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Minimum cost to make array size 1 by removing larger of pairs

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Given an array of n integers. We need to reduce size of array to one. We are allowed to select a pair of integers and remove the larger one of these two. This decreases the array size by 1. Cost of this operation is equal to value of smaller one. Find out minimum sum of costs of operations needed to convert the array into a single element.

Examples: 

Input: 4 3 2 
Output: 4
Explanation: 
Choose (4, 2) so 4 is removed, new array 
= {2, 3}. Now choose (2, 3) so 3 is removed. 
So total cost = 2 + 2 = 4

Input: 3 4
Output: 3
Explanation: choose 3, 4, so cost is 3. 

The idea is to always pick minimum value as part of the pair and remove larger value. This minimizes cost of reducing array to size 1.

Below is the implementation of the above approach: 

CPP

// CPP program to find minimum cost to
// reduce array size to 1,
#include <bits/stdc++.h>
using namespace std;
  
// function to calculate the minimum cost
int cost(int a[], int n)
{
    // Minimum cost is n-1 multiplied with
    // minimum element.
    return (n - 1) * (*min_element(a, a + n));
}
  
// driver program to test the above function.
int main()
{
    int a[] = { 4, 3, 2 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << cost(a, n) << endl;
    return 0;
}

                    

Java

// Java program to find minimum cost
// to reduce array size to 1,
import java.lang.*;
  
public class GFG {
      
    // function to calculate the
    // minimum cost
    static int cost(int []a, int n)
    {
        int min = a[0];
          
        // find the minimum using
        // for loop
        for(int i = 1; i< a.length; i++)
        {
            if (a[i] < min)
                min = a[i];
        
          
        // Minimum cost is n-1 multiplied
        // with minimum element.
        return (n - 1) * min;
    }
      
    // driver program to test the
    // above function.
    static public void main (String[] args)
    {
          
        int []a = { 4, 3, 2 };
        int n = a.length;
          
        System.out.println(cost(a, n));
    }
}
  
// This code is contributed by parashar.

                    

Python3

# Python program to find minimum 
# cost to reduce array size to 1
  
# function to calculate the 
# minimum cost
def cost(a, n):
  
    # Minimum cost is n-1 multiplied
    # with minimum element.
    return ( (n - 1) * min(a) )
  
  
# driver code
a = [ 4, 3, 2 ]
n = len(a)
print(cost(a, n))
  
# This code is contributed by
# Smitha Dinesh Semwal

                    

C#

// C# program to find minimum cost to
// reduce array size to 1,
using System;
using System.Linq;
  
public class GFG {
      
    // function to calculate the minimum cost
    static int cost(int []a, int n)
    {
          
        // Minimum cost is n-1 multiplied with
        // minimum element.
        return (n - 1) * a.Min();
    }
      
    // driver program to test the above function.
    static public void Main (){
          
        int []a = { 4, 3, 2 };
        int n = a.Length;
          
        Console.WriteLine(cost(a, n));
    }
}
  
// This code is contributed by vt_m.

                    

PHP

<?php
// PHP program to find minimum cost to
// reduce array size to 1,
  
// function to calculate
// the minimum cost
function cost($a, $n)
{
      
    // Minimum cost is n-1
    // multiplied with
    // minimum element.
    return ($n - 1) * (min($a));
}
  
    // Driver Code
    $a = array(4, 3, 2);
    $n = count($a);
    echo cost($a, $n);
  
// This code is contributed by anuj_67.
?>

                    

Javascript

<script>
  
// JavaScript program to find minimum cost
// to reduce array size to 1,
  
    // function to calculate the
    // minimum cost
    function cost(a, n)
    {
        let min = a[0];
            
        // find the minimum using
        // for loop
        for(let i = 1; i< a.length; i++)
        {
            if (a[i] < min)
                min = a[i];
        
            
        // Minimum cost is n-1 multiplied
        // with minimum element.
        return (n - 1) * min;
    }
  
// Driver code    
           
        let a = [ 4, 3, 2 ];
        let n = a.length;
            
        document.write(cost(a, n));
              
</script>

                    

Output
4

Time Complexity: O(N), as we are using a min function which will cost O(N).
Auxiliary Space: O(1), as we are not using any extra space.

 



Last Updated : 18 Sep, 2023
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