README_EN.md

908. Smallest Range I

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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.

 

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 <= 104
• 0 <= nums[i] <= 104
• 0 <= k <= 104

Solutions

Solution 1: Mathematics

According to the problem description, we can add $k$ to the maximum value in the array and subtract $k$ from the minimum value. This can reduce the difference between the maximum and minimum values in the array.

Therefore, the final answer is $\max(nums) - \min(nums) - 2 \times k$.

The time complexity is $O(n)$, where $n$ is the length of the array nums. The space complexity is $O(1)$.

class Solution:
    def smallestRangeI(self, nums: List[int], k: int) -> int:
        mx, mi = max(nums), min(nums)
        return max(0, mx - mi - k * 2)

class Solution {
    public int smallestRangeI(int[] nums, int k) {
        int mx = 0;
        int mi = 10000;
        for (int v : nums) {
            mx = Math.max(mx, v);
            mi = Math.min(mi, v);
        }
        return Math.max(0, mx - mi - k * 2);
    }
}

class Solution {
public:
    int smallestRangeI(vector<int>& nums, int k) {
        auto [mi, mx] = minmax_element(nums.begin(), nums.end());
        return max(0, *mx - *mi - k * 2);
    }
};

func smallestRangeI(nums []int, k int) int {
	mi, mx := slices.Min(nums), slices.Max(nums)
	return max(0, mx-mi-k*2)
}

function smallestRangeI(nums: number[], k: number): number {
    const max = nums.reduce((r, v) => Math.max(r, v));
    const min = nums.reduce((r, v) => Math.min(r, v));
    return Math.max(max - min - k * 2, 0);
}

impl Solution {
    pub fn smallest_range_i(nums: Vec<i32>, k: i32) -> i32 {
        let max = nums.iter().max().unwrap();
        let min = nums.iter().min().unwrap();
        (0).max(max - min - k * 2)
    }
}