Open In App
Related Articles

Minimum possible sum of array elements after performing the given operation

Improve Article
Improve
Save Article
Save
Like Article
Like

Given an array arr[] of positive integers and an integer x, the task is to minimize the sum of elements of the array after performing the given operation at most once. In a single operation, any element from the array can be divided by x (if it is divisible by x) and at the same time, any other element from the array must be multiplied by x.
Examples: 
 

Input: arr[] = {1, 2, 3, 4, 5}, x = 2 
Output: 14 
Multiply 1 by x i.e. 1 * 2 = 2 
Divide 4 by x i.e. 4 / 2 = 2 
And the updated sum will be 2 + 2 + 3 + 2 + 5 = 14
Input: arr[] = {5, 5, 5, 5, 6}, x = 3 
Output: 26 
 

 

Approach: For an optimal solution, x must be multiplied with the smallest element from the array and only the largest element divisible by x must be divided by it. Let sumAfterOperation be the sum of the array elements calculated after performing the operation and sum be the sum of all the elements of the original array then the minimized sum will be min(sum, sumAfterOperation).
Below is the implementation of the above approach:
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
 
// Function to return the minimized sum
ll minSum(int arr[], int n, int x)
{
    ll sum = 0;
 
    // To store the largest element
    // from the array which is
    // divisible by x
    int largestDivisible = -1, minimum = arr[0];
    for (int i = 0; i < n; i++) {
 
        // Sum of array elements before
        // performing any operation
        sum += arr[i];
 
        // If current element is divisible by x
        // and it is maximum so far
        if (arr[i] % x == 0 && largestDivisible < arr[i])
            largestDivisible = arr[i];
 
        // Update the minimum element
        if (arr[i] < minimum)
            minimum = arr[i];
    }
 
    // If no element can be reduced then there's no point
    // in performing the operation as we will end up
    // increasing the sum when an element is multiplied by x
    if (largestDivisible == -1)
        return sum;
 
    // Subtract the chosen elements from the sum
    // and then add their updated values
    ll sumAfterOperation = sum - minimum - largestDivisible
                           + (x * minimum) + (largestDivisible / x);
 
    // Return the minimized sum
    return min(sum, sumAfterOperation);
}
 
// Driver code
int main()
{
    int arr[] = { 5, 5, 5, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int x = 3;
    cout << minSum(arr, n, x);
 
    return 0;
}


Java




// Java implementation of the approach
class GFG
{
     
// Function to return the minimized sum
static int minSum(int arr[], int n, int x)
{
    int sum = 0;
 
    // To store the largest element
    // from the array which is
    // divisible by x
    int largestDivisible = -1,
        minimum = arr[0];
    for (int i = 0; i < n; i++)
    {
 
        // Sum of array elements before
        // performing any operation
        sum += arr[i];
 
        // If current element is divisible
        // by x and it is maximum so far
        if (arr[i] % x == 0 &&
            largestDivisible < arr[i])
            largestDivisible = arr[i];
 
        // Update the minimum element
        if (arr[i] < minimum)
            minimum = arr[i];
    }
 
    // If no element can be reduced then
    // there's no point in performing the
    // operation as we will end up increasing
    // the sum when an element is multiplied by x
    if (largestDivisible == -1)
        return sum;
 
    // Subtract the chosen elements from the
    // sum and then add their updated values
    int sumAfterOperation = sum - minimum - largestDivisible +
                            (x * minimum) + (largestDivisible / x);
 
    // Return the minimized sum
    return Math.min(sum, sumAfterOperation);
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 5, 5, 5, 5, 6 };
    int n =arr.length;
    int x = 3;
    System.out.println(minSum(arr, n, x));
}
}
 
// This code is contributed
// by Code_Mech


Python3




# Python3 implementation of the approach
 
# Function to return the minimized sum
def minSum(arr, n, x):
 
    Sum = 0
 
    # To store the largest element
    # from the array which is
    # divisible by x
    largestDivisible, minimum = -1, arr[0]
    for i in range(0, n):
 
        # Sum of array elements before
        # performing any operation
        Sum += arr[i]
 
        # If current element is divisible by x
        # and it is maximum so far
        if(arr[i] % x == 0 and
           largestDivisible < arr[i]):
            largestDivisible = arr[i]
 
        # Update the minimum element
        if arr[i] < minimum:
            minimum = arr[i]
 
    # If no element can be reduced then there's
    # no point in performing the operation as
    # we will end up increasing the sum when an
    # element is multiplied by x
    if largestDivisible == -1:
        return Sum
 
    # Subtract the chosen elements from the
    # sum and then add their updated values
    sumAfterOperation = (Sum - minimum - largestDivisible +
                        (x * minimum) + (largestDivisible // x))
 
