🔥 In this video, we solve LeetCode Problem 2413 – Smallest Even Multiple in a simple and beginner-friendly way.
📌 Problem Statement:
Given a positive integer "n", return the smallest positive integer that is a multiple of both "n" and "2".
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💡 Approach Covered:
✅ Method 1: Using Conditional Statement
- If "n" is even → answer is "n"
- If "n" is odd → answer is "2 × n"
✅ Method 2: Using Simple For Loop
- Iterate numbers until you find a number divisible by both "n" and "2"
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💻 Code Implementation (C++):
// Method 1: Conditional
int smallestEvenMultiple(int n) {
if (n % 2 == 0) return n;
return n * 2;
}
// Method 2: Loop
int smallestEvenMultiple(int n) {
for (int i = 1; ; i++) {
if (i % n == 0 && i % 2 == 0) {
return i;
}
}
}
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⏱️ Time Complexity:
- Conditional: O(1)
- Loop: O(n) (in worst case)
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🎯 Why this problem is important?
- Helps understand LCM (Least Common Multiple) concept
- Great for beginners starting with problem-solving on LeetCode
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LeetCode 2413 Smallest Even Multiple | C++ Solution | Loop vs Conditional 💡 Beginner Friendly Tips | NatokHD
