LeetCode 2413 | Smallest Even Multiple Explained 🔥 | Easy Math Trick

In this video, we solve LeetCode Problem 2413: Smallest Even Multiple in the simplest and most efficient way. This is an easy-level problem, but it helps build a strong foundation in mathematical thinking and optimization. 🔍 Problem Statement: Given a positive integer n, return the smallest positive integer that is divisible by both n and 2. 📌 Examples: Input: n = 5 Output: 10 Input: n = 6 Output: 6 💡 Intuition: We need a number that is divisible by both n and 2. Instead of checking multiples one by one, we can use a simple observation: If n is already even, it is divisible by 2, so the answer is n. If n is odd, we multiply it by 2 to make it even, so the answer becomes 2 × n. This avoids unnecessary computation and makes the solution extremely fast. ⚡ Approach: Check if n is even or odd. If even → return n. If odd → return 2 * n. ⏱️ Complexity Analysis: Time Complexity: O(1) (constant time, no loops required) Space Complexity: O(1) (no extra memory used) 💻 Code Implementation (Python): def smallestEvenMultiple(n): return n if n % 2 == 0 else n * 2 🌟 Key Takeaways: Learn to identify patterns instead of brute force Mathematical observation can simplify problems drastically Always look for O(1) solutions in easy problems 🎯 Who should watch this? Beginners in Data Structures & Algorithms Students preparing for coding interviews Anyone practicing LeetCode daily 📚 Practice more problems to strengthen your logic and speed. 👍 If you found this helpful, make sure to Like 👍, Share 🔁, and Subscribe 🔔 for more coding tutorials and LeetCode solutions! 🔥 Stay consistent and keep coding! #LeetCode #LeetCode2413 #DSA #CodingInterview #Programming #Python #Algorithms #CodeDaily