Reverse a number
Ninja is feeling very bored and wants to try something new. So, he decides to find the reverse of a given number. But he cannot do it on his own and needs your help.

Note:

If a number has trailing zeros, then its reverse will not include them. For e.g., the reverse of 10400 will be 401 instead of 00401.

Input Format:

The first line contains an integer 'T' which denotes the number of test cases or queries to be run. The first line of each test case contains one integer ‘N’ denoting the given number.

Output Format:

For each case, you need to print the reverse of the given number. The output of each test case will be printed on a separate line.

Note:

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

Constraints:

1

Question

Reverse a number

Ninja is feeling very bored and wants to try something new. So, he decides to find the reverse of a given number. But he cannot do it on his own and needs your help.

Note:

If a number has trailing zeros, then its reverse will not include them. For e.g., the reverse of 10400 will be 401 instead of 00401.

Input Format:

The first line contains an integer 'T' which denotes the number of test cases or queries to be run.

The first line of each test case contains one integer ‘N’ denoting the given number.

Output Format:

For each case, you need to print the reverse of the given number.

The output of each test case will be printed on a separate line.

Note:

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

Constraints:

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

Time Limit = 1 sec

Accepted Answer

Algorithm :
Input: num
(1) Initialize rev_num = 0
(2) Loop while num > 0
(a) Multiply rev_num by 10 and add remainder of num
divide by 10 to rev_num
rev_num = rev_num*10 + num%10;
(b) Divide num by 10
(3) Return rev_num

Accepted Answer

Number Manipulations

Firstly to solve the problem of trailing zeros, we can run a while loop that shall divide the number by 10 and the condition for the while loop shall be that the modules of the number by 10 should be 0. Once we are done with this, we will run a while loop for reversing the number. The loop will be functional till the time n is greater than 0. Inside the loop, we will be taking out the last digit of the number using the modulus operator and appending it to a new answer variable which is declared outside the loop. After this, the number will be divided by 10 as we do not need the later digit anymore.

Algorithm:

Declare a variable ‘reverseNum’ and initialize it with 0
Declare a variable ‘remainder' and initialize it with 0
Run a while loop till ‘n’ % 10 is equal to 0 to remove the trailing zeros
- Divide the number by 10 to remove the last digit.
Run a while loop till 'n' is greater than 0
- Calculate the last digit of the number by using n % 10.
- Multiply 'reverseNum' by 10 to create a new digit towards the end and add the remainder to it.
- Divide the number by 10 to remove the last digit.
Return the 'reverseNum'

Space Complexity: O(1)Explanation:

O(1)

We need constant space. So space complexity is O(1).

Time Complexity: O(logn)Explanation:

O(log N) (base 10) where N is the given input.

As we are only dealing with the digits of the given number, we can simply perform modules of the number and extract the digits.

Java (SE 1.8)


/*
	Time Complexity: O(log N to the base 10)
	Space Complexity: O(1)
	
	Where N is the given input.
*/

public class Solution {
	public static long reverseNumber(long n) {

		// Remove all trailing zeros
		while (n % 10 == 0) {
			n = n / 10;
		}

		// Declare reverseNum and remainder and initialize them with 0
		long reverseNum = 0;
		long reminder = 0;

		while (n > 0) {
			reminder = (int) (n % 10);
			reverseNum = reverseNum * 10 + reminder;
			n = n / 10;
		}

		// Return the reverse number
		return reverseNum;
	}

}

C++ (g++ 5.4)

/*
    Time Complexity: O(log N to the base 10)
    Space Complexity: O(1)

    Where N is the given input.
*/

long long int reverseNumber(long long int n)
{
    // Remove all trailing zeros
    while ( n % 10 == 0 )
    {
        n = n / 10;
    }

    // Declare reverseNum and remainder and initialize them with 0
    long long int reverseNum = 0;
    long long int reminder = 0;

    while( n > 0 )
    {
        reminder = n % 10;
        reverseNum = reverseNum * 10 + reminder;
        n = n / 10;
    }

    // Return the reverse number
    return reverseNum;
}

Python (3.5)

'''

    Time Complexity: O(log N to the base 10)

    Space Complexity: O(1)



    Where N is the given input.

'''



def reverseNumber(n):

	

	# Remove all trailing zeros.

	while ( n % 10 == 0 ):

		n = n // 10

		

	# Declare reverseNum and remainder and initialize them with 0

	reverseNum = 0

	reminder = 0

	

	while( n > 0 ):

		reminder = n % 10

		reverseNum = reverseNum * 10 + reminder

		n = n // 10

		

	# Return the reverse number

	return reverseNum

Ninja is feeling very bored and wants to try something new. So, he decides to find the reverse of a given number. But he cannot do it on his own and needs your help.

Note:

Input Format:

Output Format:

Note:

Constraints:

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