    # Return the minimized sum
    return min(Sum, sumAfterOperation)
 
# Driver code
if __name__ == "__main__":
 
    arr = [5, 5, 5, 5, 6]
    n = len(arr)
    x = 3
    print(minSum(arr, n, x))
 
# This code is contributed by Rituraj Jain


C#




// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the minimized sum
static int minSum(int[] arr, int n, int x)
{
    int sum = 0;
 
    // To store the largest element
    // from the array which is
    // divisible by x
    int largestDivisible = -1,
        minimum = arr[0];
    for (int i = 0; i < n; i++)
    {
 
        // Sum of array elements before
        // performing any operation
        sum += arr[i];
 
        // If current element is divisible
        // by x and it is maximum so far
        if (arr[i] % x == 0 &&
            largestDivisible < arr[i])
            largestDivisible = arr[i];
 
        // Update the minimum element
        if (arr[i] < minimum)
            minimum = arr[i];
    }
 
    // If no element can be reduced then
    // there's no point in performing the
    // operation as we will end up increasing
    // the sum when an element is multiplied by x
    if (largestDivisible == -1)
        return sum;
 
    // Subtract the chosen elements from the
    // sum and then add their updated values
    int sumAfterOperation = sum - minimum - largestDivisible +
                            (x * minimum) + (largestDivisible / x);
 
    // Return the minimized sum
    return Math.Min(sum, sumAfterOperation);
}
 
// Driver code
public static void Main()
{
    int[] arr = { 5, 5, 5, 5, 6 };
    int n = arr.Length;
    int x = 3;
    Console.WriteLine(minSum(arr, n, x));
}
}
 
// This code is contributed
// by Code_Mech


PHP




<?php
// PHP implementation of the approach
 
// Function to return the minimized sum
function minSum($arr, $n, $x)
{
    $sum = 0;
 
    // To store the largest element
    // from the array which is
    // divisible by x
    $largestDivisible = -1;
    $minimum = $arr[0];
    for ($i = 0; $i < $n; $i++)
    {
 
        // Sum of array elements before
        // performing any operation
        $sum += $arr[$i];
 
        // If current element is divisible
        // by x and it is maximum so far
        if ($arr[$i] % $x == 0 &&
            $largestDivisible < $arr[$i])
            $largestDivisible = $arr[$i];
 
        // Update the minimum element
        if ($arr[$i] < $minimum)
            $minimum = $arr[$i];
    }
 
    // If no element can be reduced then
    // there's no point in performing the
    // operation as we will end up increasing
    // the sum when an element is multiplied by x
    if ($largestDivisible == -1)
        return $sum;
 
    // Subtract the chosen elements from the
    // sum and then add their updated values
    $sumAfterOperation = $sum - $minimum - $largestDivisible +
                        ($x * $minimum) + ($largestDivisible / $x);
 
    // Return the minimized sum
    return min($sum, $sumAfterOperation);
}
 
// Driver code
$arr = array( 5, 5, 5, 5, 6 );
$n = sizeof($arr);
$x = 3;
 
print(minSum($arr, $n, $x));
 
// This code is contributed by Ryuga
?>


Javascript




<script>
// javascript implementation of the approach   
// Function to return the minimized sum
    function minSum(arr , n , x) {
        var sum = 0;
 
        // To store the largest element
        // from the array which is
        // divisible by x
        var largestDivisible = -1, minimum = arr[0];
        for (i = 0; i < n; i++) {
 
            // Sum of array elements before
            // performing any operation
            sum += arr[i];
 
            // If current element is divisible
            // by x and it is maximum so far
            if (arr[i] % x == 0 && largestDivisible < arr[i])
                largestDivisible = arr[i];
 
            // Update the minimum element
            if (arr[i] < minimum)
                minimum = arr[i];
        }
 
        // If no element can be reduced then
        // there's no point in performing the
        // operation as we will end up increasing
        // the sum when an element is multiplied by x
        if (largestDivisible == -1)
            return sum;
 
        // Subtract the chosen elements from the
        // sum and then add their updated values
        var sumAfterOperation = sum - minimum - largestDivisible
        + (x * minimum) + (largestDivisible / x);
 
        // Return the minimized sum
        return Math.min(sum, sumAfterOperation);
    }
 
    // Driver code
     
        var arr = [ 5, 5, 5, 5, 6 ];
        var n = arr.length;
        var x = 3;
        document.write(minSum(arr, n, x));
 
// This code contributed by aashish1995
</script>


Output: 

26

 

Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time 
Auxiliary Space: O(1), as we are not using any extra space.


Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Last Updated : 29 May, 2022
Like Article
Save Article
Previous
Next
Similar Reads
Complete Tutorials