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Leetcode 3194. Minimum Average of Smallest and Largest Elements

Problem Statement

Given an array, you need to find the minimum possible average of the smallest and largest elements in any subsequence of the array of length at least two.

Clarifying Questions

  1. Q: What is the range of the input array’s size?
    • A: The size of the array can vary, but it’s generally large enough to warrant an efficient solution.
  2. Q: Are elements in the array guaranteed to be integers?
    • A: Yes, all elements in the array are integers.
  3. Q: Can the array contain duplicate values?
    • A: Yes, the array can contain duplicate values.
  4. Q: Are there any constraints on the values of the integers in the array (e.g., positive, negative)?
    • A: The integers can be both positive and negative, without any specific constraints on their range.

Strategy

The goal is to minimize the average of the smallest and largest elements of any subsequence of length at least two. To achieve this efficiently, you can follow these steps:

  1. Sort the array: Sorting simplifies identifying the smallest and largest elements in any subsequence.
  2. Select smallest and largest elements: After sorting, the smallest element will be the first element and the largest will be the last element in the sorted array.
  3. Calculate the average: Compute the average of these two values.

Code

Here’s the C++ code to solve the problem:

#include <vector>
#include <algorithm>
#include <iostream>

double minimumSubsequenceAverage(std::vector<int>& nums) {
    // First, sort the array
    std::sort(nums.begin(), nums.end());

    // The smallest element will be the first element in the sorted array
    int smallest = nums.front();
    // The largest element will be the last element in the sorted array
    int largest = nums.back();

    // Calculate and return the average of smallest and largest
    return (smallest + largest) / 2.0;
}

int main() {
    // Example usage:
    std::vector<int> nums = {3, 1, 4, 1, 5, 9};
    double result = minimumSubsequenceAverage(nums);
    std::cout << "Minimum average of smallest and largest elements: " << result << std::endl;
    
    return 0;
}

Time Complexity

Overall, the time complexity is dominated by the sorting step, so the overall time complexity is (O(n \log n)). This is efficient and suitable for even relatively large arrays.

By following this strategy, your code will efficiently find the minimum possible average of the smallest and largest elements in any subsequence of the given array.

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