Mastering Data Structures & Algorithms with Daily Challenges

DSA Day 4 – Simple Problem, Important Insight Today I solved the “Contains Duplicate” problem on LeetCode. Problem: Check if any element appears more than once in an array. Initial Thought: -> Use nested loops to compare every pair -> Time Complexity: O(n²) ❌ Optimized Approach: -> Used HashMap to track elements -> While iterating, check if element already exists -> If yes → duplicate found Time Complexity: -> O(n) -> Space Complexity: O(n) Key Learning: -> Not every problem needs complex logic -> Choosing the right data structure makes a big difference -> Hashing is one of the most powerful tools in DSA What I Realized: Sometimes the simplest problems teach the most important concepts. Thanks to Pulkit Aggarwal sir for guiding me in DSA and helping me build strong fundamentals. Staying consistent, one day at a time. #DSA #LeetCode #Java #ProblemSolving #CodingJourney #100DaysOfCode

🚀 30 Days of DSA — Day 1 Starting today, I'm committing to solving one Data Structures & Algorithms problem every day for the next 30 days. Why this challenge? • Strengthen problem-solving skills • Build consistency • Learn in public 📌 Day 1 Problem: Bubble sort implementation 🧠 Approach: Compared adjacent elements and swapped them if they were in the wrong order. Repeated this process multiple times until the array was sorted. ⏱️ Time Complexity:  O(n²) 💡 Key Learning:  Bubble Sort is simple but inefficient for large datasets. Great for understanding sorting fundamentals. Let's see where this journey takes me. If you're also working on DSA, feel free to connect or share your approach 👇 #30DaysOfDSA #Java #SpringBoot #JavaDeveloper #DSA #DataStructures #Algorithms #CodingChallenge #ProblemSolving #SoftwareEngineer #BackendDeveloper #LearningInPublic #CodingJourney #Developers #TechCommunity

🚀 Day 22 of My DSA Journey Today I solved an important Binary Search problem: 👉 Find First and Last Position of Element in Sorted Array 🧩 Problem Understanding Given a sorted array and a target value, we need to find the starting and ending position of the target element. If the target is not present, return [-1, -1]. 🔍 Approach I Used (Binary Search x2) Instead of linear search, I used Binary Search twice: ✔️ First Binary Search → to find the first occurrence ✔️ Second Binary Search → to find the last occurrence 💡 Trick: Even after finding the target, we don’t stop — we continue searching on left/right side to get the boundary indices. ⚡ Why this approach? Because the array is sorted → Binary Search reduces time complexity drastically. ⏱ Time Complexity: O(log n) 📦 Space Complexity: O(1) 🔥 Key Learning Binary Search is not just for finding elements — it can also be used to find boundaries, ranges, and conditions. Consistency is the real game changer 💪 One problem at a time! #DSA #BinarySearch #LeetCode #Java #ProblemSolving #100DaysOfCode #CodingJourney #LearningInPublic

🚀 DSA Series #1 — Closest Target in Circular Array Today I solved an interesting problem involving circular arrays + shortest distance logic. 🧠 Key Insight: Instead of simulating movement, we can compute distance using modulo. 👉 Forward distance: (i - start + n) % n 👉 Backward distance: (start - i + n) % n Take the minimum of both — done in O(n) time ⚡ 📌 Complexity: O(n) time | O(1) space 💡 Learning: Circular problems often look tricky, but math simplifies everything. 🔥 Building consistency with: #DSA #Java #Coding #PlacementPreparation #100DaysOfCode If you’re on the same path, let’s connect and grow 🤝

Day 56 of My DSA Journey Today I solved LeetCode 162 – Find Peak Element on . 📌 Problem Given an array "nums", find a peak element and return its index. 👉 A peak element is one that is greater than its neighbors. You can assume: • "nums[-1] = -∞" and "nums[n] = -∞" --- 🧠 Approach – Binary Search (Optimized) Instead of checking every element (O(n)), I used Binary Search. Key Idea: • Calculate "mid" • Compare "nums[mid]" with "nums[mid + 1]" 👉 If "nums[mid] > nums[mid + 1]" → We are in a descending part, so peak lies on the left side (including mid) 👉 Else → We are in an ascending part, so peak lies on the right side Repeat until "start == end" → that index is a peak. --- ⏱ Time Complexity: "O(log n)" — Efficient binary search 📦 Space Complexity: "O(1)" — No extra space used --- 💡 Key Learnings ✔ Using binary search on unsorted-looking problems ✔ Understanding pattern-based decision making ✔ Learning that a problem can have multiple valid answers (any peak) This problem really strengthens intuition for advanced binary search patterns 🚀 --- Consistency continues — getting better every day 💪🔥 #100DaysOfCode #DSA #BinarySearch #LeetCode #Java #ProblemSolving #CodingJourney #DeveloperJourney #Programming #SoftwareEngineering

