Sum of array elements possible by appending arr[i] / K to the end of the array K times for array elements divisible by K

Given an array arr[] consisting of N integers and an integer K, the task is to find the sum of the array elements possible by traversing the array and adding arr[i] / K, K number of times at the end of the array, if arr[i] is divisible by K. Otherwise, stop the traversal.

Examples:

Input: arr[] = {4, 6, 8, 2}, K = 2
Output: 44
Explanation:
The following operations are performed:

1. For arr[0](= 4): arr[0](= 4) is divisible by 2, therefore append 4/2 = 2, 2 numbers of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2}.
2. For arr[1](= 6): arr[1](= 6) is divisible by 2, therefore append 6/2 = 3, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2, 3, 3}.
3. For arr[2](= 8): arr[2](= 8) is divisible by 2, therefore append 8/2 = 4, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2, 3, 3, 4, 4}.
4. For arr[3](= 2): arr[3](= 2) is divisible by 2, therefore append 2/2 = 1, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1}.
5. For arr[4](= 2): arr[4](= 2) is divisible by 2, therefore append 2/2 = 1, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1}.
6. For arr[5](= 2): arr[5](= 2) is divisible by 2, therefore append 2/2 = 1, 2 number of times at the end of the array. Now, the array modifies to {4, 6, 8, 2, 2, 2, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1}.

After completing the above steps, the sum of the array elements is = 4 + 6 + 8 + 2 + 2 + 2 + 3 + 3 + 4 + 4 + 1 + 1 + 1 + 1 + 1 + 1 = 44.

Input: arr[] = {4, 6, 8, 9}, K = 2
Output: 45

Naive Approach: The simplest approach is to solve the given problem is to traverse the given array and add the value (arr[i]/K) K a number of times at the end of the array. After completing the above steps, print the sum of the array elements.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
int sum(int arr[], int N, int K)
{
    // Stores the sum of the array
    int sum = 0;

    vector<long long> v;

    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
        v.push_back(arr[i]);
    }

    // Traverse the vector
    for (int i = 0;
         i < v.size(); i++) {

        // If v[i] is divisible by K
        if (v[i] % K == 0) {

            long long x = v[i] / K;

            // Iterate over the range
            // [0, K]
            for (int j = 0; j < K; j++) {
                // Update v
                v.push_back(x);
            }
        }
        // Otherwise
        else
            break;
    }

    // Traverse the vector v
    for (int i = 0; i < v.size(); i++)
        sum = sum + v[i];

    // Return the sum of the updated array
    return sum;
}

// Driver Code
int main()
{
    int arr[] = { 4, 6, 8, 2 };
    int K = 2;
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << sum(arr, N, K);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG{

// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
static int sum(int arr[], int N, int K)
{

    // Stores the sum of the array
    int sum = 0;

    ArrayList<Integer> v = new ArrayList<>();

    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
        v.add(arr[i]);
    }

    // Traverse the vector
    for(int i = 0; i < v.size(); i++)
    {

        // If v[i] is divisible by K
        if (v.get(i) % K == 0)
        {
            int x = v.get(i) / K;

            // Iterate over the range
            // [0, K]
            for(int j = 0; j < K; j++)
            {

                // Update v
                v.add(x);
            }
        }

        // Otherwise
        else
            break;
    }

    // Traverse the vector v
    for(int i = 0; i < v.size(); i++)
        sum = sum + v.get(i);

    // Return the sum of the updated array
    return sum;
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 4, 6, 8, 2 };
    int K = 2;
    int N = arr.length;

    System.out.println(sum(arr, N, K));
}
}

// This code is contributed by Kingash

Python3

# Python3 program for the above approach

# Function to calculate sum of array
# elements after adding arr[i] / K
# to the end of the array if arr[i]
# is divisible by K
def summ(arr, N, K):

    # Stores the sum of the array
    sum = 4

    v = [i for i in arr]

    # Traverse the vector
    for i in range(len(v)):

        # If v[i] is divisible by K
        if (v[i] % K == 0):

            x = v[i] // K

            # Iterate over the range
            # [0, K]
            for j in range(K):

                # Update v
                v.append(x)

        # Otherwise
        else:
            break

    # Traverse the vector v
    for i in range(len(v)):
        sum = sum + v[i]

    # Return the sum of the updated array
    return sum

# Driver Code
if __name__ == '__main__':

    arr = [ 4, 6, 8, 2 ]
    K = 2
    N = len(arr)

    print(summ(arr, N, K))

