Merge pull request #85 from SenthilkumarKailash/feature/231-Power-of-Two

Solution  #231 - Kailash Senthilkumar - 09/08/2025

Author: JRS296

Recursion/#231 - Power of Two - Easy/Explanation.md

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+# Power of Two
+
+**Difficulty:** Easy  
+**Category:** Bit Manipulation, Math, Recursion  
+**Leetcode Link:** [Problem Link](https://leetcode.com/problems/power-of-two/)  
+
+---
+
+## 📝 Introduction
+
+Given an integer `n`, return `true` if it is a power of two, otherwise return `false`.
+
+An integer is a power of two if there exists an integer `x` such that `n == 2^x`.
+
+**Constraints:**
+- -2^31 <= n <= 2^31 - 1
+
+---
+
+## 💡 Approach & Key Insights
+
+### Key Observation
+A power of two in binary form has exactly **one bit set to 1**.  
+Examples:
+- 1 → `0001`
+- 2 → `0010`
+- 4 → `0100`
+- 8 → `1000`
+
+For a number `n` that is a power of two:
+- `n > 0` (since negative numbers and zero cannot be powers of two)
+- `(n & (n - 1)) == 0` (removes the only set bit, leaving 0)
+
+---
+
+## 🛠️ Breakdown of Approaches
+
+### 1️⃣ Brute Force / Naive Approach
+
+- **Explanation:**  
+  Keep dividing `n` by 2 while it is even. If the result becomes 1, then it’s a power of two.
+- **Time Complexity:** O(log n) — because we keep halving the number.
+- **Space Complexity:** O(1)
+
+Example:  
+`n = 16` → 16 → 8 → 4 → 2 → 1 → return `true`
+
+---
+
+### 2️⃣ Best / Optimized Approach (Bit Manipulation)
+
+- **Explanation:**  
+  For powers of two, `n & (n - 1)` will be 0, and `n` must be positive.
+- **Time Complexity:** O(1) — single check.
+- **Space Complexity:** O(1)
+
+Example:  
+```
+n = 8 (1000)
+n-1 = 7 (0111)
+n & (n-1) = 0 → Power of two
+```
+
+---
+
+## 📊 Complexity Analysis
+
+| Approach       | Time Complexity | Space Complexity |
+| -------------- | --------------- | ---------------- |
+| Brute Force    | O(log n)        | O(1)             |
+| Bitwise Check  | O(1)            | O(1)             |
+
+---
+
+## 📌 Example Walkthroughs & Dry Runs
+
+Example:  
+```
+Input: n = 16
+Binary: 10000
+n - 1 = 01111
+n & (n - 1) = 0
+Output: true
+```
+
+Example:  
+```
+Input: n = 3
+Binary: 11
+n - 1 = 10
+n & (n - 1) = 10 (non-zero)
+Output: false
+```
+
+---
+
+## 🔗 Additional Resources
+
+- [Bit Manipulation Basics](https://www.geeksforgeeks.org/bitwise-operators-in-c-cpp/)
+- [LeetCode Discuss - Power of Two](https://leetcode.com/problems/power-of-two/discuss/)
+
+---
+
+Author: Kailash Senthilkumar  
+Date: 09/08/2025  

Recursion/#231 - Power of Two - Easy/Solution.cpp

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+class Solution {
+public:
+    bool isPowerOfTwo(int n) {
+        return n > 0 && (n & (n - 1)) == 0;
+    }
+};

Recursion/#231 - Power of Two - Easy/Solution.java

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+class Solution {
+    public boolean isPowerOfTwo(int n) {
+        // A power of two has exactly one bit set in binary representation
+        return n > 0 && (n & (n - 1)) == 0;
+    }
+}

Recursion/#231 - Power of Two - Easy/Solution.py

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+class Solution:
+    def isPowerOfTwo(self, n: int) -> bool:
+        return n > 0 and (n & (n - 1)) == 0