Skip to content
Related Articles

Related Articles

Minimize the sum calculated by repeatedly removing any two elements and inserting their sum to the Array

View Discussion
Improve Article
Save Article
  • Difficulty Level : Easy
  • Last Updated : 05 May, 2022
View Discussion
Improve Article
Save Article

Given N elements, you can remove any two elements from the list, note their sum, and add the sum to the list. Repeat these steps while there is more than a single element in the list. The task is to minimize the sum of these chosen sums in the end.
Examples: 
 

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 
 

 

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 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.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
 
# importing heapq python module
# for implementing min heap
import heapq
 
# Function to return the minimized sum
 
 
def getMinSum(arr, n):
    summ = 0
    # Heap to store the elements of the array
    # and retrieve the minimum element efficiently
    pq = arr
    # creating min heap from array pq
    heapq.heapify(pq)
    # While there are more than 1 elements
    # left in the queue
    while (len(pq) > 1):
        # storing minimum element (root of min heap)
        # into minn
        minn = pq[0]
        # replacing root with last element and
        # deleting last element from min heap
        # as per deleting procedure for HEAP
        pq[0] = pq[-1]
        pq.pop()
        # maintaining the min heap property
        heapq.heapify(pq)
        # again storing minimum element (root of min heap)
        # into secondMin
        secondMin = pq[0]
        # again replacing root with last element and
        # deleting last element as per heap procedure
        pq[0] = pq[-1]
        pq.pop()
        # Update the sum
        summ += (minn+secondMin)
        # appending the summ as last element of min heap
        pq.append(minn+secondMin)
        # again maintaining the min heap property
        heapq.heapify(pq)
    return summ
 
 
# Driver Code
if __name__ == "__main__":
    arr = [1, 3, 7, 5, 6]
    n = len(arr)
    print(getMinSum(arr, n))
'''Code is written by RAJAT KUMAR [GLAU]'''


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 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.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

 

Time Complexity : O(N * log(N))
Auxiliary Space: O(N) 


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!