Sum of Digits
Ninja is given an integer ‘N’. One day Ninja decides to do the sum of all digits and replace the ‘N’ with the sum of digits until it becomes less than 10. Ninja wants to find what will be the value of ‘N’ after applying this operation.

Help Ninja in finding out this value.

Input Format:

The first line of input contains an integer ‘T’ denoting the number of test cases. The first and only line of the test case consists of a single integer ‘N’.

Output Format:

For each test case, print a single line containing a single integer denoting the final value of ‘N’. The output of each test case will be printed in 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 Digits

Ninja is given an integer ‘N’. One day Ninja decides to do the sum of all digits and replace the ‘N’ with the sum of digits until it becomes less than 10. Ninja wants to find what will be the value of ‘N’ after applying this operation.

Help Ninja in finding out this value.

Input Format:

The first line of input contains an integer ‘T’ denoting the number of test cases. 

The first and only line of the test case consists of a single integer ‘N’.

Output Format:

For each test case, print a single line containing a single integer denoting the final value of ‘N’.

The output of each test case will be printed in 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’ <= 11
1 <= ‘N’ <= 10 ^ 9

Time Limit: 1 sec.

Accepted Answer

Algorithm :
1. Declare a variable to store the sum and set it to 0
2. Repeat the next two steps till the number is not 0
3. Get the rightmost digit of the number with help of the remainder ‘%’ operator by dividing it by 10 and add it to sum.
4. Divide the number by 10 with help of ‘/’ operator to remove the rightmost digit.
5. Return the sum

Accepted Answer

Brute Force Approach

The main idea is to do the sum of digits until it becomes less than 10.

Algorithm:

Add a condition if N is less than 10 then return N.
Do the sum of all digits.
Make a recursive call and pass the sum of digits as N.
Return the answer which you get from a recursive call.

Space Complexity: O(1)Explanation:

O(1).

We are using constant space to solve this.

Time Complexity: OtherExplanation:

O(log10(N)) where ‘N’ is the given number.

We are traversing the digits of a number until it becomes less than 10. The total number of maximum digits is log10(N). Hence time complexity will be O(log10(N)).

C++ (g++ 5.4)

/*

    Time Complexity: O(log10(N))

    Space Complexity: O(1)



    where 'N' is the given number.

*/



int sumofDigits(int n)

{

	// If n less than 10 simply return n.

	if (n < 10)

	{

		return n;

	}



	// Sum of digits.

	int sum = 0;

	while (n > 0)

	{

		sum += n % 10;

		n /= 10;

	}

    

	// Returning the answer from recursion.

	return sumofDigits(sum);



}

Python (3.5)

"""

    Time Complexity: O(log10(N))

    Space Complexity: O(1)



    where 'N' is the given number.

"""



def sumofDigits(n):



    # If n less than 10 simply return n.

    if (n < 10):



        return n



    # Sum of digits.

    summ = 0

    while (n > 0):

        summ += n % 10

        n //= 10



    # Returning the answer from recursion.

    return sumofDigits(summ)

Java (SE 1.8)

/*

    Time Complexity: O(log10(N))

    Space Complexity: O(1)

    

    where 'N' is the given number.

*/



public class Solution {

    public static int sumofDigits(int n) {



        // If N less than 10 simply return n.

        if (n < 10){

            return n;

        }



        // Sum of digits.

        int sum = 0;

        while (n > 0){

            sum += n % 10;

            n /= 10;

        }



        // Returning the answer from recursion.

        return sumofDigits(sum);

    }

}

Accepted Answer

Divisibility Rule of 9 Approach

Express the given number as 9 * K +T. The divisibility rule of 9 states that if the given number is divisible by 9, then the sum of all digits is also divisible by 9. Express N as 9 * K + T. After calculating the sum of digits ‘N’ will become 9 * K + T since T is less than 9. After applying the operation only T will be left in the end. Thus answer is T. If the number is divisible by 9 then in the end 9 will be left. Thus answer will be 9 in that case.

Algorithm:

If N is less than 9 return N.
If N % 9 == 0 then return 9.
Else return N % 9.

Space Complexity: O(1)Explanation:

O(1).

We are using constant space to solve this.

Time Complexity: O(1)Explanation:

O(1).

We are using constant time to solve this.

C++ (g++ 5.4)

/*

    Time Complexity: O(1)

    Space Complexity: O(1)

*/



int sumofDigits(int n)

{

	// If n less than equal to 9 simply return n.

	if (n <= 9)

    {

		return n;

    }



	// After applying operation the final value will be 9.

	if (n % 9 == 0)

    {

		return 9;

    }



	// After applying operation the value will be n % 9.

	return n % 9;



}

Python (3.5)

"""

    Time Complexity: O(1)

    Space Complexity: O(1)

"""



def sumofDigits(n):



    # If n less than equal to 9 simply return n.

    if (n <= 9):

        return n

        

    # After applying operation the final value will be 9.

    if (n % 9 == 0):

        return 9



    # After applying operation the value will be n % 9.

    return n % 9

Java (SE 1.8)

/*

    Time Complexity: O(1)

    Space Complexity: O(1)

*/



public class Solution {

    public static int sumofDigits(int n) {



        // If N less than equal to 9 simply return n.

        if (n <= 9){

            return n;

        }



        // After applying operation the final value will be 9.

        if (n % 9 == 0){

            return 9;

        }



        // After applying operation the value will be n % 9.

        return n % 9;

    }



}

Ninja is given an integer ‘N’. One day Ninja decides to do the sum of all digits and replace the ‘N’ with the sum of digits until it becomes less than 10. Ninja wants to find what will be the value of ‘N’ after applying this operation.

Help Ninja in finding out this value.

Input Format:

Output Format:

Note:

Constraints:

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