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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

• Java
• C++
• Python
• Go
• TypeScript

• 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;
  }
  
  

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