# Mastering the Reverse Integer Algorithm for Interviews

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## Chapter 1: Understanding the Problem

In the realm of technical interviews, the Reverse Integer challenge is a common question. The task is straightforward: given a signed 32-bit integer x, your goal is to return x with its digits reversed. However, if the reversal results in a value that exceeds the signed 32-bit integer range of [-2^31, 2^31 - 1], you should return 0. Additionally, it's essential to note that the environment does not permit the use of 64-bit integers (neither signed nor unsigned).

**Example Cases:**

**Input:**x = 123**Output:**321**Input:**x = -123**Output:**-321**Input:**x = 120**Output:**21

**Constraints:**

- The integer x must satisfy: -2^31 <= x <= 2^31 - 1

## Optimized Approach

To reverse the integer effectively, we can construct the reversed number digit by digit while ensuring we don’t encounter overflow conditions.

### Section 1.1: Algorithm Overview

Reversing an integer can be approached similarly to reversing a string. The idea is to "pop" the last digit from x and "push" it to a new integer rev. Ultimately, rev will hold the reversed value of x.

To accomplish this without using an auxiliary stack or array, we can employ basic arithmetic operations.

#### Subsection 1.1.1: Implementation in C++

Here’s a C++ solution for the Reverse Integer problem:

class Solution {

public:

int reverse(int x) {

int rev = 0;

while (x != 0) {

int pop = x % 10;

x /= 10;

if (rev > INT_MAX / 10 || (rev == INT_MAX / 10 && pop > 7)) return 0;

if (rev < INT_MIN / 10 || (rev == INT_MIN / 10 && pop < -8)) return 0;

rev = rev * 10 + pop;

}

return rev;

}

};

### Section 1.2: Implementation in Java

Here's how you can implement the same logic in Java:

class Solution {

public int reverse(int x) {

int rev = 0;

while (x != 0) {

int pop = x % 10;

x /= 10;

if (rev > Integer.MAX_VALUE / 10 || (rev == Integer.MAX_VALUE / 10 && pop > 7)) return 0;

if (rev < Integer.MIN_VALUE / 10 || (rev == Integer.MIN_VALUE / 10 && pop < -8)) return 0;

rev = rev * 10 + pop;

}

return rev;

}

}

## Chapter 2: Python Solution

For those who prefer Python, here's a simple implementation:

class Solution:

def reverse(self, x: int) -> int:

if x > 0:

ans = int(str(x)[::-1])else:

ans = int(str(x * -1)[::-1]) * -1

mi = 2 ** 31 * (-1)

ma = 2 ** 31 - 1

if ans > ma or ans < mi:

return 0return ans

The first video titled "GOOGLE Coding Interview Question - Reverse Integer | LeetCode" provides a detailed explanation of the Reverse Integer problem and demonstrates how to solve it effectively.

Another insightful video titled "[Interview Question] Reverse Integer (with Binary Number Explanation)" delves deeper into the concept, offering a unique perspective on this coding challenge.

Stay tuned for more engaging interview questions and coding challenges, as I continue to share valuable insights from my experience as a senior software engineer at MANNG.