Reverse Digits of a Number

Given an integer N, return the number formed by reversing its digits.

If the number ends with zeroes, those zeroes disappear in the reversed value because a number does not keep leading zeroes.

Brute Force Approach

The first observation is that the last digit of a number can be taken out very easily using % 10.

For example, if the number is 1234, the last digit is 4. After taking it out, the remaining number becomes 123. Then the next last digit is 3, and this keeps going until no digits are left.

That gives the natural idea: keep picking the last digit and attach it to a new number from the right side. To attach a new digit properly, first shift the current reversed number one place left by multiplying it by 10, then add the new digit.

Algorithm

• If N is negative, remember that sign first, because the digits should be reversed without changing their order logic.

• Work with the absolute value of the number so digit extraction becomes simple and clean.

• Keep a variable revNum as 0. This variable will store the reversed number as it is built.

• While the current number is greater than 0, take its last digit using % 10.

• Shift revNum one place left by multiplying it by 10, then add the extracted digit so that digit becomes the new last digit of the reversed number.

• Remove the last digit from the current number using integer division by 10, because that digit has already been used.

• After the loop finishes, put the original sign back on the reversed number.

Key Points

• If N = 0, the reversed number is also 0.

• Trailing zeroes in the original number disappear after reversal. For example, 1200 becomes 21.

• If negative numbers are part of the problem, reverse the digits and keep the sign separately.

Dry Run

Solution

Complexity Analysis

Time Complexity: O(log₁₀ N), because one digit is processed in each iteration.

Space Complexity: O(1), because only a few variables are used.

Optimal Approach

A number like 1234 can naturally be treated as a string "1234". The key observation is that reversing a number is exactly the same as reversing its string representation:

1234 → "1234" → "4321" → 4321

So instead of manually extracting digits using % 10 and / 10, we can let string operations handle the reversal.

Algorithm

• Convert the number to a string

  • We use to_string(n) because reversing characters is simpler than repeatedly extracting digits.

• Reverse the string

  • reverse(s.begin(), s.end())

  • After this, the digits are in the required reverse order.

• Convert the reversed string back to an integer

  • Use stoi(s) because the final answer needs to be a number, not a string.

Dry Run

Solution

Complexity Analysis

Time Complexity: O(d) where d is the number of digits since each digit is processed only twice.

Space Complexity: O(d) for the string created using to_string.