🚀 SolveVerse-Problem-Solving Hub 🌍

Initiative on September 12, 2024, to solve a LeetCode problem daily in order to steadily build and refine my problem-solving skills. https://leetcode.com/problemset/

LeetCode Heatmap Building

Welcome to my repository of coding challenges and solutions in different programming languages! This repo is constantly evolving as I add more problems and solutions over time. Whether you're learning a new language, preparing for coding interviews, or just love problem-solving, this is the place for you. Feel free to explore, contribute, or give feedback!

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🧠 What You'll Find Here

This repository consists of a growing collection of algorithmic and data structure problems, each solved in various languages to demonstrate versatility and approach.

📚 Categories Covered:

• Array & String Manipulation
• Dynamic Programming
• Graph Algorithms
• Linked Lists
• Recursion & Backtracking
• Search & Sorting Algorithms
• Tree Traversals
• Greedy Algorithms
• Mathematical Problems
• Bit Manipulation

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🌐 Languages

Solutions are available in a range of programming languages to offer flexibility and variety, including:

• C++
• Python
• Java
• JavaScript

More languages to be added soon!

Bit Problems:

XOR Gate Truth Table

Input A	Input B	A ⊕ B (XOR)
0	0	0
0	1	1
1	0	1
1	1	0

AND Gate Truth Table

Input A	Input B	A ∧ B (AND)
0	0	0
0	1	0
1	0	0
1	1	1

Bit Manipulation Concepts

Bit manipulation involves performing operations directly on bits. Below is a table summarizing common bit manipulation techniques and their corresponding operations.

Operation	Description	Formula	Example Code	Result (Decimal)	Result (Binary)
Checking if a bit is set	Check if the n-th bit is set (1)	x & (1 << n)	x = 5 (0101 in binary), n = 2 (x & (1 << n))	4	0001
Setting a bit	Set the n-th bit to 1	`x	= (1 << n)`	x = 5 (0101 in binary), n = 1 `x	= (1 << n)`
Clearing a bit	Clear (set to 0) the n-th bit	x &= ~(1 << n)	x = 5 (0101 in binary), n = 2 x &= ~(1 << n)	1	0001
Toggling a bit	Flip the n-th bit	x ^= (1 << n)	x = 5 (0101 in binary), n = 1 x ^= (1 << n)	7	0111
Checking if a number is power of 2	Check if the number is a power of 2	x & (x - 1) == 0	x = 4 (0100 in binary) (x & (x - 1)) == 0	true	0100
Counting set bits	Count the number of set bits (Hamming Weight)	__builtin_popcount(x)	x = 5 (0101 in binary) __builtin_popcount(x)	2	0101
Right shift (>>)	Shift bits to the right	x >> n	x = 8 (1000 in binary), n = 2 x >> n	2	0010
Left shift (<<)	Shift bits to the left	x << n	x = 3 (0011 in binary), n = 2 x << n	12	1100
Swapping two numbers using XOR	Swap two numbers without using a temporary variable	a ^= b; b ^= a; a ^= b;	a = 5, b = 7 a ^= b; b ^= a; a ^= b;	a = 7, b = 5	0111, 0101