946. Smallest Range Ii¶

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946. Smallest Range II

Medium

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

Solution¶

class Solution {
    public int smallestRangeII(int[] nums, int k) {
        int n = nums.length;
        Arrays.sort(nums);
        int maxi = nums[n - 1] , mini = nums[0], res = maxi - mini;
        for (int i = 0; i < n - 1; i++) {
            maxi = Math.max(maxi, nums[i] + 2 * k);
            mini = Math.min(nums[i + 1], nums[0] + 2 * k);
            res = Math.min(res, maxi - mini);
        }
        return res;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here