908. Smallest Range I

1. Description

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.

2. Example

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.

3. Constraints

• 1 <= nums.length <= $10^4$
• 0 <= nums[i] <= $10^4$
• 0 <= k <= $10^4$

4. Solutions

Loop

n = nums.size()
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    int smallestRangeI(const vector<int>& nums, int k) {
        int max_value = INT_MIN;
        int min_value = INT_MAX;
        for(auto num : nums) {
            max_value = max(max_value, num);
            min_value = min(min_value, num);
        }

        return max(max_value - min_value - 2 * k, 0);
    }
};