🔥 DSA Challenge – Day 119/360 🚀 📌 Topic: Recursion / Math 🧩 Problem: Pow(x, n) Problem Statement: Find x raised to the power n (xⁿ) efficiently, even for large values of n. 🔍 Example: Input: x = 2, n = 5 Output: 32 💡 Approach: Optimized (Binary Exponentiation) 1️⃣ Step 1 – If n is negative, convert → x = 1/x and n = -n 2️⃣ Step 2 – Recursively calculate half = power(x, n/2) 3️⃣ Step 3 – • If n is even → return half × half • If n is odd → return half × half × x ⏱ Complexity: Time: O(log n) Space: O(log n) 📚 Key Learning: Breaking a problem into halves can drastically improve performance 🚀 #DSA #Java #Coding #InterviewPrep #ProblemSolving #TechJourney #119DaysOfCode #LeetCode

🚀 DSA Preparation 💪 Solved an advanced Binary Search problem on a rotated sorted array. Learned how to identify the sorted half and efficiently narrow down the search space 🔥 🧠 Problem 🔎 Search in Rotated Sorted Array Given a sorted array nums that is rotated at an unknown index, and a target value: 👉 Return the index of the target if found, otherwise return -1. 👉 The solution must run in O(log n) time complexity. Example Input: nums = [4,5,6,7,0,1,2], target = 0 Output: 4 Input: nums = [4,5,6,7,0,1,2], target = 3 Output: -1 Input: nums = [1], target = 0 Output: -1 ⚡ Key Learning 📌 Even after rotation, one half of the array is always sorted 📌 Use Binary Search to decide which half to explore 📌 Time Complexity: O(log n) → reduces search space by half in each step, making it highly efficient even for large inputs 📌 Space Complexity: O(1) Improving DSA with advanced searching techniques 🚀 #DSA #LeetCode #BinarySearch #Algorithms #Java #ProblemSolving #Consistency

🚀 Day 90 – DSA Journey | Binary Tree Paths Continuing my daily DSA practice, today I explored how to track and construct paths in a binary tree. 📌 Problem Practiced: Binary Tree Paths (LeetCode 257) 🔍 Problem Idea: Return all root-to-leaf paths in a binary tree as strings. 💡 Key Insight: While traversing the tree, we can build the path step by step. Once we reach a leaf node, we add the complete path to the result. 📌 Approach Used: • Use DFS traversal • Maintain a string to track the current path • At each node, append its value to the path • If it’s a leaf node → add the path to result • Continue exploring left and right subtrees 📌 Concepts Strengthened: • Tree traversal (DFS) • Recursion • Path tracking • String manipulation ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(h) (recursion stack) 🔥 Today’s takeaway: Tracking state (like a path) during recursion is a powerful technique for solving tree problems. On to Day 91! 🚀 #Day90 #DSAJourney #LeetCode #BinaryTree #DFS #Java #ProblemSolving #Coding #LearningInPublic #Consistency

#CodeEveryday — My DSA Journey | Day 13 🧩 Problem Solved: Kth Missing Positive Number (LeetCode #1539) 💭 What I Learned: Used binary search to efficiently find the k-th missing positive number. Key idea: 👉 At any index mid, the number of missing elements before it is: arr[mid] - (mid + 1) At each step: ✔️ Calculated how many numbers are missing till mid ✔️ Adjusted search space based on whether it’s less than or greater than k Finally used low + k to compute the answer. This helped me understand how to transform a problem into a mathematical observation + binary search. ⏱ Time Complexity: O(log n) 🧠 Space Complexity: O(1) ⚡ Key Takeaways: ✔️ Binary search can be applied on derived values, not just indices ✔️ Identifying patterns like “missing count till index” simplifies problems ✔️ Mathematical insight + binary search = optimal solution 💻 Language Used: Java ☕ 📘 Concepts: Binary Search · Math Insight · Arrays · Optimization #CodeEveryday #DSA #LeetCode #Java #BinarySearch #ProblemSolving #Algorithms #CodingJourney #Consistency 🚀

🚀 3-Month DSA Challenge | Week 4 Continuing my journey of consistent DSA practice and sharing my weekly progress. 📚 Week 4 Topic: Recursion & Backtracking (Deep Practice) 📌 Problems I practiced: 🔹 Reverse String 🔹 Fibonacci Number 🔹 Binary Watch 🔹 Power of Two / Pow(x, n) 🔹 Print N to 1 (without loop) 🔹 Generate Parentheses 🔹 Combinations 🔹 Combination Sum I & III 🔹 Subsets & Subsets II (Power Set) 🔹 Permutations II 🔹 Word Break 🔹 Restore IP Addresses 🔹 Predict the Winner 🔹 Count Good Numbers 🔹 Rat in a Maze (Multiple Jumps) 🔹 N-Queens 🔹 Handshakes Problem 🔹 Bracket Challenge 💭 My Learning: Recursion became much clearer when I started visualizing recursion trees. Backtracking helped me understand how to explore all possibilities and undo choices efficiently. ✅ Solved 20+ problems to strengthen recursion and backtracking patterns. This week was challenging but very important for improving problem-solving skills 💪 #DataStructures #DSA #Recursion #Backtracking #Java #CodingJourney #Learning #InterviewPreparation