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {

    // Function to calculate sum of array
    // elements after adding arr[i] / K
    // to the end of the array if arr[i]
    // is divisible by K
    static int sum(int[] arr, int N, int K)
    {

        // Stores the sum of the array
        int sum = 0;

        List<int> v = new List<int>();

        // Traverse the array arr[]
        for (int i = 0; i < N; i++) {
            v.Add(arr[i]);
        }

        // Traverse the vector
        for (int i = 0; i < v.Count; i++) {

            // If v[i] is divisible by K
            if (v[i] % K == 0) {
                int x = v[i] / K;

                // Iterate over the range
                // [0, K]
                for (int j = 0; j < K; j++) {

                    // Update v
                    v.Add(x);
                }
            }

            // Otherwise
            else
                break;
        }

        // Traverse the vector v
        for (int i = 0; i < v.Count; i++)
            sum = sum + v[i];

        // Return the sum of the updated array
        return sum;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int[] arr = { 4, 6, 8, 2 };
        int K = 2;
        int N = arr.Length;

        Console.WriteLine(sum(arr, N, K));
    }
}

// This code is contributed by ukasp.

Javascript

<script>

// Javascript program for the above approach

// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
function sum(arr, N, K)
{
    // Stores the sum of the array
    var sum = 0;

    var v = [];

    // Traverse the array arr[]
    for (var i = 0; i < N; i++) {
        v.push(arr[i]);
    }

    // Traverse the vector
    for (var i = 0;
         i < v.length; i++) {

        // If v[i] is divisible by K
        if (v[i] % K == 0) {

            var x = v[i] / K;

            // Iterate over the range
            // [0, K]
            for (var j = 0; j < K; j++) {
                // Update v
                v.push(x);
            }
        }
        // Otherwise
        else
            break;
    }

    // Traverse the vector v
    for (var i = 0; i < v.length; i++)
        sum = sum + v[i];

    // Return the sum of the updated array
    return sum;
}

// Driver Code
var arr = [4, 6, 8, 2];
var K = 2;
var N = arr.length;
document.write( sum(arr, N, K));

</script>

Output:

44

Time Complexity: O(N * K * log N)
Auxiliary Space: O(M), M is the maximum element of the array.

Efficient Approach: The above approach can also be optimized based on the following observations:

• If arr[i] is divisible by K, then adding arr[i] / K, K times increases the sum by arr[i].
• Therefore, the idea is to only push the arr[i] / K only once at the end of the vector.

Follow the steps below to solve the problem:

• Initialize a variable, say sum as 0 that stores the sum of all the array elements array A[].
• Initialize an array, say A[] and store all the array elements arr[] in A[].
• Initialize a variable, say flag as 0 that stores whether the element is to be added at the end of the array or not.
• Traverse the array A[] and perform the following steps:
  • If the value flag is 0 and A[i] is divisible by K, then push A[i] at the end of V.
  • Otherwise, update the value of flag as 1.
  • Increment the value of the sum by V[i % N].
• After completing the above steps, print the value of the sum as the resultant sum.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
int sum(int arr[], int N, int K)
{
    // Stores the sum of the array
    int sum = 0;

    // Stores the array elements
    vector<int> v;

    // Traverse the array
    for (int i = 0; i < N; i++) {
        v.push_back(arr[i]);
    }

    // Stores if the operation
    // should be formed or not
    bool flag = 0;

    // Traverse the vector V
    for (int i = 0; i < v.size(); i++) {

        // If flag is false and if
        // v[i] is divisible by K
        if (!flag && v[i] % K == 0)
            v.push_back(v[i] / K);

        // Otherwise, set flag as true
        else {
            flag = 1;
        }

        // Increment the sum by v[i % N]
        sum = sum + v[i % N];
    }

    // Return the resultant sum
    return sum;
}

// Driver Code
int main()
{
    int arr[] = { 4, 6, 8, 2 };
    int K = 2;
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << sum(arr, N, K);

    return 0;
}

Java

// Java program for the above approach

import java.io.*;
import java.util.*;

class GFG {

      // Function to calculate sum of array
    // elements after adding arr[i] / K
    // to the end of the array if arr[i]
    // is divisible by K
    static int sum(int arr[], int N, int K)
    {
        // Stores the sum of the array
        int sum = 0;

