Solution Explanation: Minimum Difference Between Highest and Lowest of K Scores

The problem asks to find the minimum difference between the highest and lowest scores among any k students selected from a given array of scores.

Approach:

The most efficient approach involves these steps:

1. Sorting: Sort the nums array in ascending order. This allows us to easily find the minimum and maximum scores within a window of size k.

2. Sliding Window: Use a sliding window of size k to traverse the sorted array. For each window:

   • The minimum score is the first element of the window (nums[i]).
   • The maximum score is the last element of the window (nums[i + k - 1]).
   • Calculate the difference between the maximum and minimum scores (nums[i + k - 1] - nums[i]).
   • Update the minimum difference found so far.

3. Return Minimum Difference: After iterating through all possible windows, return the minimum difference found.

Time Complexity Analysis:

• Sorting the array takes O(n log n) time, where n is the length of the nums array.
• The sliding window iteration takes O(n) time.
• Therefore, the overall time complexity is dominated by the sorting step, resulting in O(n log n).

Space Complexity Analysis:

• In-place sorting algorithms can be used, meaning we don't need extra space proportional to the input size.
• The space used for the ans variable and the window is constant, O(1).
• Thus, the space complexity is O(log n) (due to the recursive calls in some sorting algorithms or O(1) if using in-place algorithms).

Code Implementation (Python):

class Solution:
    def minimumDifference(self, nums: List[int], k: int) -> int:
        nums.sort()  # Sort the array in ascending order
        min_diff = float('inf')  # Initialize min_diff to a large value
 
        for i in range(len(nums) - k + 1):
            diff = nums[i + k - 1] - nums[i]  # Calculate the difference
            min_diff = min(min_diff, diff)  # Update min_diff if necessary
 
        return min_diff

Code Implementation (Java):

import java.util.Arrays;
 
class Solution {
    public int minimumDifference(int[] nums, int k) {
        Arrays.sort(nums); // Sort the array
        int minDiff = Integer.MAX_VALUE; // Initialize minDiff to a large value
 
        for (int i = 0; i <= nums.length - k; i++) {
            int diff = nums[i + k - 1] - nums[i]; // Calculate the difference
            minDiff = Math.min(minDiff, diff); // Update minDiff
        }
 
        return minDiff;
    }
}

The other language implementations follow a very similar structure, adapting the sorting and min/max functions to the specific language. The core algorithm remains the same.