[i]` to `nums[i] + x` where `x` is an integer from the range `[-k, k]`. You can apply this operation at most once for each index `i`.
The score of `nums` is the difference between the maximum and minimum elements in `nums`.
Return the minimum score of `nums` after applying the mentioned operation at most once for each index in it.
Example 1:
Input: nums = , k = 0
Output: 0
Explanation: The score is max(nums) - min(nums) = 1 - 1 = 0.
Example 2:
Input: nums = [0,10], k = 2
Output: 6
Explanation: Change nums to be [2, 8]. The score is max(nums) - min(nums) = 8 - 2 = 6.
Example 3:
Input: nums = [1,3,6], k = 3
Output: 0
Explanation: Change nums to be [4, 4, 4]. The score is max(nums) - min(nums) = 4 - 4 = 0.
Constraints:
• `1 <= nums.length <= 10^4`
• `0 <= nums[i] <= 10^4`
• `0 <= k <= 10^4`

### 解題

func smallestRangeI(nums []int, k int) int {
max := nums
min := nums
for _, n := range nums {
if n > max { max = n }
if n < min { min = n }
}
if k*2 > max - min {
return 0
}
return max - min - k*2
}