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Hello Friends! I Have Migrated My Old Channel To A New Brand Channel. Please Subscribe To My New Channel For Learning And Understanding - "The Light Bulb!" 😊💙 "The Light Bulb!" URL : https://lnkd.in/g5J32Xum -------------------------------------------------------------------------------- LeetCode Problem Solving : Generate All Subsets : Video 5 : PseudoCode Time Complexity - O(2^N) Space Complexity - O(N) Analysis Of Call Stack: Given an integer array nums of unique elements, return all possible subsets (the power set) The Unification Of Philosophy Of Reason And Understanding + Philosophy Of Judgment + Math + Problem Solving. #datastructuresandalgorithms #leetcodechallenge #leetcode #computerscience #linkedincommunity #techcommunity #bigtech. https://lnkd.in/gEzn5kxj

🌟Day 3 : Transformative Tree Magic: Binary to Doubly Linked List! 🌳💫 Hello Guyz! Today, let's explore a fascinating binary tree transformation using C++! 🚀 **Logic Behind the Transformation:** 1. **Recursive Approach:** - Utilize a recursive approach to traverse the binary tree. - Start from the right subtree, then process the current node, and finally, traverse the left subtree. 2. **Node Conversion:** - For each node encountered during the traversal: - Link the current node's right pointer to the previously processed node. - If the previously processed node exists, link its left pointer to the current node. - Update the head to the current node. 3. **Result:** - The result is a doubly linked list where each node's right pointer points to the next node, and the left pointer points to the previous node. 4. **Return Result:** - The head of the doubly linked list is returned as the final result. Feel free to share your thoughts, ask questions, or share your experiences with tree transformations! Let's continue our coding journey together! 💡🌐 #BinaryTree #LinkedListTransformation #CodingMagic #DSAchallenge

#day44 problem:- Forming the Largest Number from an Array of Integers 🚀. Interesting problem: forming the largest possible number from an array of integers. 💡 Approach: I started by converting each integer in the array into a string. This allows us to manipulate and compare the numbers based on string concatenation. I then defined a custom comparator function to sort the strings. The idea is to compare two strings str1 and str2 by checking if str1 + str2 is greater than str2 + str1. This ensures that the combination yielding the larger number comes first in the sorted order. After sorting, I concatenated the sorted strings to form the final result. If the result starts with a '0', it means the entire number is zero becuase i have sortedd the array in descending order,even though i sorted in descending order if the first number is 0,then entire number is 0 (e.g., for an input array of all zeros), and I return "0" in that case. 📈 Time and Space Complexity: Time Complexity: The time complexity is dominated by the sort operation, which is O(nlog⁡n), where nnn is the number of integers. The comparator function operates in constant time for each comparison. Space Complexity: The space complexity is O(n) due to the storage of the string representations of the numbers. Let me know your thoughts or any other efficient approaches you've tried! 💬 #coding #algorithms #leetcode #programming #problemsolving #developer #c++

Day 22/100 - #100DaysOfCode Problem: Combination Sum II Difficulty: Medium Today, I tackled the Combination Sum II problem. Unlike the original Combination Sum, this version requires each candidate to be used only once, and the combinations must be unique. I implemented a recursive backtracking approach where I carefully skipped over duplicates to ensure only unique combinations were added. Sorting the candidates helped handle the duplicate elements efficiently. This problem really tested my understanding of recursion and backtracking techniques while teaching me how to manage duplicate elements in a sorted array! #LeetCode #CodingChallenge #LearningEveryday #KeepCoding #ConsistencyMatters 4o

DAY 35 of problem solving in leetcode PROBLEM : Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value. If target is not found in the array, return [-1, -1]. You must write an algorithm with O(log n) runtime complexity. Example 1: Input: nums = [5,7,7,8,8,10], target = 8 Output: [3,4] Example 2: Input: nums = [5,7,7,8,8,10], target = 6 Output: [-1,-1] Example 3: Input: nums = [], target = 0 Output: [-1,-1] Constraints: 0 <= nums.length <= 105 -109 <= nums[i] <= 109 nums is a non-decreasing array. -109 <= target <= 109 #C++ #CODING #LEETCODE