        // Stores the array elements
        ArrayList<Integer> v = new ArrayList<Integer>();

        // Traverse the array
        for (int i = 0; i < N; i++) {
            v.add(arr[i]);
        }

        // Stores if the operation
        // should be formed or not
        boolean flag = false;

        // Traverse the vector V
        for (int i = 0; i < v.size(); i++) {

            // If flag is false and if
            // v[i] is divisible by K
            if (!flag && v.get(i) % K == 0)
                v.add(v.get(i) / K);

            // Otherwise, set flag as true
            else {
                flag = true;
            }

            // Increment the sum by v[i % N]
            sum = sum + v.get(i % N);
        }

        // Return the resultant sum
        return sum;
    }

    // Driver Code
    public static void main (String[] args) {
        int arr[] = { 4, 6, 8, 2 };
        int K = 2;
        int N = arr.length;
        System.out.println(sum(arr, N, K));
    }
}

// This code is contributed by Dharanendra L V.

Python3

# Python  program for the above approach

# Function to calculate sum of array
# elements after adding arr[i] / K
# to the end of the array if arr[i]
# is divisible by K
def Sum(arr, N, K):

    # Stores the sum of the array
    sum = 0

    # Stores the array elements
    v = []

    # Traverse the array
    for i in range(N):
        v.append(arr[i])

    # Stores if the operation
    # should be formed or not
    flag = False

    i = 0
    lenn = len(v)

    # Traverse the vector V
    while(i < lenn):

        # If flag is false and if
        # v[i] is divisible by K
        if( flag == False and (v[i] % K == 0)):
            v.append(v[i]//K)

        # Otherwise, set flag as true
        else:
            flag = True

        # Increment the sum by v[i % N]
        sum += v[i % N]
        i += 1
        lenn = len(v)

    # Return the resultant sum
    return sum

# Driver Code
arr = [ 4, 6, 8, 2]
K = 2
N = len(arr)
print(Sum(arr, N, K))

# This code is contributed by avanitrachhadiya2155

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
static int sum(int []arr, int N, int K)
{

    // Stores the sum of the array
    int sum = 0;

    // Stores the array elements
    List<int> v = new List<int>();

    // Traverse the array
    for(int i = 0; i < N; i++)
    {
        v.Add(arr[i]);
    }

    // Stores if the operation
    // should be formed or not
    bool flag = false;

    // Traverse the vector V
    for(int i = 0; i < v.Count; i++)
    {

        // If flag is false and if
        // v[i] is divisible by K
        if (!flag && v[i] % K == 0)
            v.Add(v[i] / K);

        // Otherwise, set flag as true
        else
        {
            flag = true;
        }

        // Increment the sum by v[i % N]
        sum = sum + v[i % N];
    }

    // Return the resultant sum
    return sum;
}

// Driver Code
static void Main()
{
    int[] arr = { 4, 6, 8, 2 };
    int K = 2;
    int N = arr.Length;

    Console.WriteLine(sum(arr, N, K));
}
}

// This code is contributed by SoumikMondal

Javascript

<script>

// javascript program for the above approach

// Function to calculate sum of array
// elements after adding arr[i] / K
// to the end of the array if arr[i]
// is divisible by K
function sum(arr, N, K)
{
    // Stores the sum of the array
    var sum = 0;
    var i;
    // Stores the array elements
    var v = [];

    // Traverse the array
    for (i = 0; i < N; i++) {
        v.push(arr[i]);
    }

    // Stores if the operation
    // should be formed or not
    var flag = 0;

    // Traverse the vector V
    for (i = 0; i < v.length; i++) {

        // If flag is false and if
        // v[i] is divisible by K
        if (!flag && v[i] % K == 0)
            v.push(v[i] / K);

        // Otherwise, set flag as true
        else {
            flag = 1;
        }

        // Increment the sum by v[i % N]
        sum = sum + v[i % N];
    }

    // Return the resultant sum
    return sum;
}

// Driver Code
    var arr = [4, 6, 8, 2];
    var K = 2;
    var N = arr.length;
    document.write(sum(arr, N, K));

</script>

Output:

44

Time Complexity: O(N * log N) 
Auxiliary Space: O(N * log N)