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Adding solution of strassen algorithm in C++
## Fixing Issue codeharborhub#3624 ## Description This pull request adds the implementation of Strassen's algorithm in C++ to the DSA folder. Strassen's algorithm is an efficient algorithm for matrix multiplication that is faster than the conventional matrix multiplication algorithm. It reduces the time complexity from O(n^3) to approximately O(n^2.81), making it suitable for large matrices. ## Type of PR - [ ] Bug fix - [x] Feature enhancement - [ ] Documentation update - [ ] Security enhancement - [ ] Other (specify): _______________ ## Checklist - [x] I have performed a self-review of my code. - [x] I have read and followed the Contribution Guidelines. - [x] I have tested the changes thoroughly before submitting this pull request. - [x] I have provided relevant issue numbers, screenshots, and videos after making the changes. - [x] I have commented my code, particularly in hard-to-understand areas. - [x] I have followed the code style guidelines of this project. - [x] I have checked for any existing open issues that my pull request may address. - [x] I have ensured that my changes do not break any existing functionality. - [x] Each contributor is allowed to create a maximum of 4 issues per day. This helps us manage and address issues efficiently. - [x] I have read the resources for guidance listed below. - [x] I have followed security best practices in my code changes. ## Additional Context Strassen's algorithm implementation includes: Efficient matrix multiplication using a divide-and-conquer approach. Reduction in time complexity compared to the conventional algorithm. Clear and well-commented code for easy understanding and maintenance. ## Resources for Guidance Please read the following resources before submitting your contribution: - [x] [Code Harbor Hub Community Features](https://www.codeharborhub.live/community/features) - [x] [Markdown Guide](https://www.markdownguide.org/)
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‎dsa-solutions/gfg-solutions/Easy problems/power-of-two.md

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---
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id: power-of-2
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title: Power of 2
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tags: [Easy, Bit Manipulation, GeeksforGeeks, CPP, Python, DSA]
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description: "This tutorial covers the solution to the Power of 2 problem from the GeeksforGeeks website, featuring implementations in Python and C++."
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---
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## Problem Description
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Given a non-negative integer `n`, the task is to check if it is a power of 2.
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## Examples
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**Example 1:**
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```
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Input: n = 1
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Output: true
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Explanation: 2^0 = 1
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```
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**Example 2:**
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```
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Input: n = 16
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Output: true
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Explanation: 2^4 = 16
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```
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**Example 3:**
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```
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Input: n = 3
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Output: false
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Explanation: 3 is not a power of 2
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```
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## Your Task
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You don't need to read input or print anything. Your task is to complete the function `isPowerofTwo()` which takes an integer `n` as input and returns `true` if `n` is a power of 2, otherwise `false`.
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Expected Time Complexity: $O(1)$
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Expected Auxiliary Space: $O(1)$
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## Constraints
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- `0 ≤ n ≤ 10^18`
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## Problem Explanation
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A number is a power of 2 if there exists an integer `x` such that `n = 2^x`.
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## Code Implementation
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<Tabs>
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<TabItem value="Python" label="Python" default>
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<SolutionAuthor name="@YourUsername"/>
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```python
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class Solution:
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def isPowerofTwo(self, n: int) -> bool:
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if n <= 0:
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return False
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# A power of two has exactly one bit set in binary representation
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return (n & (n - 1)) == 0
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# Example usage
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if __name__ == "__main__":
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solution = Solution()
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print(solution.isPowerofTwo(16)) # Expected output: True
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print(solution.isPowerofTwo(3)) # Expected output: False
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```
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</TabItem>
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<TabItem value="C++" label="C++">
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<SolutionAuthor name="@YourUsername"/>
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```cpp
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#include <bits/stdc++.h>
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using namespace std;
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class Solution {
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public:
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// Function to check if given number n is a power of two.
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bool isPowerofTwo(long long n) {
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if (n <= 0) return false;
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// A power of two has exactly one bit set in binary representation
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return (n & (n - 1)) == 0;
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}
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};
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int main() {
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int t;
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cin >> t; // testcases
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for (int i = 0; i < t; i++) {
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long long n; // input a number n
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cin >> n;
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Solution ob;
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if (ob.isPowerofTwo(n))
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cout << "true" << endl;
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else
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cout << "false" << endl;
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}
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return 0;
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}
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```
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</TabItem>
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</Tabs>
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## Solution Logic:
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1. A number `n` is a power of 2 if it has exactly one bit set in its binary representation.
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2. To check this, you can use the property: `n & (n - 1) == 0`.
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3. This expression is true only for powers of 2.
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## Time Complexity
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- The function operates in constant time, $O(1)$.
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## Space Complexity
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- The function uses constant space, $O(1)$.
