7. Reverse Integer | LeetCode Medium | C++ Solution | Step-by-Step Explanation

Problem Name: 7. Reverse Integer In this video, I solved LeetCode 7 “Reverse Integer” using C++ with an easy step-by-step explanation. 📌 Problem Statement: Given a signed 32-bit integer x, reverse its digits and return the reversed number. If reversing x causes the value to go outside the 32-bit signed integer range, return 0. 📌 Approach Used: - Create a variable rev to store reversed number - Extract last digit using x % 10 - Add digit into rev - Remove last digit using x / 10 - Check integer overflow - Return reversed number 📌 Example: Input: x = 123 Output: 321 📌 Concepts Covered: ✅ Math ✅ Modulo Operator ✅ Division Operator ✅ Integer Overflow ✅ 32-bit Integer Range ✅ Time & Space Complexity 📌 Time Complexity: O(log n) 📌 Space Complexity: O(1) 📌 Language Used: C++ 🔗 GitHub: https://github.com/amankumar1508 🔗 LinkedIn: https://www.linkedin.com/in/aman-kumar-067825379 🔗 Portfolio: https://aman-kumar-dev-portfolio.vercel.app 🔗 LeetCode Profile: https://leetcode.com/u/OVzm6rcAP2/ #leetcode #cplusplus #cpp #leetcode7 #reverseinteger #dsa #coding #programming