Data Structures Essentials: Arrays, Linked Lists, Stacks, Queues, Hash Tables, Trees, Graphs

Data Structures Cheatsheet — Everything You Need in One Place! Whether you're preparing for technical interviews or brushing up on CS fundamentals, understanding data structures is non-negotiable. Here's a quick breakdown of the 7 essential ones every developer should know: 📦 Array → Contiguous memory, index-based access. Simple but powerful. 🔗 Linked List → Dynamic size, efficient insertions & deletions. Great when memory flexibility matters. 📚 Stack → Last In, First Out (LIFO). Think function calls & undo operations. 🚶 Queue → First In, First Out (FIFO). Perfect for task scheduling & message passing. #️⃣ Hash Table → Key-value pairs with O(1) average lookup. Every backend dev's best friend. 🌳 Tree → Hierarchical structure with parent-child relationships. Used in searching, sorting & databases. 🕸️ Graph → Vertices + Edges. Models networks, social connections & real-world relationship #DataStructures #DSA #Java #Programming #TechInterview #ComputerScience #100DaysOfCode #BackendDevelopment #CodingTips #SoftwareEngineering

🚀 Day 86 — Prefix Sum Pattern (Introduction) Today I started learning the Prefix Sum pattern — a fundamental technique for array problems where decisions at a given index depend on the sum of previous elements. Unlike sliding window, prefix sum handles negative numbers gracefully. 📌 What is Prefix Sum? At index i, the prefix sum is the sum of all elements from the start up to i-1. It stores “historical information” to make efficient decisions. 🧠 Why Prefix Sum > Sliding Window for Negatives? Sliding window fails with negatives because removing a negative actually increases the sum — breaking the logic of expanding/shrinking. Prefix sum remains effective. 📊 Four Categories of Prefix Sum Problems: 1️⃣ Left‑Right Comparisons (e.g., Pivot Index) → Can often be solved with simple variables (no extra arrays). 2️⃣ Exact Sums (Subarray Sum = K) → Requires a HashMap to store and lookup previous prefix sums. 3️⃣ Shortest Window with Sum ≥ K (with negatives) → Needs a Deque (double‑ended queue). 4️⃣ Range Sum Queries → Advanced techniques like Merge Sort on prefix array. 🔧 General Template: Initialize data structure (variable, array, or HashMap). Loop through the array, updating prefix information. Check problem‑specific condition. ⚡ Optimization Insight (Pivot Index Example): Instead of storing both prefix and suffix arrays, use: Total Sum = Left Sum + arr[i] + Right Sum → Right Sum = Total Sum - Left Sum - arr[i] This finds the pivot using only O(1) space. 💡 Takeaway: Prefix sum is a pattern, not just an algorithm. Recognizing when to use it — especially with negative numbers — unlocks efficient solutions for subarray sum problems. No guilt about past breaks — just adding another powerful pattern to the toolkit. #DSA #PrefixSum #ArrayPatterns #CodingJourney #Revision #Java #ProblemSolving #Consistency #GrowthMindset #TechCommunity #LearningInPublic

Most people don’t struggle with SQL… 🧠 They struggle with how they think about it. I used to write SQL like this: SELECT → JOIN → WHERE → Done. Then I learned to ask one simple question: “Can SQL solve this before I write more code?” Everything changed. 🚀 JOINs replaced manual loops GROUP BY replaced tedious manual calculations Window Functions replaced complex application logic SQL stopped being just a query language… it became a superpower. ⚡ Whether you are navigating DQL (Querying), DML (Manipulation), or DDL (Definition), this chart (attached) is the roadmap to thinking like a Senior Developer. #SQL #DataEngineering #SoftwareDevelopment #Database #DataScience #Coding #TechTrends2026 #Programming #BigData #Analytics #BackendDevelopment #Python #WebDev #EngineeringMindset #TechCareer

𝐃𝐚𝐲 𝟓𝟖 – 𝐃𝐒𝐀 𝐉𝐨𝐮𝐫𝐧𝐞𝐲 | 𝐀𝐫𝐫𝐚𝐲𝐬 🚀 Today’s problem focused on finding two numbers in a sorted array that add up to a target. 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐒𝐨𝐥𝐯𝐞𝐝 • Two Sum II – Input Array Is Sorted 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡 • Used two pointers: • One at the beginning (left) • One at the end (right) • Calculated sum of both elements Logic: • If sum == target → return indices • If sum < target → move left pointer forward • If sum > target → move right pointer backward This works because the array is already sorted. 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠𝐬 • Sorting enables two-pointer optimization • Two pointers reduce time complexity from O(n²) to O(n) • Direction of movement depends on comparison with target • Index-based problems often become simpler with sorted data 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲 • Time: O(n) • Space: O(1) 𝐓𝐚𝐤𝐞𝐚𝐰𝐚𝐲 When data is sorted, two pointers can turn a complex problem into a simple one. 58 days consistent 🚀 On to Day 59. #DSA #Arrays #TwoPointers #LeetCode #Java #ProblemSolving #DailyCoding #LearningInPublic #SoftwareDeveloper