I'm thrilled to share my latest coding milestone. I've just cracked another challenging problem on LeetCode: Problem #111 - Minimum Depth of Binary Tree. Each step forward in this journey keeps pushing my problem-solving skills to the next level. The approach used in this problem is Depth-First Search (DFS) to find the minimum depth of a binary tree. It can be solved by recursively traversing the tree, starting from the root and exploring each subtree. Time Complexity: O(N) Space Complexity: O(H) #MinimumDepth #DepthFirstSearch #Recursion #ProblemSolving #LeetCode #Algorithms

Hello LinkedIn family! Day (64/150) - Today's Highlights: Today, I attempted an easy-level problem from leetcode focused on the stack concept. I encourage everyone to take part in the challenge and give it a shot! Share your solutions, and let's learn together. Here's a detailed explanation of the problem statement. Problem discription is as follows: Backspace String Compare: Given two strings s and t, return true if they are equal when both are typed into empty text editors. '#' means a backspace character. Note that after backspacing an empty text, the text will continue empty. Example 1: Input: s = "ab#c", t = "ad#c" Output: true Explanation: Both s and t become "ac". Example 2: Input: s = "ab##", t = "c#d#" Output: true Explanation: Both s and t become "". Example 3: Input: s = "a#c", t = "b" Output: false Explanation: s becomes "c" while t becomes "b". Constraints: 1 <= s.length, t.length <= 200 s and t only contain lowercase letters and '#' characters. Today's key takeaways! Stay tuned for tomorrow's updates, and let the learning journey continue! #LeetCodeChallenge #CodingJourney #DailyCoding #AlgorithmLearning #ProblemSolving #ProgrammingLogic

🚀 Just tackled the "Find All Duplicates in an Array" problem on Leetcode with C++! 💡 My Approach: To find duplicates in the array, I utilized the negation method. I iterated through the array and marked visited elements by negating their values. If an element was encountered twice, I added it to the result vector. Here's a brief explanation of the code: 1. Iterate through the input array nums. 2. For each element nums[i], find its absolute value num and calculate its index idx by subtracting 1. 3.Check if the value at index idx is negative. If it is, it means that num has already been encountered, so it is a duplicate and is added to the result vector result. 4. If the value at index idx is not negative, negate it to mark it as encountered 💬 Let's Discuss: How would you approach solving this problem? Share your thoughts and strategies in the comments below! #Coding #LeetCode #Algorithm #ProblemSolving #CPP #programmingchallenges #LinkedInlearning

🚀 Today’s LeetCode Daily Problem: Permutation in String Today, I tackled an interesting challenge on LeetCode: Permutation in String. The task? Determine if any permutation of string s1 exists as a substring within string s2. Example: Input: s1 = "abc", s2 = "eidbaooo" Output: true (because "ba" is a permutation of "abc" found in s2). My Journey: From Brute Force to Optimization 💡 I started off with the Brute Force approach: Idea: Generate all permutations of s1 and check if any of them are substrings of s2. Outcome: While it worked for smaller inputs, I quickly hit a wall with larger strings, facing Time Limit Exceeded (TLE) errors due to the factorial complexity (O(n! * m)). 🚫 Realizing the limitations, I shifted and explored the Sliding Window technique: How It Works: Create a frequency count of characters in s1. Initialize a sliding window of the same length in s2 and maintain its frequency count. As the window slides across s2, update the counts by adding the new character and removing the old one. Check for equality between the frequency counts. If they match, we have a valid permutation! Time Complexity: O(n + m), making it efficient even for large strings. 💨 Key Takeaways: Switching from a brute-force mindset to an optimized approach . The Sliding Window technique not only enhanced performance but also deepened my understanding of substring searches through character frequency management. This experience reminded me of the importance of adaptability in problem-solving. 🌱 I’d love to hear your thoughts! Have you faced similar challenges? What strategies do you use for optimizing your solutions? Let’s learn together! 🔗 #LeetCode #Algorithms #DataStructures #Coding #ProblemSolving #Programming #Java #SlidingWindow #Optimization #LearnInPublic #TechCommunity #SoftwareDevelopment #DailyCoding