LeetCode #1984 — EASY

Minimum Difference Between Highest and Lowest of K Scores

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a 0-indexed integer array nums, where nums[i] represents the score of the i^th 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:

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:

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Sliding Window

Example 1

[90]
1

Example 2

[9,4,1,7]
2

Core Insight

What unlocks the optimal approach

For the difference between the highest and lowest element to be minimized, the k chosen scores need to be as close to each other as possible.
What if the array was sorted?
After sorting the scores, any contiguous k scores are as close to each other as possible.
Apply a sliding window solution to iterate over each contiguous k scores, and find the minimum of the differences of all windows.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1984: Minimum Difference Between Highest and Lowest of K Scores
class Solution {
    public int minimumDifference(int[] nums, int k) {
        Arrays.sort(nums);
        int ans = 100000;
        for (int i = 0; i < nums.length - k + 1; ++i) {
            ans = Math.min(ans, nums[i + k - 1] - nums[i]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1984: Minimum Difference Between Highest and Lowest of K Scores
func minimumDifference(nums []int, k int) int {
	sort.Ints(nums)
	ans := 100000
	for i := 0; i < len(nums)-k+1; i++ {
		ans = min(ans, nums[i+k-1]-nums[i])
	}
	return ans
}

# Accepted solution for LeetCode #1984: Minimum Difference Between Highest and Lowest of K Scores
class Solution:
    def minimumDifference(self, nums: List[int], k: int) -> int:
        nums.sort()
        return min(nums[i + k - 1] - nums[i] for i in range(len(nums) - k + 1))

// Accepted solution for LeetCode #1984: Minimum Difference Between Highest and Lowest of K Scores
impl Solution {
    pub fn minimum_difference(mut nums: Vec<i32>, k: i32) -> i32 {
        nums.sort();
        let k = k as usize;
        let mut res = i32::MAX;
        for i in 0..=nums.len() - k {
            res = res.min(nums[i + k - 1] - nums[i]);
        }
        res
    }
}

// Accepted solution for LeetCode #1984: Minimum Difference Between Highest and Lowest of K Scores
function minimumDifference(nums: number[], k: number): number {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    let ans = nums[n - 1] - nums[0];
    for (let i = 0; i + k - 1 < n; i++) {
        ans = Math.min(nums[i + k - 1] - nums[i], ans);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.