标签：#datastructures

We all know the Big-O complexity of basic data structures. Arrays are O(n) for search. Hash maps are O(1). Linked lists are... well, complicated. But when I set out to build hashbrowns — a C++17 benchmarking suite comparing arrays, linked lists, and hash maps — I discovered that theory and practice are very different beasts. Here's what I learned building this project from scratch, and why you should probably benchmark before you optimize. 🎯 The Goal Was Simple (Ha!) I wanted a clean, educational project that would: Implement dynamic arrays, linked lists, and hash maps from scratch Benchmark insert, search, and remove operations Find the "crossover points" where one structure beats another Export everything to CSV for analysis Sounds straightforward, right? Four months later, I had written a custom memory tracker, implemented multiple hash map strategies, added statistical bootstrapping for confidence intervals, and learned more about CPU caches than I ever wanted to know. 📚 Lesson 1: Polymorphism Has a Price (But It's Worth It) My first architectural decision was creating a common DataStructure interface: class DataStructure { public: virtual void insert ( int key , const std :: string & value ) = 0 ; virtual bool search ( int key , std :: string & value ) const = 0 ; virtual bool remove ( int key ) = 0 ; virtual size_t memory_usage () const = 0 ; virtual std :: string type_name () const = 0 ; // ... }; This made benchmarking elegant — I could write generic code that tested any data structure: for ( auto & structure : structures ) { timer . start (); structure -> insert ( key , value ); timer . stop (); } But virtual function calls have overhead. In tight loops, that vtable lookup adds up. I spent a whole weekend convinced my hash map was slower than expected... until I realized I was measuring the cost of polymorphism, not the data structure itself. The fix? I kept the clean interface for the benchmarking harness but used templates internally where performance-cri

The Quest Begins (The "Why") I still remember the first time I faced the “Next Greater Element” interview question. The array looked innocent enough, but every brute‑force attempt felt like I was hammering a nail with a sponge— O(n²) time, nested loops, and a sinking feeling that I was missing something elegant. I spent an hour sketching out the problem on a whiteboard, muttering, “There has to be a way to look ahead without looking back every single time.” That frustration is a rite of passage for many developers. We’re taught to think in terms of scanning left‑to‑right, but some array puzzles scream for a different perspective: we need to remember what we’ve seen in a way that lets us answer questions about the future elements instantly. Enter the monotonic stack—a deceptively simple data structure that turns those scary “look‑ahead” problems into straight‑line walks. The Revelation (The Insight) So what’s the secret sauce? A monotonic stack is just a stack that maintains its elements in strictly increasing or strictly decreasing order. Why does that help? Consider the Next Greater Element problem: for each index i , we want the first element to its right that’s larger than arr[i] . If we walk from left to right and keep a stack of indices whose next greater element we haven’t found yet, the stack will naturally be decreasing in value. Why decreasing? Imagine the stack holds indices [i₁, i₂, …, i_k] where arr[i₁] > arr[i₂] > … > arr[i_k] . When we encounter a new value arr[j] , any element on the stack that is smaller than arr[j] has just found its next greater element—namely arr[j] . We pop those indices, record the answer, and stop when we hit a value that’s not smaller (or the stack empties). Then we push j onto the stack. Because each index is pushed once and popped at most once , the total work is linear: O(n) . No nested loops, no repeated scans—just a single pass with a stack that does the heavy lifting. The same invariant works for other “first bigger/smal

The Quest Begins (The "Why") I still remember the first time I saw LeetCode 323 “Number of Connected Components in an Undirected Graph”. I stared at the adjacency list, thought “I’ll just run a DFS from every node”, and coded it up in ten minutes. The solution passed the easy tests, but when the hidden test cases hit a graph with 10⁵ nodes and 10⁵ edges, my DFS started to choke—stack overflows, repeated visits, and a sinking feeling that I was brute‑forcing a problem that deserved a smarter tool. That night, after a few too many coffees, I stumbled upon a tiny comment in a discussion thread: “Union‑Find can do this in almost O(1) per operation”. My curiosity sparked like a power‑up in a retro arcade game. I had to know why this seemingly simple data structure could turn a nightmare into a breeze. The Revelation (The Insight) At its heart, Union‑Find (aka Disjoint Set Union, DSU) maintains a collection of elements partitioned into disjoint subsets. It supports two operations: Find(x) – returns the representative (root) of the set containing x . Union(x, y) – merges the sets containing x and y . The magic lies in two simple heuristics: Path Compression – when we walk up the tree to find a root, we make every node on that path point directly to the root. Future finds become flat, almost constant‑time. Union by Rank/Size – we always attach the smaller tree under the root of the larger one, keeping the overall tree shallow. Why does this give us near‑O(1) amortized time? Think of each Find as paying a small “tax” to flatten the path. The tax is paid only a few times per node before it becomes a direct child of the root. Over a sequence of m operations, the total work is bounded by O(m α(n)) , where α is the inverse Ackermann function—so slow‑growing it’s practically a constant for any realistic n . In plain English: every time we climb up, we leave a shortcut behind. The next climber benefits from that shortcut, and the structure keeps getting better. It’s like building

I was watching The Mandalorian the other day when it struck me that I don't know Pedro Pascal, which...

