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Minimum sum by choosing minimum of pairs from array
• Difficulty Level : Easy
• Last Updated : 24 Feb, 2021

Given an array A[] of n-elements. We need to select two adjacent elements and delete the larger of them and store smaller of them to another array say B[]. We need to perform this operation till array A[] contains only single element. Finally, we have to construct the array B[] in such a way that total sum of its element is minimum. Print the total sum of array B[].
Examples:

```Input : A[] = {3, 4}
Output : 3

Input : A[] = {2, 4, 1, 3}
Output : 3```

There is an easy trick to solve this question and that is always choose the smallest element of array A[] and its adjacent, delete the adjacent element and copy smallest one to array B[]. Again for next iteration we have same smallest element and any random adjacent element which is to be deleted. After n-1 operations all of elements of A[] got deleted except the smallest one and at the same time array B[] contains “n-1” elements and all are equal to smallest element of array A[].
Thus total sum of array B[] is equal to smallest * (n-1).

## C++

 `// CPP program to minimize the cost``// of array minimization``#include ``using` `namespace` `std;` `// Returns minimum possible sum in``// array B[]``int` `minSum(``int` `A[], ``int` `n)``{``    ``int` `min_val = *min_element(A, A+n);``    ``return` `(min_val * (n-1));``}` `// driver function``int` `main()``{``    ``int` `A[] = { 3, 6, 2, 8, 7, 5};``    ``int` `n = ``sizeof``(A)/ ``sizeof` `(A);``    ``cout << minSum(A, n);``    ``return` `0;``}`

## Java

 `// Java program to minimize the``// cost of array minimization``import` `java.util.Arrays;` `public` `class` `GFG {` `// Returns minimum possible``// sum in array B[]``    ``static` `int` `minSum(``int``[] A, ``int` `n) {``        ``int` `min_val = Arrays.stream(A).min().getAsInt();``        ``return` `(min_val * (n - ``1``));``    ``}` `    ``// Driver Code``    ``static` `public` `void` `main(String[] args) {``        ``int``[] A = {``3``, ``6``, ``2``, ``8``, ``7``, ``5``};``        ``int` `n = A.length;``        ``System.out.println((minSum(A, n)));` `    ``}``}``// This code is contributed by Rajput-Ji`

## Python

 `# Python code for minimum cost of``# array minimization` `# Function defintion for minCost``def` `minSum(A):` `    ``# find the minimum element of A[]``    ``min_val ``=` `min``(A);` `    ``# return the answer``    ``return` `min_val ``*` `(``len``(A)``-``1``)` `# driver code``A ``=` `[``7``, ``2``, ``3``, ``4``, ``5``, ``6``]``print` `(minSum(A))`

## C#

 `// C# program to minimize the``// cost of array minimization``using` `System;``using` `System.Linq;` `public` `class` `GFG``{` `// Returns minimum possible``// sum in array B[]``static` `int` `minSum(``int` `[]A, ``int` `n)``{``    ``int` `min_val = A.Min();``    ``return` `(min_val * (n - 1));``}``    ` `    ``// Driver Code``    ``static` `public` `void` `Main()``    ``{``        ``int` `[]A = {3, 6, 2, 8, 7, 5};``        ``int` `n = A.Length;``        ``Console.WriteLine(minSum(A, n));``        ` `    ``}``}` `// This code is contributed by vt_m.`

## PHP

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## Javascript

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Output:

`10`

Time Complexity : O(n) in finding the smallest element of the array.
This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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