# 910. Smallest Range II

## Description

You are given an integer array nums and an integer k.

For each index i where 0 <= i < nums.length, change nums[i] to be either nums[i] + k or nums[i] - k.

The score of nums is the difference between the maximum and minimum elements in nums.

Return the minimum score of nums after changing the values at each index.

Example 1:

Input: nums = [1], 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: 3
Explanation: Change nums to be [4, 6, 3]. The score is max(nums) - min(nums) = 6 - 3 = 3.


Constraints:

• 1 <= nums.length <= 104
• 0 <= nums[i] <= 104
• 0 <= k <= 104

## Solutions

• class Solution {
public int smallestRangeII(int[] nums, int k) {
Arrays.sort(nums);
int n = nums.length;
int ans = nums[n - 1] - nums[0];
for (int i = 1; i < n; ++i) {
int mi = Math.min(nums[0] + k, nums[i] - k);
int mx = Math.max(nums[i - 1] + k, nums[n - 1] - k);
ans = Math.min(ans, mx - mi);
}
return ans;
}
}

• class Solution {
public:
int smallestRangeII(vector<int>& nums, int k) {
sort(nums.begin(), nums.end());
int n = nums.size();
int ans = nums[n - 1] - nums[0];
for (int i = 1; i < n; ++i) {
int mi = min(nums[0] + k, nums[i] - k);
int mx = max(nums[i - 1] + k, nums[n - 1] - k);
ans = min(ans, mx - mi);
}
return ans;
}
};

• class Solution:
def smallestRangeII(self, nums: List[int], k: int) -> int:
nums.sort()
ans = nums[-1] - nums[0]
for i in range(1, len(nums)):
mi = min(nums[0] + k, nums[i] - k)
mx = max(nums[i - 1] + k, nums[-1] - k)
ans = min(ans, mx - mi)
return ans


• func smallestRangeII(nums []int, k int) int {
sort.Ints(nums)
n := len(nums)
ans := nums[n-1] - nums[0]
for i := 1; i < n; i++ {
mi := min(nums[0]+k, nums[i]-k)
mx := max(nums[i-1]+k, nums[n-1]-k)
ans = min(ans, mx-mi)
}
return ans
}

• function smallestRangeII(nums: number[], k: number): number {
nums.sort((a, b) => a - b);
let ans = nums.at(-1)! - nums[0];
for (let i = 1; i < nums.length; ++i) {
const mi = Math.min(nums[0] + k, nums[i] - k);
const mx = Math.max(nums.at(-1)! - k, nums[i - 1] + k);
ans = Math.min(ans, mx - mi);
}
return ans;
}