How do you solve Minimum Average of Smallest and Largest Elements on LeetCode?

Minimum Average of Smallest and Largest Elements (LeetCode #3194) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting the array allows for efficient pairing of elements.

What is the time complexity of Minimum Average of Smallest and Largest Elements?

The optimal solution for Minimum Average of Smallest and Largest Elements runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the averages.

What is the brute force solution for Minimum Average of Smallest and Largest Elements?

The brute force approach for Minimum Average of Smallest and Largest Elements is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves repeatedly finding and removing the smallest and largest elements from the array until it's empty. This is straightforward but inefficient due to the repeated searches. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Average of Smallest and Largest Elements?

Minimum Average of Smallest and Largest Elements uses the following concepts: Array, Two Pointers, Sorting. The recommended patterns to study are: Two Pointers, Sorting.

What are the interview tips for Minimum Average of Smallest and Largest Elements?

Always consider edge cases, such as arrays with duplicate values. Explain your thought process clearly during the interview. Optimize your solution step-by-step and justify each improvement.

What are common mistakes when solving Minimum Average of Smallest and Largest Elements?

Not sorting the array before processing pairs. Confusing the indices while accessing elements from both ends.

#3194

Minimum Average of Smallest and Largest Elements

Easy

ArrayTwo PointersSortingTwo PointersSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

By sorting the array first, we can directly calculate the averages of pairs of elements from both ends of the sorted array. This eliminates the need for repeated searches for min and max.

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Algorithm

4 steps

1Step 1: Sort the array nums.
2Step 2: Initialize an empty list called averages.
3Step 3: For each index i from 0 to n/2 - 1, calculate the average of nums[i] and nums[n - i - 1] and add it to averages.
4Step 4: Return the minimum value from the averages list.

solution.py9 lines

1# Full working Python code
2
3def min_average(nums):
4    nums.sort()
5    averages = []
6    n = len(nums)
7    for i in range(n // 2):
8        averages.append((nums[i] + nums[n - i - 1]) / 2)
9    return min(averages)

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the averages.

1Sorting the array allows for efficient pairing of elements.
2The average of the smallest and largest elements converges towards the overall average.

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