The Quest Begins (The "Why") I still remember the first time I got hit with a sorting question in an interview. The interviewer slid a whiteboard marker across the table and said, “Sort this array of a million integers – and tell me why you chose your method.” My brain went straight to the trusty old bubble sort I’d learned in CS101. I started writing nested loops, feeling like Neo dodging bullets in slow motion, only to realize the runtime was creeping toward O(n²). After a few painful minutes, I could see the interviewer’s eyes glaze over – not because I was wrong, but because I was using a sledgehammer to crack a nut. That moment sparked a quest: What makes a sorting algorithm truly efficient, and how do I know when to reach for it? I dove into textbooks, blog posts, and late‑night YouTube deep dives. The answer kept pointing back to one algorithm that felt like discovering a hidden cheat code: Merge Sort . The Revelation (The Insight) So why does Merge Sort work so well? It’s not just about splitting and merging; it’s about guaranteeing that each level of recursion does a linear amount of work, no matter how the input is arranged. Think of an unsorted array as a messy pile of LEGO bricks. Merge Sort first divides the pile into two halves, then halves again, until each sub‑pile contains a single brick – which is, by definition, sorted. The magic happens in the merge step: we take two already‑sorted sub‑arrays and walk through them with two pointers, always picking the smaller front element and appending it to the result. Because each sub‑array is sorted, we never need to look back; we simply advance one pointer at a time. That walk is O(n) for the merge: each element is examined exactly once as it gets placed into the output array. Since we split the array log₂ n times (each level halves the size), we perform an O(n) merge at each of those log₂ n levels. Multiply them together and you get O(n log n) worst‑case time, with O(n) extra space for the temporary buffer

The Quest Begins (The "Why") Honestly, I still remember the first time I tried to build an autocomplete widget for a side‑project. I had a list of 200 k product names, a simple filter that ran on every keystroke, and the UI felt like wading through molasses. Each keypress triggered a full scan of the list, and with a few users typing at once the browser would start to lag. I was stuck in a loop that felt like the infamous “boss fight” where you keep hitting the same pattern over and over, hoping for a different outcome. I kept asking myself: There has to be a smarter way. Why am I re‑checking the same prefixes again and again? If ten users type “tea”, why do I walk through the whole dictionary ten separate times? That question turned into a mini‑quest, and the treasure at the end was the trie data structure. The Revelation (The Insight) Look, the magic of a trie isn’t that it’s some exotic tree; it’s that it stores words by their shared prefixes . Imagine you have the words “cat”, “car”, “cart”, and “dog”. In a trie you’d have a root node, then a c branch that splits into a → t (for “cat”) and a → r → t (for “cart”), while “dog” lives on its own d → o → g path. Every common prefix is stored once , and you can walk down the tree following the characters of a query to land exactly at the node that represents all words with that prefix. Why does this give us O(L + K) time for autocomplete, where L is the length of the prefix and K is the number of results? Walking the trie follows the prefix character‑by‑character → O(L). From that node we just need to collect all words in its subtree. If we keep a list of words at each node (or run a DFS), we touch each result once → O(K). No extra work for words that don’t share the prefix. Contrast that with the naive filter approach: O(N × L) where N is the total dictionary size. For a large N, the trie is a game‑changer—it’s like switching from swinging a blunt sword to wielding a lightsaber that cuts through the prefix forest in

The Quest Begins (The "Why") I still remember the first time I stared at a LeetCode problem that asked me to count the number of islands in a grid. My initial instinct? Run a BFS/DFS from every unvisited land cell, mark everything reachable, and repeat. It worked, but each query felt like I was re‑exploring the same territory over and over again—like walking the same hallway in a dungeon every time I wanted to open a new door. Then a friend tossed me another problem: “Given a list of friendships, tell me if two people are in the same social circle.” Again, the naive solution was to rebuild the whole graph for every query. I felt like I was stuck in a grind‑fest, repeating the same low‑level work while the real challenge—understanding the structure of the connections—remain. That frustration sparked a question: Is there a way to remember what we’ve already discovered about connectivity, so future queries are instant? The answer, as many of you have guessed, lives in a humble but mighty data structure called Union‑Find (also known as Disjoint Set Union, DSU). The Revelation (The Insight) At its core, Union‑Find is about two simple ideas : Each element starts in its own set – think of every person as a lone adventurer. When we learn that two elements belong together, we merge their sets – we call that a union . The magic isn’t just in merging; it’s in how we find the representative (or “root”) of a set later on. If we naïvely walked up a chain of parents every time, we could end up with O(n) per find—still a grind. Two optimizations turn this into near‑constant time: Union by rank (or size) – always attach the smaller tree under the root of the larger one. This keeps the overall tree shallow, guaranteeing that the height never exceeds log n. Path compression – during a find operation, we make every node we pass point directly to the root. It’s like handing every traveler a map that instantly shows the shortest route to the campfire, so next time they don’t need to trek

