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Given an Integer n, find the reverse of its digits.
Examples:
Input: n = 122
Output: 221
Explanation: By reversing the digits of number, number will change into 221.Input: n = 200
Output: 2
Explanation: By reversing the digits of number, number will change into 2.Input: n = 12345
Output: 54321
Explanation: By reversing the digits of number, number will change into 54321.
This idea for this approach is to repeatedly extract the last digit of
nusing the modulus operator (n % 10) and appending it to the reverse number (revNum). After extracting the digit, the numbernis reduced by dividing it by 10 (n = n / 10). This process continues untilnbecomes 0. Finally, the reversed number (revNum) is returned.
2654
Time Complexity - O(log n)
Space Complexity - O(1)
This approach reverses a number by converting it into a string, reversing the string, and then converting it back into an integer. This avoids manual digit manipulation by leveraging string operations. The string reversal is done using a built-in function, and the result is then converted back to an integer and returned. This method is straightforward but requires extra space for the string conversion.
2654
Time Complexity - O(log n)
Space Complexity - O(1)
Note: The above program doesn't consider leading zeroes. For example, for 100 programs will print 1. If you want to print 001
The approach used is "Using Slicing". This technique involves converting the number into a string, then reversing that string by using slicing operations. After reversing, the string is converted back into a number. This method is simple and efficient but requires additional space for storing the string representation of the number.
2654
Time Complexity - O(n)
Space Complexity - O(n)