Tarun Khandagare’s Post

"Let's dive into a fascinating problem in string manipulation! 🧵 In problem 1021, 'Remove Outermost Parentheses,' we're given a valid parentheses string and asked to remove the outermost parentheses of every primitive string in its decomposition. But what is a primitive string in this context? A valid parentheses string is considered primitive if it's nonempty and cannot be split into two nonempty valid parentheses strings. For instance: - For the string "(())", the primitive decomposition is "(())". - For "(()())", the decomposition is "()" + "()", where both parts are primitive strings. To solve this problem, we need to traverse the string and identify the primitive strings. Then, we remove the outermost parentheses of each primitive string and concatenate the results. Here's an example: - For the input "(()())(())", the primitive decomposition is "(()())" + "(())". - Removing the outermost parentheses gives us "()" + "()", resulting in "()()". Are you ready to embark on this exciting journey of string manipulation? Let's delve into the code and uncover the beauty of solving this problem! #StringManipulation #CodingChallenge #Programming #Algorithms"

Leetcode Problem Of The Day: Smallest String Starting From Leaf Find the smallest string formed by traversing from any leaf node to the root in a binary tree. Approach: 1. smallestFromLeaf Method: This method takes a TreeNode named root as input and returns the lexicographically smallest string. It initializes a StringBuilder named smallest to store the smallest string found. It calls the dfs method to perform depth-first search traversal of the tree, passing a new StringBuilder for the path and the smallest StringBuilder. Finally, it returns the string representation of the smallest StringBuilder. 2. dfs Method: This method performs a depth-first search traversal of the binary tree. It takes three parameters: the current TreeNode (root), a StringBuilder for the current path (path), and the StringBuilder for the smallest string found (smallest). If the root is null, it returns. It appends the character representation of the node's value to the path. If the current node is a leaf node (i.e., it has no left or right child), it converts the path StringBuilder to a string, compares it with the smallest string found so far, and updates smallest if the current path is lexicographically smaller. It then proceeds to traverse the left and right subtrees recursively. After visiting the children, it removes the last character from the path StringBuilder to backtrack. Time Complexity : O(nlogn) Space Complexity : O(n) Problem Link : https://lnkd.in/ghKgrkpM #Coding #CompetitiveProgramming #dsa #DatastructuresAndAlgorithms #Leetcode #Consistency #Programming #CodingChallenge

#BinaryTree algorithms #LevelOrder traversal #BreadthFirstSearch... A tree is a connected, acyclic graph that consists of n nodes and n - 1 edges. In a rooted tree, all nodes are beneath the root of the tree. In a rooted tree, the descendants of a node are below the node, and the ancestors of the node are above the node. Direct descendants are children. In a rooted tree, the root has no direct ancestor and all other nodes has one, parent, direct ancestor. Generally, graph traversal algorithms can be used to traverse the nodes of a tree. There are simpler tree traversal algorithms for rooted trees.  Breadth-first search (BFS) starts with the insertion of the root of the rooted binary tree into a queue. BFS processes the queue as follows: take out the front most vertex u from the queue, enqueue the child vertices of the removed vertex u, and loop the previous steps until all vertices have been processes. Let's look at the level-order traversal. #LeetCode102 Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level). Solution: This is a direct implementation of breadth-first search. See photo for code in #C++. #SoftwareEngineering #DataStructures #Algorithms Resources [1] Halim, S. and Halim, F. (2013). Competitive programming 3: The new lower bound of programming contests. Springer

Dynamic programming is a method for solving complex problems by breaking them into smaller subproblems. Instead of repeatedly solving the same subproblems, it saves their results in a table. The process of storing the result in a table is known as memoization. We can apply dynamic programming to a problem if it satisfies two properties: 1. It can be represented as a recurrence relation, indicating optimal substructure. 2. It exhibits overlapping subproblems, where the same subproblem recurs multiple times. For example, if two subproblems with the same values occur, like (1,2) and (1,2), dynamic programming can store and reuse the results. #DynamicProgramming #Memoization #Algorithm #ProgrammingTips #ComputerScience #Recursion

✨ DSA Challenge Update: Solved problem 166/251 - "Interleaving Two Strings" ✨ 🚀 Today, I tackled the "Interleaving Two Strings" problem, which tests whether a third string 'C' can be formed by interleaving characters from two other strings 'A' and 'B' while preserving the order of characters. 📌 Problem 166/251 Overview: Given three strings A, B, and C, the task was to determine if C is an interleaving of A and B. This means C should contain all characters from A and B in the same order as they appear in their respective strings. 💡 Approach: - I implemented a dynamic programming solution to efficiently check if C can be formed by interleaving A and B. By using a 2D array `dp`, where `dp[i][j]` represents whether the first `i` characters of A and the first `j` characters of B can interleave to form the first `i+j` characters of C. - The solution involved handling base cases and transitions for insertion of characters from A or B into C, ensuring that the conditions for interleaving were met. 🔍 Key Insights: - Understanding the problem's requirements and breaking it down into manageable subproblems was crucial. Dynamic programming provided an optimal solution with a time complexity of O(N * M), where N and M are the lengths of A and B respectively. - This problem underscores the importance of sequence preservation and efficient string manipulation in algorithmic challenges. #DSA #DynamicProgramming #Algorithms #CodingChallenge #TechInterviewPrep

