Question

908. Smallest Range I

中文文档

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)
  }
}