Minimum sum by choosing minimum of pairs from array

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++

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// CPP program to minimize the cost
// of array minimization
#include <bits/stdc++.h>
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[0]);
    cout << minSum(A, n);
    return 0;
}

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Java














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// 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 



















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Python














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# 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))  



















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C#














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// 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. 



















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PHP














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<?php 


// PHP program to minimize the  


// cost of array minimization 


  


// Returns minimum possible  


// sum in array B[] 


function minSum($A, $n) 


{ 


    $min_val = min($A); 


    return ($min_val * ($n - 1)); 


} 


  


    // Driver Code 


    $A = array(3, 6, 2, 8, 7, 5); 


    $n = count($A); 


    echo minSum($A, $n); 


  


// This code is contributed by vt_m. 


?> 



















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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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