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Leetcode 2465. Number of Distinct Averages

Problem Statement:

You are given an array of nums consisting of integers. You need to compute the number of distinct averages that can be obtained by repeatedly picking and removing the highest and lowest elements from the array until no elements are left. The average is defined as the sum of the picked elements divided by 2.

Clarifying Questions:

  1. Q: Are the numbers in the array positive only or can they be negative as well?
    • A: The numbers can be both positive and negative.
  2. Q: Can the array contain duplicate elements?
    • A: Yes, the array can contain duplicate elements.
  3. Q: What should be returned when the array is empty?
    • A: Since no elements are present to form any averages, the count of distinct averages should be 0.

Strategy:

To solve this problem, we can use the following strategy:

  1. Sort the array: First, we need to sort the array so that the smallest and largest elements are easily accessible.
  2. Two-pointer technique: Use two pointers: one starting at the beginning of the array (smallest element) and the other at the end of the array (largest element).
  3. Calculate averages: Compute the average of the elements at the two pointers, move the pointers accordingly (start pointer moves forward, end pointer moves backward), and store these averages in a hash set to track distinct values.
  4. Count distinct values: The size of the hash set at the end will give the number of distinct averages.

Code:

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

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

    // Hash set to store distinct averages
    std::set<double> distinctAvg;

    // Two pointer technique
    int left = 0;
    int right = nums.size() - 1;

    while (left < right) {
        double avg = (nums[left] + nums[right]) / 2.0;
        distinctAvg.insert(avg);
        left++;
        right--;
    }

    // Return the number of distinct averages
    return distinctAvg.size();
}

int main() {
    std::vector<int> nums = {4, 1, 4, 0, 3, 5}; // Example input
    std::cout << "Number of distinct averages: " << distinctAverages(nums) << std::endl;
    return 0;
}

Time Complexity:

Therefore, the overall time complexity is (O(n \log n)).

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