#leetcode# 1984. Minimum Difference Between Highest and Lowest of K Scores

1984. Minimum Difference Between Highest and Lowest of K Scores1984. Minimum Difference Between Highest and Lowest of K Scores1984. Minimum Difference Between Highest and Lowest of K Scores

Easy

You are given a 0-indexed integer array nums, where nums[i] represents the score of the ith student. You are also given an integer k.

Pick the scores of any k students from the array so that the difference between the highest and the lowest of the k scores is minimized.

Return the minimum possible difference.

Example 1:

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Input: nums = [90], k = 1
Output: 0
Explanation: There is one way to pick score(s) of one student:
- [90]. The difference between the highest and lowest score is 90 - 90 = 0.
The minimum possible difference is 0.

Example 2:

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Input: nums = [9,4,1,7], k = 2
Output: 2
Explanation: There are six ways to pick score(s) of two students:
- [9,4,1,7]. The difference between the highest and lowest score is 9 - 4 = 5.
- [9,4,1,7]. The difference between the highest and lowest score is 9 - 1 = 8.
- [9,4,1,7]. The difference between the highest and lowest score is 9 - 7 = 2.
- [9,4,1,7]. The difference between the highest and lowest score is 4 - 1 = 3.
- [9,4,1,7]. The difference between the highest and lowest score is 7 - 4 = 3.
- [9,4,1,7]. The difference between the highest and lowest score is 7 - 1 = 6.
The minimum possible difference is 2.

Constraints:

1 <= k <= nums.length <= 1000
0 <= nums[i] <= 105

06年看看之前写的解题文章，居然一共有10w+浏览量了， 2026年了，跳槽还得刷题，不只是该作何感想，多想无益，这次用Python，AI时代，换主力语言

这个题，暴力解法就是遍历所有的组合可能，然后分别求最小的difference，太久不做算法提了，其他什么都想不出来。。。big O 也想不出来，直接看答案学习了

官方的hint已经很精辟了

先排序，然后维护一个长度为k的sliding window，遍历一遍即可找到答案

python 复制代码

class Solution:
    def minimumDifference(self, nums: List[int], k: int) -> int:
        sortedNums = sorted(nums)
        start = 0
        l = len(sortedNums)
        res = sortedNums[l - 1] - sortedNums[0]
        for start in range(0, l - k + 1):
            res = min(res, sortedNums[start + k - 1] - sortedNums[start])
        return res

我的代码如上，对Python还不是特别熟悉，不够简练，另外range，以及start end index的设置也费了点脑子，用一个例子比用脑子想容易很多

比如

python 复制代码

0, 1, 2, 3 # index

1, 2, 3, 4 # size

k = 2

start from 0 to 2, range (0, 3)
2 = 4 - 2 
when start = 2, end  = 3, end  = start + k - 1

让GPT评价了一下，优化如下

Your code is correct and idiomatic for the standard solution (sort + fixed-size window).

What's good

Correct logic: after sorting, the minimum difference among any size-k subset is the minimum of a $i+k-1$ - a $i$ .
Complexity: O(n log n) time for sorting + O(n) scan; O(n) extra space because sorted() returns a new list.
Clean window indexing; no off-by-one issues.

Small improvements

start = 0 before the loop is unused (can remove).
Use clearer names + handle edge cases explicitly (optional).

A slightly cleaner version:

python 复制代码

class Solution:
    def minimumDifference(self, nums: List[int], k: int) -> int:
        if k <= 1:
            return 0

        nums.sort()  # in-place, saves memory vs sorted(nums)
        res = float("inf")

        for i in range(len(nums) - k + 1):
            res = min(res, nums[i + k - 1] - nums[i])

        return res

Opinion: Prefer nums.sort() over sorted(nums) in LeetCode unless you need to preserve the original list.

确实更简洁