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## References
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- **GeeksforGeeks Problem:** [Power of 2](https://www.geeksforgeeks.org/problems/power-of-2-1587115620/1)
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- **Solution Author:** [arunimad6yuq](https://www.geeksforgeeks.org/user/arunimad6yuq/)
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---
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id: power-of-2
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title: Power of 2
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tags:
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- Easy
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- Bit Manipulation
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- GeeksforGeeks
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- CPP
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- Python
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- DSA
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description: "This tutorial covers the solution to the Power of 2 problem from the GeeksforGeeks."
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---
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## Problem Description
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Given a non-negative integer `n`, the task is to check if it is a power of 2.
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## Examples
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**Example 1:**
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```
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Input: n = 1
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Output: true
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Explanation: 2^0 = 1
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```
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**Example 2:**
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```
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Input: n = 16
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Output: true
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Explanation: 2^4 = 16
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```
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**Example 3:**
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```
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Input: n = 3
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Output: false
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Explanation: 3 is not a power of 2
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```
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## Your Task
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You don't need to read input or print anything. Your task is to complete the function `isPowerofTwo()` which takes an integer `n` as input and returns `true` if `n` is a power of 2, otherwise `false`.
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Expected Time Complexity: $O(1)$
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Expected Auxiliary Space: $O(1)$
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## Constraints
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- `0 ≤ n ≤ 10^18`
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## Problem Explanation
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A number is a power of 2 if there exists an integer `x` such that `n = 2^x`.
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## Code Implementation
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<Tabs>
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<TabItem value="Python" label="Python" default>
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<SolutionAuthor name="@Ishitamukherjee2004"/>
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```python
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class Solution:
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def isPowerofTwo(self, n: int) -> bool:
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if n <= 0:
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return False
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# A power of two has exactly one bit set in binary representation
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return (n & (n - 1)) == 0
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# Example usage
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if __name__ == "__main__":
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solution = Solution()
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print(solution.isPowerofTwo(16)) # Expected output: True
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print(solution.isPowerofTwo(3)) # Expected output: False
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```
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</TabItem>
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<TabItem value="C++" label="C++">
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<SolutionAuthor name="@Ishitamukherjee2004"/>
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```cpp
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#include <bits/stdc++.h>
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using namespace std;
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class Solution {
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public:
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// Function to check if given number n is a power of two.
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bool isPowerofTwo(long long n) {
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if (n <= 0) return false;
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// A power of two has exactly one bit set in binary representation
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return (n & (n - 1)) == 0;
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}
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};
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int main() {
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int t;
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cin >> t; // testcases
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for (int i = 0; i < t; i++) {
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long long n; // input a number n
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cin >> n;
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Solution ob;
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if (ob.isPowerofTwo(n))
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cout << "true" << endl;
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else
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cout << "false" << endl;
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}
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return 0;
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}
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```
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</TabItem>
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<TabItem value="Javascript" label="Javascript">
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<SolutionAuthor name="@Ishitamukherjee2004"/>
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```javascript
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class Solution {
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isPowerofTwo(n) {
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if (n <= 0) return false;
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return (n & (n - 1)) == 0;
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}
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}
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let t = parseInt(prompt("Enter number of testcases"));
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for (let i = 0; i < t; i++) {
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let n = parseInt(prompt("Enter a number"));
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let sol = new Solution();
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console.log(sol.isPowerofTwo(n));
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}
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```
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</TabItem>
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<TabItem value="Typescript" label="Typescript">
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<SolutionAuthor name="@Ishitamukherjee2004"/>
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```typescript
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class Solution {
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isPowerofTwo(n: number): boolean {
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if (n <= 0) return false;
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return (n & (n - 1)) == 0;
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}
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}
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let t: number = parseInt(prompt("Enter number of testcases"));
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for (let i: number = 0; i < t; i++) {
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let n: number = parseInt(prompt("Enter a number"));
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let sol: Solution = new Solution();
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console.log(sol.isPowerofTwo(n));
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}
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```
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</TabItem>
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<TabItem value="Java" label="Java">
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<SolutionAuthor name="@Ishitamukherjee2004"/>
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```java
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import java.util.Scanner;
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class Solution {
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boolean isPowerofTwo(long n) {
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if (n <= 0) return false;
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return (n & (n - 1)) == 0;
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}
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}
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public class Main {
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public static void main(String[] args) {
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Scanner sc = new Scanner(System.in);
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int t = sc.nextInt();
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for (int i = 0; i < t; i++) {
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long n = sc.nextLong();
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Solution sol = new Solution();
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System.out.println(sol.isPowerofTwo(n));
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}
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}
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}
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```
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</TabItem>
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</Tabs>
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## Solution Logic:
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1. A number `n` is a power of 2 if it has exactly one bit set in its binary representation.
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2. To check this, you can use the property: `n & (n - 1) == 0`.
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3. This expression is true only for powers of 2.
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## Time Complexity
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- The function operates in constant time, $O(1)$.
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## Space Complexity
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- The function uses constant space, $O(1)$.

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