Минимизируйте сумму, вычисленную путем многократного удаления любых двух элементов и вставки их суммы в массив

Опубликовано: 14 Января, 2022

Учитывая N элементов, вы можете удалить любые два элемента из списка, отметить их сумму и добавить сумму в список. Повторите эти шаги, пока в списке более одного элемента. Задача состоит в том, чтобы в итоге минимизировать сумму этих выбранных сумм.
Примеры:

Input: arr[] = {1, 4, 7, 10} 
Output: 39 
Choose 1 and 4, Sum = 5, arr[] = {5, 7, 10} 
Choose 5 and 7, Sum = 17, arr[] = {12, 10} 
Choose 12 and 10, Sum = 39, arr[] = {22}
Input: arr[] = {1, 3, 7, 5, 6} 
Output: 48 
 

Рекомендуется: сначала попробуйте свой подход в {IDE}, прежде чем переходить к решению.

Approach: In order to minimize the sum, the elements that get chosen at every step must the minimum elements from the list. In order to do that efficiently, a priority queue can be used. At every step, while there is more than a single element in the list, choose the minimum and the second minimum, remove them from the list add their sum to the list after updating the running sum.
Below is the implementation of the above approach: 
 

C++

// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to return the minimized sum
int getMinSum(int arr[], int n)
{
    int i, sum = 0;
 
    // Priority queue to store the elements of the array
    // and retrieve the minimum element efficiently
    priority_queue<int, vector<int>, greater<int> > pq;
 
    // Add all the elements
    // to the priority queue
    for (i = 0; i < n; i++)
        pq.push(arr[i]);
 
    // While there are more than 1 elements
    // left in the queue
    while (pq.size() > 1)
    {
 
        // Remove and get the minimum
        // element from the queue
        int min = pq.top();
 
        pq.pop();
 
        // Remove and get the second minimum
        // element (currently minimum)
        int secondMin = pq.top();
         
        pq.pop();
 
        // Update the sum
        sum += (min + secondMin);
 
        // Add the sum of the minimum
        // elements to the queue
        pq.push(min + secondMin);
    }
 
    // Return the minimized sum
    return sum;
}
 
// Driver code
int main()
{
 
    int arr[] = { 1, 3, 7, 5, 6 };
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << (getMinSum(arr, n));
}
 
// This code is contributed by mohit

Java

// Java implementation of the approach
import java.util.PriorityQueue;
 
class GFG
{
 
    // Function to return the minimized sum
    static int getMinSum(int arr[], int n)
    {
        int i, sum = 0;
 
        // Priority queue to store the elements of the array
        // and retrieve the minimum element efficiently
        PriorityQueue<Integer> pq = new PriorityQueue<>();
 
        // Add all the elements
        // to the prioriry queue
        for (i = 0; i < n; i++)
            pq.add(arr[i]);
 
        // While there are more than 1 elements
        // left in the queue
        while (pq.size() > 1)
        {
 
            // Remove and get the minimum
            // element from the queue
            int min = pq.poll();
 
            // Remove and get the second minimum
            // element (currently minimum)
            int secondMin = pq.poll();
 
            // Update the sum
            sum += (min + secondMin);
 
            // Add the sum of the minimum
            // elements to the queue
            pq.add(min + secondMin);
        }
 
        // Return the minimized sum
        return sum;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 3, 7, 5, 6 };
        int n = arr.length;
        System.out.print(getMinSum(arr, n));
    }
}

Python3

# Python3 implementation of the approach
 
# Function to return the minimized sum
def getMinSum(arr, n):
 
    sum = 0
  
    # Priority queue to store the elements of the array
    # and retrieve the minimum element efficiently
    pq = []
  
    # Add all the elements
    # to the priority queue
    for i in range( n ):
        pq.append(arr[i])
      
    # While there are more than 1 elements
    # left in the queue
    while (len(pq) > 1) :
         
        pq.sort(reverse=True)
   
        # Remove and get the minimum
        # element from the queue
        min = pq[-1];
        
        pq.pop();
  
        # Remove and get the second minimum
        # element (currently minimum)
        secondMin = pq[-1];
          
        pq.pop();
  
        # Update the sum
        sum += (min + secondMin);
  
        # Add the sum of the minimum
        # elements to the queue
        pq.append(min + secondMin)
     
    # Return the minimized sum
    return sum
  
# Driver code
if __name__ == "__main__":
  
    arr = [ 1, 3, 7, 5, 6 ]
    n = len(arr)
    print(getMinSum(arr, n))
 
# This code is contributed by chitranayal

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
 
  // Function to return the minimized sum
  static int getMinSum(int[] arr, int n)
  {
    int i, sum = 0;
 
    // Priority queue to store the elements of the array
    // and retrieve the minimum element efficiently
    List<int> pq = new List<int>();
 
    // Add all the elements
    // to the priority queue
    for (i = 0; i < n; i++)
    {
      pq.Add(arr[i]);
    }
 
    // While there are more than 1 elements
    // left in the queue
    while(pq.Count > 1)
    {
      pq.Sort();
 
      // Remove and get the minimum
      // element from the queue
      int min = pq[0];
      pq.RemoveAt(0);
 
      // Remove and get the second minimum
      // element (currently minimum)
      int secondMin = pq[0];
      pq.RemoveAt(0);
 
      // Update the sum
      sum += (min + secondMin);
 
      // Add the sum of the minimum
      // elements to the queue
      pq.Add(min + secondMin);
    }
 
    // Return the minimized sum
    return sum;
  }
 
  // Driver code
  static public void Main ()
  {
    int[] arr = { 1, 3, 7, 5, 6 };
    int n = arr.Length;
    Console.WriteLine(getMinSum(arr, n));
  }
}
 
// This code is contributed by avanitrachhadiya2155

Javascript

<script>
 
// JavaScript implementation of the approach
 
    // Function to return the minimized sum
    function getMinSum(arr,n)
    {
        let i, sum = 0;
  
        // Priority queue to store the elements of the array
        // and retrieve the minimum element efficiently
        let pq = [];
  
        // Add all the elements
        // to the prioriry queue
        for (i = 0; i < n; i++)
            pq.push(arr[i]);
  
        // While there are more than 1 elements
        // left in the queue
        while (pq.length > 1)
        {
  
            // Remove and get the minimum
            // element from the queue
            let min = pq.shift();
  
            // Remove and get the second minimum
            // element (currently minimum)
            let secondMin = pq.shift();
  
            // Update the sum
            sum += (min + secondMin);
  
            // Add the sum of the minimum
            // elements to the queue
            pq.push(min + secondMin);
        }
  
        // Return the minimized sum
        return sum;
    }
     
    // Driver code
    let arr=[1, 3, 7, 5, 6];
    let n = arr.length;
    document.write(getMinSum(arr, n));
 
     
 
// This code is contributed by rag2127
 
</script>
Output: 
48

 

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