𝐃𝐚𝐲 𝟓𝟓 – 𝐃𝐒𝐀 𝐉𝐨𝐮𝐫𝐧𝐞𝐲 | 𝐀𝐫𝐫𝐚𝐲𝐬 🚀 Today’s problem extended the previous one by introducing duplicates in a rotated sorted array. 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝐒𝐨𝐥𝐯𝐞𝐝 • Find Minimum in Rotated Sorted Array II 𝐀𝐩𝐩𝐫𝐨𝐚𝐜𝐡 • Used modified binary search • Compared nums[mid] with nums[right] Logic: • If nums[mid] > nums[right] → minimum is in right half • If nums[mid] < nums[right] → minimum is in left half (including mid) • If equal → cannot decide, so shrink search space (right--) 𝐊𝐞𝐲 𝐋𝐞𝐚𝐫𝐧𝐢𝐧𝐠𝐬 • Duplicates can break standard binary search logic • When unsure, safely reduce the search space • Edge cases often define the complexity of the problem • Small changes in conditions can affect performance 𝐂𝐨𝐦𝐩𝐥𝐞𝐱𝐢𝐭𝐲 • Average Time: O(log n) • Worst Case: O(n) (due to duplicates) • Space: O(1) 𝐓𝐚𝐤𝐞𝐚𝐰𝐚𝐲 When the data becomes ambiguous, focus on reducing the problem safely instead of forcing a decision. 55 days consistent 🚀 On to Day 56. #DSA #Arrays #BinarySearch #LeetCode #Java #ProblemSolving #DailyCoding #LearningInPublic #SoftwareDeveloper

🚀 08/04/26 — Stack Foundations: Balancing Parentheses with Precision Today was a productive session where I successfully implemented the Valid Parentheses (LeetCode 20) challenge. This problem is a fundamental exercise in using the Stack data structure to manage nested relationships and maintain order-based logic. 🧱 The Valid Parentheses Logic The goal is to determine if an input string containing (, ), {, }, [ and ] is valid based on whether every open bracket is closed by the correct type and in the correct order. The Stack Strategy: Pushing: As I iterate through the string, whenever I encounter an opening bracket—(, {, or [—I push it onto the stack. Popping and Matching: When I encounter a closing bracket, I first check if the stack is empty. If it is, the string is invalid. Otherwise, I pop the top element from the stack and compare it to the current closing bracket. Validation: If the brackets don't match (e.g., a ) following a {), the function immediately returns false. Final Check: After the loop finishes, the string is only valid if the stack is completely empty, ensuring all open brackets were properly closed. Complexity Metrics: Time Complexity: O(n) where n is the length of the string, as we perform a single linear pass. Space Complexity: O(n) in the worst case where the string contains only opening brackets, requiring them all to be stored in the stack. 📈 Consistency Report Coming off my 50-day streak milestone, today's focus on stacks feels like a solid pivot from the sliding window and array patterns I've been mastering recently. The logic used here is remarkably similar to the structural checks I used in the "String Search" and "Mountain Array" problems earlier this month, where maintaining a specific state across iterations was key. Huge thanks to Anuj Kumar (a.k.a CTO Bhaiya on YouTube) for the continuous inspiration. Every new data structure I master adds another layer to my problem-solving toolkit! My tested O(n) stack-based implementation is attached below! 📄👇 #DSA #Java #Stack #ValidParentheses #DataStructures #Complexity #Consistency #LearningInPublic #CTOBhaiya

Backend Interviews: What Really Matters (2026) Coding – clean logic, edge cases Problem Solving – clear approach Backend Basics – APIs, DB, caching System Design – tradeoffs, scaling Debugging – find root cause Concurrency – async basics Data Modeling – tables, indexes Communication – explain clearly

Day 80/100 | #100DaysOfDSA 🧩🚀 Today’s problem: Convert Sorted Array to Binary Search Tree A classic divide-and-conquer problem that builds intuition for balanced trees. Problem idea: Convert a sorted array into a height-balanced BST. Key idea: Recursion + choosing the middle element as root. Why? • The array is already sorted • Picking the middle ensures balance • Left half forms left subtree, right half forms right subtree How it works: • Find the middle index of the array • Create a node with that value • Recursively build left subtree using left half • Recursively build right subtree using right half Time Complexity: O(n) Space Complexity: O(log n) (recursion stack) Big takeaway: Whenever you need a balanced BST from sorted data, think divide & conquer with mid as root. 🔥 This pattern is widely used in tree construction problems. Day 80 done. 🚀 #100DaysOfCode #LeetCode #DSA #Algorithms #BinarySearchTree #DivideAndConquer #Recursion #Java #CodingJourney #ProblemSolving #InterviewPrep #TechCommunity

Just dropped: Top 50 SQL Q&A to help you crack your next data or backend interview 🗄️ From JOINs and indexes to window functions and CTEs — everything you need in one place. Save this. You'll thank yourself later. #SQL #DataEngineering #TechInterview #Programming #DatabaseDesign

Day 67/100 | #100DaysOfDSA 🧩🚀 Today’s problem: Subsets II Another classic backtracking problem with a twist (duplicates). Problem idea: Generate all possible subsets (power set), but avoid duplicate subsets. Key idea: Backtracking + sorting to handle duplicates. Why? • We need to explore all subset combinations • Duplicates in input can lead to duplicate subsets • Sorting helps us skip repeated elements efficiently How it works: • Sort the array first • At each step, add current subset to result • Iterate through elements • Skip duplicates using condition: 👉 if (i > start && nums[i] == nums[i-1]) continue • Choose → recurse → backtrack Time Complexity: O(2^n) Space Complexity: O(n) recursion depth Big takeaway: Handling duplicates in backtracking requires careful skipping logic, not extra data structures. This pattern appears in many problems (subsets, permutations, combinations). 🔥 Day 67 done. 🚀 #100DaysOfCode #LeetCode #DSA #Algorithms #Backtracking #Recursion #Java #CodingJourney #ProblemSolving #InterviewPrep #TechCommunity