README.md

910. Smallest Range II

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 (Rust)

1. Solution

impl Solution {
    pub fn smallest_range_ii(nums: Vec<i32>, k: i32) -> i32 {
        let n = nums.len();
        let mut nums = nums.into_iter().map(|x| x + k).collect::<Vec<_>>();
        let mut ret = 0;

        nums.sort_unstable();

        if nums[n - 1] - 2 * k < nums[0] {
            ret = nums[n - 1] - nums[0];
        } else {
            for i in 1..nums.len() {
                if nums[i] - 2 * k >= nums[0] {
                    ret = nums[i - 1].max(nums[n - 1] - 2 * k) - nums[0];
                    break;
                }
            }
        }

        for i in 1..nums.len() {
            if nums[i] - 2 * k > nums[0] {
                break;
            }

            ret = ret.min(nums[i - 1].max(nums[n - 1] - 2 * k) - nums[i] + 2 * k);
        }

        ret
    }
}