Recursion and Backtracking: Generating Parentheses with Constraints

✅ Day 69 of 100 Days LeetCode Challenge Problem: 🔹 #1009 – Complement of Base 10 Integer 🔗 https://lnkd.in/geVPugvi Learning Journey: 🔹 Today’s challenge involved finding the bitwise complement of a base-10 integer. 🔹 I first converted the integer into its binary representation using bin(n)[2:] to remove the 0b prefix. 🔹 Then I iterated through each bit and flipped it: • 0 becomes 1 • 1 becomes 0 🔹 After constructing the flipped binary string, I converted it back to a decimal integer using int(s, 2). Concepts Used: 🔹 Binary Representation 🔹 Bit Manipulation 🔹 String Traversal 🔹 Base Conversion (Binary → Decimal) Key Insight: 🔹 Converting the number to binary makes it easy to flip bits directly. 🔹 After inversion, converting the binary string back to base-10 produces the required complement. Complexity: 🔹 Time: O(b) where b is the number of bits in n 🔹 Space: O(b) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

🚀 Day 41 / 200 Days of Code Solved the “Majority Element” problem on LeetCode today! 🔹 Problem: Find the element that appears more than ⌊n/2⌋ times in an array. 🔹 Constraint: Majority element is guaranteed to exist. 🔹 Approach Used: Python built-in optimization (max(set(nums), key=nums.count)) 🔹 Status: ✅ Accepted 🔹 Runtime: 0 ms 🔹 Memory: 21.07 MB (Beats 84.28%) While a simple built-in solution works, I also explored the Boyer–Moore Voting Algorithm, which solves it in O(n) time and O(1) space — a powerful technique frequently asked in interviews. 💡 Key Learning: Efficient problem-solving isn’t just about passing test cases — it’s about understanding optimized approaches and improving algorithmic thinking. Consistency > Motivation. One problem a day. Steady progress toward mastery. #Day41 #200DaysOfCode #LeetCode #Python #DataStructures #Algorithms #CodingJourney #ProblemSolving

✅ Day 64 of 100 Days LeetCode Challenge Problem: 🔹 #1784 – Check if Binary String Has at Most One Segment of Ones 🔗 https://lnkd.in/gpJYed9J Learning Journey: 🔹 Today’s problem focused on determining whether a binary string contains only one contiguous segment of '1's. 🔹 Since the string has no leading zeros, the first character is always '1'. 🔹 If a '1' appears again after a '0', it indicates the start of a second segment. 🔹 A single pass through the string is enough to detect this pattern. Concepts Used: 🔹 String Traversal 🔹 Pattern Detection 🔹 Conditional Logic Key Insight: 🔹 The pattern "01" followed by another '1' reveals multiple segments of ones. 🔹 Detecting this transition allows us to solve the problem efficiently. Complexity: 🔹 Time: O(n) 🔹 Space: O(1) 💬 What’s your favorite trick for solving binary string problems? #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

✅ Day 74 of 100 Days LeetCode Challenge Problem: 🔹 #653 – Two Sum IV: Input is a BST 🔗 https://lnkd.in/g9vDFKMA Learning Journey: 🔹 Today’s problem required checking whether there exist two nodes in a Binary Search Tree whose values sum to k. 🔹 I performed a Depth-First Search (DFS) traversal using a stack to visit all nodes of the tree. 🔹 While traversing, I stored each node’s value in a list. 🔹 Then I used a hash set to track visited values and checked whether the complement (k - value) already exists. 🔹 If the complement is found, it means two numbers sum to k, so the function returns True. Concepts Used: 🔹 Depth-First Search (DFS) 🔹 Stack-based Tree Traversal 🔹 Hash Set for Fast Lookup 🔹 Two Sum Pattern Key Insight: 🔹 The classic Two Sum approach works well after collecting values from the BST. 🔹 Using a set allows O(1) lookup, making it efficient to check complements during iteration. Complexity: 🔹 Time: O(n) 🔹 Space: O(n) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

🚀 Day 18 of #100DaysOfCode 📌 LeetCode 977 – Squares of a Sorted Array Today’s problem was about transforming a sorted array (with possible negative numbers) into a new array of squared values that remains sorted. 💡 Approach Used: Two Pointers Instead of squaring all elements and sorting again (O(n log n)), I applied the Two Pointer technique to compare absolute values from both ends and filled the result array from the back. 🧠 Key Learnings: ✔️ Importance of optimal thinking over brute force ✔️ How negative numbers affect sorting after squaring ✔️ Efficient use of Two Pointers in array problems ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(n) 🔍 Problem Pattern: Array + Two Pointers Consistency is the key — showing up every day and improving step by step! 💪 #Day18 #100DaysOfCode #LeetCode #DSA #Python #CodingJourney #BTech #DataStructures

