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Leetcode 3065. Minimum Operations to Exceed Threshold Value I

Problem Statement

You are given an array nums consisting of positive integers. You must perform operations to reduce the size of the array by repeatedly removing the largest element until the average of the elements in the array becomes strictly greater than a given threshold x. You need to return the minimum number of operations required to achieve this.

The average of an array is the sum of its elements divided by its size.

Clarifying Questions

  1. What are the constraints on the input?
    • Each element in nums is a positive integer.
    • The threshold x is a positive integer or float.
    • There is at least one element in the array.
  2. What should be returned if the average is initially greater than x?
    • If the average of the array is already greater than x, zero operations should be performed, so the function should return 0.
  3. Are there any constraints on the size of the array?
    • There are no explicit size constraints provided here, but common constraints (like 1 ≤ length ≤ 10^5) will be assumed unless otherwise specified.
  4. Should the elements be removed in a specific manner (e.g., remove the largest elements first)?
    • Yes, you should remove the largest elements first to more efficiently reduce the average.

Strategy

  1. Calculate the Initial Average: Compute the initial average of the array.
  2. Check Threshold: If this initial average already exceeds the threshold x, return 0.
  3. Use a Max Heap: Utilize a max heap (priority queue) to continuously extract the largest element until the average exceeds x.
  4. Track Operations: Keep a count of the number of operations performed.

Time Complexity

The primary operations involve repeatedly removing the maximum element and updating the sum, both of which can be handled efficiently with a max heap. The time complexity will be O(n log n) due to the max heap operations.

Code

Here is the C++ code to solve the problem:

#include <iostream>
#include <vector>
#include <queue>
#include <numeric>

using namespace std;

int minOperationsToExceedThreshold(vector<int>& nums, double x) {
    // Calculate initial sum and average
    double sum = accumulate(nums.begin(), nums.end(), 0.0);
    double average = sum / nums.size();
    
    // If the initial average is already greater than x, return 0
    if (average > x) {
        return 0;
    }
    
    // Create a max heap for the numbers
    priority_queue<int> maxHeap(nums.begin(), nums.end());
    int operations = 0;
    
    while (!maxHeap.empty() && average <= x) {
        // Remove the largest element
        int maxElement = maxHeap.top();
        maxHeap.pop();
        
        // Update sum and size of array
        sum -= maxElement;
        operations++;
        
        // Recalculate the average
        average = sum / (nums.size() - operations);
    }
    
    return operations;
}

int main() {
    vector<int> nums = {10, 20, 5, 3, 15};
    double x = 7.0;
    cout << "Minimum operations: " << minOperationsToExceedThreshold(nums, x) << endl;
    return 0;
}

Explanation

  1. Calculate Initial Sum and Average: The function first calculates the initial sum of the array elements and their average.
  2. Threshold Check: It checks if the initial average is already above the threshold. If yes, it returns 0 immediately.
  3. Max Heap Initialization: It uses a max heap to keep the largest elements accessible.
  4. Iterative Removal: It iteratively removes the largest element from the heap, updates the sum, and recalculates the average until the average exceeds x.
  5. Return Operations Count: Finally, it returns the number of operations performed to achieve the desired condition.

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