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Sum of array elements possible by appending arr[i] / K to the end of the array K times for array elements divisible by K
  • Last Updated : 11 May, 2021

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:
Following operations are performed:

  1. For arr[0](= 4): arr[0](= 4) is divisible by 2, therefore append 4/2 = 2, 2 number 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 element 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 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.
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.
Output: 
44

 

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

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