Welcome to Subscribe On Youtube
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 <= 1040 <= nums[i] <= 1040 <= 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; }