What is an Algorithm? An algorithm is a step-by-step procedure, similar to a recipe, used to solve a given problem. What are the properties of an algorithm? ◾ An algorithm must have 𝐢𝐧𝐩𝐮𝐭 values. ◾ An algorithm must produce 𝐨𝐮𝐭𝐩𝐮𝐭 values. ◾ An algorithm must terminate after a finite number of steps (𝐟𝐢𝐧𝐢𝐭𝐞𝐧𝐞𝐬𝐬). ◾ All steps of the algorithm must be precisely defined (𝐝𝐞𝐟𝐢𝐧𝐢𝐭𝐞𝐧𝐞𝐬𝐬). ◾ An algorithm must accurately produce meaningful and correct results (𝐞𝐟𝐟𝐞𝐜𝐭𝐢𝐯𝐞𝐧𝐞𝐬𝐬). Some of the Algorithms Tools 𝐅𝐥𝐨𝐰𝐜𝐡𝐚𝐫𝐭𝐬 Is a graphical representation using symbols to depict the flow of control in an algorithm. Flowcharts are excellent for visualizing the structure and flow of an algorithm. 𝐏𝐬𝐞𝐮𝐝𝐨𝐜𝐨𝐝𝐞 Is textual representation that describes the steps of an algorithm in a language-agnostic way. Pseudocode is often written in a format that resembles programming languages but is more abstract and human-readable. 𝐃𝐞𝐜𝐢𝐬𝐢𝐨𝐧 𝐓𝐚𝐛𝐥𝐞𝐬 Is a tabular method to represent complex algorithms, especially those involving multiple conditions and decision points. Each row of the table typically represents a rule or a condition. #SoftwareEngineering #TechLeadership

🚀 Find the Length of the Longest Common Prefix 🚀 Excited to have solved an interesting problem on finding the longest common prefix between integers in two arrays using efficient algorithms! The key was improving performance through sorting and using a two-pointer technique to reduce unnecessary comparisons. Problem Overview: Given two arrays of integers, I needed to find the longest common prefix between any pair of numbers, optimizing for large datasets. Key Highlights: Transformed integers to strings for easier prefix comparison. Leveraged sorting and two-pointer traversal to reduce time complexity to O(n log n), avoiding brute force comparisons. Optimized the solution to pass all test cases, including those with input sizes up to 100,000+. 💻 Technologies Used: C++, Sorting Algorithms, String Manipulation, Time Complexity Optimization Proud to continue honing my algorithmic skills with challenges like this. Looking forward to applying these optimizations and problem-solving techniques in real-world applications. #Programming #Algorithms #Optimization #ProblemSolving #TechSkills #Cplusplus #CodingJourney LeetCode

🚀 **Day 14 of #100DaysOfCode: Tackling Binary Strings with No Consecutive 1s** 🚀 Today, I delved into the fascinating world of binary strings, specifically focusing on the challenge of generating binary strings without any consecutive 1s. It's been an enriching experience as I explored various approaches and honed my problem-solving skills. 🔍 **Problem Statement:** Generate all binary strings of length N that do not contain consecutive 1s. 🔧 **Approach:** 1. **Recursive Method:** Using recursion to build the string step-by-step while ensuring the no-consecutive-1s constraint is maintained. 2. **Dynamic Programming:** Leveraging DP to optimize and store intermediate results, thus improving efficiency. 📈 **Key Takeaways:** - Understanding the importance of base cases in recursion. - Appreciating the power of dynamic programming in solving complex problems efficiently. - Enhancing my debugging skills by thoroughly testing edge cases. 💡 **Why This Matters:** Binary strings with constraints are not just an academic exercise but have practical applications in error detection and correction, data compression, and more. This challenge has provided a deeper insight into these concepts, preparing me for real-world scenarios. Onward to Day 15! 🏆 #Coding #Programming #BinaryStrings #DynamicProgramming #ProblemSolving #CodeChallenge #TechJourney

🚀 Solving Isomorphic Strings Problem with C++ I recently tackled the Isomorphic Strings problem on LeetCode, and I'm excited to share my solution with you! This problem challenges us to determine whether two given strings s and t are isomorphic, meaning that characters in s can be replaced to get t. 💡 Approach: I utilized a mapping technique to efficiently determine if the strings are isomorphic. Here's how the code works: ✨ Explanation: • Two vectors, mp1 and mp2, are initialized to store mappings from characters in s to characters in t, and vice versa. • The function iterates through each character of both strings, establishing mappings and checking for conflicts along the way. • If at any point a conflict arises, indicating non-isomorphism, the function returns false. Otherwise, it returns true once all characters have been processed. 📊 Time Complexity: The time complexity of this solution is O(n), where n is the length of the input strings s and t. This complexity arises from iterating through both strings once. 💾 Space Complexity: The space complexity of this solution is O(1), as the space used by the vectors mp1 and mp2 remains constant regardless of the input size. #Coding #LeetCode #Algorithm #ProblemSolving #CPP #LinkedInlearning #programming #strings

🚀 Daily Challenge Solution: Longest Happy String Today, I tackled the problem of constructing the longest string using the characters 'a', 'b', and 'c', ensuring no character appears more than twice consecutively. Here's how I approached it: 🔍 Intuition: I built the string greedily by always appending the character with the highest remaining count, ensuring it doesn't violate the consecutive character limit. This approach guarantees a valid string at all times. 🛠 Approach: - Tracked consecutive occurrences of 'a', 'b', and 'c' to avoid repeating any character more than twice. - Appended the character with the largest remaining count in each step, maintaining the consecutive limit. - Reset the consecutive count for the other characters after appending one. ⏳ Time Complexity: The time complexity is O(a + b + c), where a, b, and c are the counts of the respective characters. This linear time complexity results from making decisions for each character. 💾 Space Complexity: With a space complexity of O(a + b + c), the resulting string is stored in the StringBuilder, ensuring efficient memory usage. #Programming #Algorithm #Coding #DailyChallenge #LinkedInLearning