#3194

Easy

auto_awesomeTwo-pointer scanning with invariant tracking

LeetCode Problem Workspace

Minimum Average of Smallest and Largest Elements

Solve the Minimum Average of Smallest and Largest Elements problem using two-pointer scanning to track averages effectively.

array two pointers sorting

Problem Statement

Given an even-length array of integers, the task is to pair up the smallest and largest elements repeatedly, then compute the average of each pair. After performing this operation n / 2 times, the goal is to return the minimum average from these pairs. The array must first be sorted to ensure proper pairing of smallest and largest elements.

You are required to repeatedly pair up the smallest and largest elements, calculate their average, and repeat the process n / 2 times. The result will be the smallest average among all pairs. The array has a length constraint of 2 to 50, ensuring an even number of elements.

Examples

Example 1

Input: nums = [7,8,3,4,15,13,4,1]

Output: 5.5

Example 2

Input: nums = [1,9,8,3,10,5]

Output: 5.5

Example 3

Input: nums = [1,2,3,7,8,9]

Output: 5.0

Constraints

2 <= n == nums.length <= 50
n is even.
1 <= nums[i] <= 50

Solution Approach

Sort the Array

Start by sorting the array to ensure the smallest and largest elements are easily paired. This step will help us efficiently access the smallest and largest values during each iteration.

Two-pointer Scanning

Use two pointers: one starting from the beginning (left) and the other from the end (right). Pair elements pointed to by these two pointers, compute their averages, and track the minimum average.

Track Minimum Average

During the two-pointer scanning, continuously update the minimum average found. At the end of the process, return the smallest of all the calculated averages.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is dominated by the sorting step, O(n log n), where n is the length of the array. The space complexity is O(1) if sorting is done in-place, otherwise O(n) for additional storage if necessary.

What Interviewers Usually Probe

Tests the candidate's understanding of sorting and its role in pairing elements.
Evaluates familiarity with two-pointer techniques and invariant tracking.
Checks if the candidate can optimize average calculations by focusing on sorting and pairing.

Common Pitfalls or Variants

Common pitfalls

Failing to sort the array before pairing the elements.
Not properly updating the minimum average during the two-pointer scanning.
Ignoring edge cases, such as arrays with the smallest possible length.

Follow-up variants

Change the problem to return the maximum average instead of the minimum.
Modify the problem to handle arrays of odd length.
Adjust the problem to calculate the sum of all averages instead of the minimum.

FAQ

How do I solve the Minimum Average of Smallest and Largest Elements problem?

Start by sorting the array, then use two pointers to pair the smallest and largest elements. Calculate the average for each pair and track the minimum average.

Why is sorting important for this problem?

Sorting ensures that the smallest and largest elements are paired optimally, making the calculation of averages more efficient and correct.

Can I solve this problem without sorting the array?

No, sorting is essential to optimally pair the smallest and largest elements for correct average calculation.

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

The time complexity is O(n log n) due to the sorting step, where n is the length of the array.

What is the role of the two-pointer technique in this problem?

The two-pointer technique allows efficient pairing of the smallest and largest elements by scanning from both ends of the sorted array.

terminal

Solution

Solution 1: Sorting

First, we sort the array $\textit{nums}$. Then, we start taking elements from both ends of the array, calculating the sum of the two elements, and taking the minimum value. Finally, we return the minimum value divided by 2 as the answer.

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class Solution:
    def minimumAverage(self, nums: List[int]) -> float:
        nums.sort()
        n = len(nums)
        return min(nums[i] + nums[-i - 1] for i in range(n // 2)) / 2

Continue Practicing

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