908. Smallest Range I

Problem

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 = [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: 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

Solution (Java)

class Solution {
    public int smallestRangeI(int[] nums, int k) {
        int min = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;
        for (int num : nums) {
            min = Math.min(min, num);
            max = Math.max(max, num);
        }
        if (min + k >= max - k) {
            return 0;
        }
        return (max - k) - (min + k);
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).