README_EN.md

910. Smallest Range II

中文文档

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

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

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