Sum of even & odd
Ninja got a test problem in which he was given an integer and he had to find the sum of even digits and odd digits separately.

Ninja is very weak in studies and didn’t know much about odd and even concepts. But he said if he got the solution he can easily understand the concepts of even and odd very clearly on his own.

Ninja asks for your help to find the solution. Can you help Ninja out?

Note:

Digits mean numbers, not the places. That is if the given integer is "13245", even digits are 2 & 4 and odd digits are 1, 3 & 5.

Input format :

The first line of input contains an integer ‘T’ denoting the number of test cases. Then each test case follows: The first and only line of each test case contains an integer ‘N’ denoting the input integer.

Output format :

The only line of output contains two single space-separated integers denoting the sum of even digits and sum of odd digits respectively. Output for each test cases will be printed on a separate line

Note:

You do not need to print anything, it has already been taken care of. Just implement the given function.

Constraints

1

Question

Sum of even & odd

Ninja got a test problem in which he was given an integer and he had to find the sum of even digits and odd digits separately.

Ninja is very weak in studies and didn’t know much about odd and even concepts. But he said if he got the solution he can easily understand the concepts of even and odd very clearly on his own.

Ninja asks for your help to find the solution. Can you help Ninja out?

Note:

Digits mean numbers, not the places. That is if the given integer is "13245",  even digits are 2 & 4 and odd digits are 1, 3 & 5.

Input format :

The first line of input contains an integer ‘T’ denoting the number of test cases. Then each test case follows:

The first and only line of each test case contains an integer ‘N’ denoting the input integer.

Output format :

The only line of output contains two single space-separated integers denoting the sum of even digits and sum of odd digits respectively.

Output for each test cases will be printed on a separate line

Note:

You do not need to print anything, it has already been taken care of. Just implement the given function.

Constraints

 1 <= T <= 5
 0 <= N <= 10 ^ 18


Time Limit: 1 sec.

Accepted Answer

Mathematical Operations

Use modulo operation to find digits of numbers.

Algorithm:

First, make an iteration till the given integer ‘num’ is greater than 0.
Store the one’s place digit in a variable.
Check whether the digit is even or odd
- If the digit is even, add the digit in the ‘evenSum’.
- If the digit is odd, add the digit in the ‘oddSum’.
Now, remove the one’s place digit from ‘num’.

Space Complexity: O(1)Explanation:

O(1)

This approach uses constant space. Therefore, the overall space complexity will be O(1).

Time Complexity: O(n)Explanation:

O(N), where ‘N’ is the number of digits in the given input.

Since we are iterating through all digits of the given input. Therefore, overall time complexity will be O(N).

C++ (g++ 5.4)

/*

    Time Complexity: O(N)

    Space Complexity: O(1)



    Where 'N' is the number of digits in the given input.

*/



vector sumOfEvenOdd(long long num) 

{

	int evenSum = 0;

    int oddSum = 0;

    vector ans;



	while(num > 0)

	{



        // Storing the one's place digit in 'last'.

		int last = num % 10;



        // Digit is even.

		if(last % 2 == 0) 

		{

			evenSum = evenSum + last;

		}



        // Digit is odd.

		else 

		{

			oddSum = oddSum + last;

		}



        // Removing the one's place digit from 'num'.

		num = num / 10;

	}



    ans.push_back(evenSum);

    ans.push_back(oddSum);



    return ans;

}

Java (SE 1.8)


/*
	Time Complexity: O(N)
	Space Complexity: O(1)
	
	Where 'N' is the number of digits in the given input.
*/

public class Solution {
	public static int[] sumOfEvenOdd(long num, int evenSum, int oddSum) {

		while (num > 0) {

			// Storing the one's place digit in 'last'.
			int last = (int) (num % 10);

			// Digit is even.
			if (last % 2 == 0) {
				evenSum = evenSum + last;
			}

			// Digit is odd.
			else {
				oddSum = oddSum + last;
			}

			// Removing the one's place digit from 'num'.
			num = num / 10;
		}
		int[] ans= {evenSum,oddSum};
		return ans;
	}

}

Python (3.5)

'''

    Time Complexity: O(N)

    Space Complexity: O(1)



    Where 'N' is the number of digits in the given input.

'''



def sumOfEvenOdd(num):



	# Declare two variables to store results.

	evenSum = 0

	oddSum = 0

	

	while(num > 0):

		# Storing the one's place digit in 'last'.

		last = num % 10



		# Digit is even.

		if(last % 2 == 0):

			evenSum = evenSum + last

			

		# Digit is odd.

		else:

			oddSum = oddSum + last

			

		# Removing the one's place digit from 'num'.

		num = num // 10



	return [evenSum, oddSum]

Accepted Answer

We can also follow this approach: first perform mergeSort

Then do run a loop on sorted array and and for every index value find the required value by target - arr[i] ,

Then for evry required value -You can use binary search to find number.

Ninja got a test problem in which he was given an integer and he had to find the sum of even digits and odd digits separately.

Ninja is very weak in studies and didn’t know much about odd and even concepts. But he said if he got the solution he can easily understand the concepts of even and odd very clearly on his own.

Ninja asks for your help to find the solution. Can you help Ninja out?

Note:

Input format :

Output format :

Note:

Constraints

Algorithm:

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