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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 <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;
}

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




<?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.
?>

Javascript




<script>
// JavaScript program to minimize the cost
// of array minimization
 
 
// Returns minimum possible sum in
// array B[]
function minSum(A, n)
{
    let min_val = Math.min(...A);
    return (min_val * (n-1));
}
 
// driver function
  
    let A = [3, 6, 2, 8, 7, 5];
    let n = A.length;
    document.write(minSum(A, n));
     
// This code is contributed by Mayank Tyagi
   
</script>

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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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