Minimum Difference Between Highest and Lowest of K Scores

Minimum Difference Between Highest and Lowest of K Scores - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [9,4,1,7], k = 3

› Output: 5

💡 Note: After sorting: [1,4,7,9]. Check windows: [1,4,7] has difference 7-1=6, [4,7,9] has difference 9-4=5. Minimum is 5.

Example 2 — All Elements

$ Input: nums = [90], k = 1

› Output: 0

💡 Note: When k=1, we select only one student, so the difference between highest and lowest is 0.

Example 3 — Large Range

$ Input: nums = [41,56,58,75,88], k = 3

› Output: 17

💡 Note: Already sorted. Windows: [41,56,58] diff=17, [56,58,75] diff=19, [58,75,88] diff=30. Minimum is 17.

Constraints

1 ≤ k ≤ nums.length ≤ 1000
0 ≤ nums[i] ≤ 10⁵

Visualization

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Understanding the Visualization

Input

Array of student scores and k value

Sort

Sort scores to group similar values together

Window

Find k consecutive scores with minimum range

Key Takeaway

🎯 Key Insight: After sorting, the optimal k students are always consecutive in the array

Asked in

a Amazon 8 G Google 5

The key insight is that the optimal k students must be consecutive in a sorted array. After sorting, we use a sliding window of size k to find the minimum difference. Best approach is sliding window on sorted array. Time: O(n log n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Sort First - Check All Windows	O(n log n + n)	O(1)	Sort the array first, then check every possible window of k consecutive elements
Brute Force - Try All Combinations	O(C(n,k) × k)	O(k)	Generate all possible combinations of k students and find the one with minimum difference
Optimized Sliding Window	O(n log n)	O(1)	Sort first, then use single pass sliding window to find minimum difference

Sort First - Check All Windows — Algorithm Steps

Sort the array in ascending order
Use a window of size k and slide it through the sorted array
For each window, calculate difference between last and first element
Return the minimum difference found

Visualization

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Step-by-Step Walkthrough

Sort Array

Sort scores in ascending order

Slide Window

Check each window of k consecutive elements

Find Minimum

Return the smallest difference found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

int compare(const void* a, const void* b) {
    return (*(int*)a - *(int*)b);
}

int solution(int* nums, int numsSize, int k) {
    // Sort the array
    qsort(nums, numsSize, sizeof(int), compare);
    
    int minDiff = INT_MAX;
    
    // Try all windows of size k
    for (int i = 0; i <= numsSize - k; i++) {
        // Difference between max and min in this window
        int diff = nums[i + k - 1] - nums[i];
        if (diff < minDiff) {
            minDiff = diff;
        }
    }
    
    return minDiff;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[100];
    int numsSize = 0;
    
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, "[],\n");
    }
    
    int k;
    scanf("%d", &k);
    
    int result = solution(nums, numsSize, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n + n)

O(n log n) for sorting plus O(n) to check all windows

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Sort First - Check All Windows — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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