What is the time complexity of Minimum Difference Between Highest and Lowest of K Scores?

The optimal solution for Minimum Difference Between Highest and Lowest of K Scores runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(1) since we are using a constant amount of extra space.

What is the brute force solution for Minimum Difference Between Highest and Lowest of K Scores?

The brute force approach for Minimum Difference Between Highest and Lowest of K Scores is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible combinations of k scores from the array and calculating the difference between the maximum and minimum scores for each combination. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Difference Between Highest and Lowest of K Scores?

Minimum Difference Between Highest and Lowest of K Scores uses the following concepts: Array, Sliding Window, Sorting. The recommended patterns to study are: Sorting, Two Pointers.

What are the interview tips for Minimum Difference Between Highest and Lowest of K Scores?

Always consider sorting as a first step for problems involving ranges or differences. Explain your thought process clearly; it shows your understanding of the problem. Practice writing clean and efficient code, as interviewers often look for both correctness and clarity.

What are common mistakes when solving Minimum Difference Between Highest and Lowest of K Scores?

Not considering edge cases, like when k equals the length of the array. Overcomplicating the problem by trying to generate all combinations instead of leveraging sorting.

#1984

Minimum Difference Between Highest and Lowest of K Scores

Easy

ArraySliding WindowSortingSortingTwo Pointers

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

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Intuition

Time O(n log n)Space O(1)

By sorting the array first, we can ensure that the k scores we choose are adjacent in value. This allows us to minimize the difference between the highest and lowest scores efficiently.

⚙️

Algorithm

4 steps

1Step 1: Sort the nums array.
2Step 2: Initialize min_diff to a large value.
3Step 3: Iterate through the sorted array, checking the difference between the i-th score and the (i+k-1)-th score for all valid i.
4Step 4: Update min_diff with the smallest difference found.

solution.py9 lines

1# Full working Python code
2
3def minDifference(nums, k):
4    nums.sort()
5    min_diff = float('inf')
6    for i in range(len(nums) - k + 1):
7        current_diff = nums[i + k - 1] - nums[i]
8        min_diff = min(min_diff, current_diff)
9    return min_diff

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(1) since we are using a constant amount of extra space.

1Sorting the array allows us to efficiently find the minimum difference by only checking adjacent elements.
2The problem can be visualized as finding the smallest range in a sorted list.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.