Multiples of 2 and 3
Ninja is bored with his previous game of numbers, so now he is playing with divisors.

He is given 'N' numbers, and his task is to return the sum of all numbers which is divisible by 2 or 3.

Let the number given to him be - 1, 2, 3, 5, 6. As 2, 3, and 6 is divisible by either 2 or 3 we return 2 + 3 + 6 = 11.

Input Format:

The first line contains ‘T’, denoting the number of test cases. Each test case contains a single integer ‘N’, denoting the number of elements given to Ninja. The next line contains the ‘N’ elements given to Ninja.

Output Format:

For each test case, return an integer denoting the sum of numbers divisible by 2 or 3.

Note:

You are not required to print the expected output. It has already been taken care of. Just implement the function.

Constraints:

1

Question

Multiples of 2 and 3

Ninja is bored with his previous game of numbers, so now he is playing with divisors.

He is given 'N' numbers, and his task is to return the sum of all numbers which is divisible by 2 or 3.

Let the number given to him be - 1, 2, 3, 5, 6. As 2, 3, and 6 is divisible by either 2 or 3 we return 2 + 3 + 6 = 11.

Input Format:

The first line contains ‘T’, denoting the number of test cases.

Each test case contains a single integer ‘N’, denoting the number of elements given to Ninja.

The next line contains the ‘N’ elements given to Ninja.

Output Format:

For each test case, return an integer denoting the sum of numbers divisible by 2 or 3.

Note:

You are not required to print the expected output. It has already been taken care of. Just implement the function.

Constraints:

1 <= T <= 10
1 <= N <= 10^3
0 <= input[i] <= 10^3

Where ‘T’ denotes the number of test cases and ‘N’ is the elements given to Ninja and input[i] denotes theith input.

Time Limit: 1 sec

Accepted Answer

I followed brute force approach to solve this question.

Step 1 : Iterate the array

Step 2 : Check the condition if each iteration integer is completely divisible by 2 or 3 or not. If divisible add to sum

Step 3 : Print total sum,.

Accepted Answer

Brute force

The key here is to traverse all the elements in the given array and check whether it is divisible by 2 or 3, and simultaneously add the value to the sum.

The steps are as follows:

We will initialize a variable ‘sum’ to 0, which will store the sum of numbers divisible by 2 or 3.
We will iterate from i = 0 to N - 1:
If the element at ith index is divisible by 2 or 3 then add it to ‘sum’.
We will return the sum as the final answer.

Space Complexity: O(1)Explanation:

O(1)

As we are not using any extra space. Hence, the overall space complexity is O(1).

Time Complexity: O(n)Explanation:

O(N), where N is the number of lines.

As we are checking for all the elements in the given array

Hence, the overall complexity is O(N).

C++ (g++ 5.4)

/*

    Time Complexity: O(N)

    Space Complexity: O(1)



Where N is the number of elements in the given array 'arr'



*/



int findSum(vector < int > & arr) {

  // Initialize sum which will store sum of elements

  int sum = 0;



  // For loop to iterate over all the elements

  for (int i = 0; i < arr.size(); i++) {



    // checking the given number is divisible by 2 or 3

    if ((arr[i] % 2 == 0) || (arr[i] % 3 == 0)) {

      sum = sum + arr[i];

    }

  }

  // Return sum of elements

  return sum;

}

Python (3.5)

'''

    Time Complexity: O(N)

    Space Complexity: O(1)



    Where N is the number of elements in the given array 'arr'



'''





def findSum(arr):



    # Initialize sum which will store sum of elements

    sum = 0



    # For loop to iterate over all the elements

    for i in range(len(arr)):



        # checking the given number is divisible by 2 or 3

        if (arr[i] % 2 == 0) or (arr[i] % 3 == 0):

            sum = sum + arr[i]



    # Return sum of elements

    return sum

Java (SE 1.8)

/*

	Time Complexity: O(N)

	Space Complexity: O(1)



	Where 'N' is the number of elements in the given array 'arr'.

*/





import java.util.ArrayList;



public class Solution {



	public static int findSum(ArrayList arr) {



		// Initialize sum which will store sum of elements

		int sum = 0;



		// For loop to iterate over all the elements

		for (int i = 0; i < arr.size(); i++) {



			// checking the given number is divisible by 2 or 3

			if ((arr.get(i) % 2 == 0) || (arr.get(i) % 3 == 0)) {

				sum = sum + arr.get(i);

			}

		}

		// Return sum of elements

		return sum;



	}

}