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

• Difficulty Level : Easy
• Last Updated : 19 Jan, 2023

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

Recommended Problem

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.

Steps to solve the problem:

1. initialize two variable i and sum to store the minimized sum.
2. initialize the priority queue with min heap.
3. iterate through the array and push all elements in the queue.
4. while size of queue is greater than one:
• initialize the min variable and store the top element in the queue.
• pop the top element from the queue.
• initialize the secondMin variable store the top element in the queue.
• update the sum variable with min and secondMin.
• push the sum of min and secondMin in the queue.

5. return the sum.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include` `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);` `    ``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 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;` `      ``pq.RemoveAt(0);`   `      ``// Remove and get the second minimum ` `      ``// element (currently minimum) ` `      ``int` `secondMin = pq;` `      ``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

 ``

Output

`48`

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

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