Reverse Integer Leetcode Problem 7 [Python Solution]

Reverse Integer LeetCode problem, while it’s categorized as a medium difficulty problem, it has some tricky aspects due to the constraints imposed on the solution.

We’ll not only provide a Python solution but also thoroughly explain the thought process and the reasoning behind our approach.

This guide is designed for beginners, so we’ll cover every little detail to ensure a clear understanding of the problem and its solution.

Problem Overview

The Reverse Integer problem is quite straightforward on the surface.

Given a signed 32-bit integer x, our task is to reverse its digits.

However, there’s a catch: if reversing x causes the value to go outside the signed 32-bit integer range (-2^31 to 2^31 - 1), we must return 0. The problem description emphasizes the importance of not exceeding the 32-bit integer constraints, which makes it more challenging than it may seem at first glance.

Example:

Let’s illustrate the problem with a few examples:

Example 1:

Input: x = 123
Output: 321

Example 2:

Input: x = -123
Output: -321

Example 3:

Input: x = 120
Output: 21

Understanding the Constraints

To understand the constraints better, let’s break them down:

• The input integer x is a signed 32-bit integer, which means it can represent values from -2^31 to 2^31 - 1.

This 32-bit constraint means that we must take extra precautions when reversing the integer to avoid integer overflow.

We’ll dive deeper into how to handle this aspect in our Python solution.

Reverse Integer LeetCode Problem Solution

Now, let’s discuss the Python solution for this problem.

The solution involves a careful reversal of digits while considering the constraints and possible integer overflow scenarios.

1. Bruteforce Approach

One approach to solve this problem would be to convert the integer to a string, reverse the string, and then convert it back to an integer.

While this approach is straightforward, it’s not efficient and doesn’t take the 32-bit integer constraint into account.

Here’s a simple Python code snippet for this approach:

def reverse(self, x: int) -> int:
    if x >= 0:
        reversed_str = str(x)[::-1]
    else:
        reversed_str = '-' + str(x)[1:][::-1]

    reversed_x = int(reversed_str)

    if reversed_x < -2**31 or reversed_x > 2**31 - 1:
        return 0

    return reversed_x

This approach converts the integer to a string, reverses it, and then converts it back to an integer.

However, it doesn’t efficiently handle the integer overflow cases, and it’s not the recommended way to solve the problem.

2. Efficient Approach with Constraints Handling

For an efficient solution that takes into account the 32-bit integer constraints, we can reverse the integer while checking for overflow at each step.

Here’s the optimized Python code for this approach:

class Solution:
    def reverse(self, x: int) -> int:
        MAX = 2**31 - 1 # Maximum 32-bit integer

        res = 0
        sign = 1
        if x < 0:
            sign = -1
            x = abs(x)

        while x:
            digit, x = x % 10, x // 10
            if res > MAX // 10 or (res == MAX // 10 and digit > MAX % 10):
                return 0
            res = (res * 10) + digit

        return res * sign

In this efficient solution, we iteratively reverse the digits while continuously checking for overflow.

Here’s a step-by-step breakdown:

1. We define the maximum (MAX) 32-bit integer value.
2. Initialize a variable res to store the reversed integer.
3. Enter a while loop to reverse the digits of the input integer x.
4. Inside the loop, we extract the ones place digit using x % 10.
5. We perform integer division on x to remove the ones place digit.
6. We then check if adding the current digit to res would cause an overflow.
7. This is done by comparing res with MAX // 10 and the ones place digit with MAX % 10. If it would overflow, we return 0.
8. If the overflow condition isn’t met, we add the current digit to res.
9. Finally, once the loop is done, we return the reversed integer stored in res.

This efficient approach handles the constraints effectively and provides a correct solution to the Reverse Integer problem.

Time and Space Complexity

Now, let’s analyze the time and space complexity of our efficient solution.

Time Complexity: The time complexity of our solution is O(log(x)) because we iterate through the digits of the input integer x.

The number of iterations is determined by the number of digits in x, which is roughly log(x).

Space Complexity: The space complexity is O(1) since we use a constant amount of extra space for variables like res, MIN, and MAX.

We do not use any additional data structures that grow with the input size.

Reasoning Behind Our Approach

Our approach is based on careful digit reversal while considering the 32-bit integer constraints.

Here’s the reasoning behind our steps:

1. We start by defining the minimum and maximum 32-bit integer values to check for overflows and underflows.
2. We use a while loop to reverse the digits one by one.
3. At each step, we check if adding the current digit would cause an overflow or underflow, and we return 0 if it does.
4. If neither condition is met, we add the current digit to the reversed result.
5. Finally, we return the reversed integer.

Our approach is efficient and guarantees that the reversed integer does not violate the 32-bit integer constraints.

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Conclusion

In this blog post, we’ve tackled the Reverse Integer problem on LeetCode.

We provided an efficient Python solution that reverses an integer while taking into account the constraints of 32-bit integers.

We explained the reasoning behind our approach and discussed the time and space complexity of the solution.

We hope this guide has been helpful to you, especially if you’re a beginner looking to understand this problem and its solution.

If you have any questions, suggestions, or comments, please feel free to share them.

We encourage you to engage and continue learning.

Happy coding!

Question Link: Reverse Integer – LeetCode

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