✅ Day 75 of 100 Days LeetCode Challenge Problem: 🔹 #415 – Add Strings 🔗 https://lnkd.in/gHVN5VfP Learning Journey: 🔹 Today’s problem required adding two numbers represented as strings without directly using built-in conversion functions. 🔹 I implemented a helper function to convert a string into its numeric value by iterating through digits and multiplying them with the correct powers of 10. 🔹 After converting both strings to integers, I performed the addition. 🔹 Then, I converted the result back into a string by extracting digits using modulo (%) and reversing the constructed list. Concepts Used: 🔹 String to Number Conversion 🔹 Mathematical Digit Handling 🔹 Modulo & Division 🔹 String Construction Key Insight: 🔹 Manually converting between string and integer helps understand how numbers are built and processed internally. 🔹 Constructing the result in reverse and then reversing it ensures correct digit order. Complexity: 🔹 Time: O(n + m) 🔹 Space: O(n + m) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

🚀 LeetCode 41 – First Missing Positive (Hard) Solved the classic First Missing Positive problem using two different approaches. 🧩 Problem: Given an unsorted integer array, find the smallest missing positive integer in O(n) time and O(1) space. 🔗 Try the problem here:https://lnkd.in/gCAGSftq Approach 1: Using Hash Set 💡 Idea: Store all elements in a set and check from 1 upward to find the first missing number. ⏱ Time Complexity: O(n) 📦 Space Complexity: O(n) ❌ (extra space used) 🔎 Why Not Optimal? Uses extra memory, so it doesn’t satisfy the O(1) space requirement. Approach 2: Cyclic Sort (Optimal Solution) 💡 Idea: Place each number x at index x - 1 if 1 ≤ x ≤ n. Then scan the array to find the first index where nums[i] != i+1. ⏱ Time Complexity: O(n) 📦 Space Complexity: O(1) ✅ 🔥 Why This Works? We use the array itself as a hash map by placing numbers in their correct indices. Each element is swapped at most once → overall linear time. If you're preparing for interviews, this is a must-know hard-level pattern 💪 #LeetCode #Coding #DataStructures #Algorithms #DSA #Python #CodingInterview #InterviewPreparation #ProblemSolving #SoftwareDeveloper #100DaysOfCode #TechCareers #ProgrammersLife #CodeNewbie #CompetitiveProgramming

✨ Day 8 of #GeekStreak60 Today’s Problem of the Day focused on determining whether two strings are Isomorphic. 🔍 Problem Statement Given two strings s1 and s2 of equal length, check whether they are isomorphic. Two strings are isomorphic if: • Each character in s1 maps to exactly one character in s2 • The mapping is consistent and preserves order • No two characters in s1 map to the same character in s2 💡 Approach ✔️ Compared the pattern of first occurrences in both strings ✔️ Used list comprehension with index() to capture structural patterns ✔️ Verified if both patterns match — ensuring isomorphism 🚀 Key Insight Instead of directly mapping characters, comparing structural patterns simplifies the solution and keeps the code clean and efficient. 🎯 Outcome Clean logic ➝ Minimal code ➝ Maximum clarity 📈 Progress Update Day 8 completed successfully! Consistency is the key to building strong problem-solving skills 💪 #GeeksforGeeks #GeekStreak60 #Day8 #ProblemSolving #Python #CodingJourney #NPCI

Weekly challenge 5: Recursion To understand recursion, you must first understand recursion. For Week 5 of my algorithm challenge, I decided to tackle a concept that trips up many beginners: Recursive Functions, using the classic Factorial problem. What is Recursion? Instead of using a standard `for` or `while` loop, a recursive function calls **itself** to solve a smaller piece of the problem. It keeps digging deeper until it hits a "Base Case" (the bottom), and then it passes the answers back up the chain. Think of it like a set of Russian nesting dolls. The Trade-off:** > While recursive code is extremely clean and mathematical, it uses more memory. Every time the function calls itself, it adds a new layer to the computer's **Call Stack**. If you forget your Base Case, your program crashes with a "Stack Overflow"! > > I added a visual trace to my Python script so you can literally see the Call Stack growing and shrinking in the console. Check the full code and console output on GitHub: https://lnkd.in/es5TzCUg #Python #Recursion #Algorithms #CodingChallenge #SoftwareEngineering #DataScience 

🚀 Day 8/30 | LeetCode Problem: Merge Sorted Array (88) Problem: You are given two sorted arrays nums1 and nums2, and two integers m and n representing the number of initialized elements in each array. Merge nums2 into nums1 as one sorted array. The final result should be stored inside nums1. 💡 Approach Since nums1 already has enough space at the end (filled with 0s), Steps: Copy elements of nums2 into the empty positions of nums1 Sort the entire nums1 array This ensures the final array remains sorted. ⏱ Complexity Time Complexity: O((m+n) log(m+n)) → due to sorting Space Complexity: O(1) → in-place modification 🧠 Python Code class Solution: def merge(self, nums1, m, nums2, n): for i in range(n): nums1[m + i] = nums2[i] nums1.sort() return nums1 📌 Example Input: nums1 = [1,2,3,0,0,0], m = 3 nums2 = [2,5,6], n = 3 Output: [1,2,2,3,5,6] 🎯 Key Takeaway Sometimes a straightforward solution works perfectly. Optimization (like using two pointers from the end) can further improve time complexity — but clarity first, then optimization. ✅ Accepted | 100% Runtime 🔖 Hashtags #LeetCode #30DaysOfLeetCode #Day8 #Python #Arrays #Sorting #DataStructures #Algorithms #ProblemSolving #CodingJourney