JavaScript ships with Array , Set , and Map — but nothing that keeps its elements sorted as you insert. If you've ever built a leaderboard, an order book, or anything that answers "give me the items between X and Y", you know the workaround: push into an array and .sort() after every insertion. It works, until scale punishes you — you're paying O(n log n) over and over for data that was already 99.9% sorted. Python solved this years ago with sortedcontainers , built on an elegant "list of lists" design instead of balanced trees. I just published sorted-collections , which brings that idea to TypeScript — with full credit to the original as its inspiration. What you get SortedList, SortedSet, SortedMap — always sorted, no manual re-sorting, range queries built in. O(log n) insertions, O(√n) positional access via sqrt-decomposition into buckets. Zero runtime dependencies , ~2KB gzip, types included, dual ESM/CJS. Package quality gated in CI with publint , arethetypeswrong , and size-limit . The API in 30 seconds import { SortedList , SortedSet , SortedMap } from " sorted-collections " ; // SortedList: stays sorted on every insert const list = new SortedList ([ 5 , 1 , 4 , 2 , 3 ]); list . add ( 0 ); console . log ([... list ]); // [0, 1, 2, 3, 4, 5] console . log ( list . at ( 2 )); // 2 — positional access on sorted order // SortedSet: no duplicates, plus set algebra const a = new SortedSet ([ 1 , 2 , 3 , 4 ]); const b = new SortedSet ([ 3 , 4 , 5 ]); console . log ([... a . intersection ( b )]); // [3, 4] // SortedMap: keys always in order, range queries built in const prices = new SortedMap < number , string > ([ [ 104.5 , " order-3 " ], [ 99.2 , " order-1 " ], [ 101.0 , " order-2 " ], ]); for ( const [ price , id ] of prices . irange ( 100 , 105 )) { console . log ( price , id ); // 101.0 order-2, then 104.5 order-3 } Custom comparators are fully typed: number and string get natural ordering for free; for your own types, TypeScript requires a comparator at compile

The Quest Begins (The "Why") I still remember the first time I faced a problem that asked for the sum of numbers in a sub‑array, over and over again, with updates sprinkled in between. It felt like I was stuck in a never‑ending loop of for i in range(l, r+1): total += arr[i] – O(n) per query, and with up to 10⁵ queries the solution timed out every single time. I was staring at the screen, thinking, “There has to be a smarter way to answer these range questions without scanning the whole array each time.” That moment was my dragon: a seemingly simple problem that kept biting me because I kept reaching for the brute‑force sword. I needed a data structure that could give me the answer in logarithmic time while still supporting point updates. Enter the segment tree – the tool that turned my O(n·q) nightmare into an O((n+q)·log n) victory. The Revelation (The Insight) So why does a segment tree work? Imagine you have an array and you want to know the sum of any interval [l, r] . If you could break that interval into a handful of pre‑computed chunks, you’d only need to add those chunk values together instead of touching every element. A segment tree is exactly that: a binary tree where each node stores the aggregate (sum, min, max, etc.) of a segment of the original array. The root covers the whole array [0, n‑1] . Its two children cover the left half and the right half, and this keeps splitting until the leaves represent single elements. The magic lies in two facts: Every node’s value is a function of its children. If you know the sum of the left child and the sum of the right child, the parent’s sum is just their addition. This means we can build the tree bottom‑up in O(n) time. Any interval can be represented as O(log n) disjoint nodes. When you walk down the tree to answer [l, r] , you either take a whole node (if its segment lies completely inside the query) or you recurse further. Because the tree’s height is log₂n, you’ll visit at most 2·log₂n nodes. Thus, building

The Quest Begins (The "Why") Honestly, I still remember the first time I stared at the Daily Temperatures problem on LeetCode and felt like I was trying to crack a vault with a toothpick. The brute‑force solution — two nested loops, checking every future day for a warmer temperature — was simple to write, but it timed out on the larger test cases. I spent an hour tweaking loops, adding early breaks, and even trying to memoize results, only to watch the same red “Time Limit Exceeded” banner flash again. I was frustrated, but more than that, I was curious. Why did this problem feel so repetitive ? Every element seemed to be asking the same question: “What’s the next greater value to my right?” If I could answer that for each index in a single pass, the whole thing would collapse into O(n). That’s when I remembered a weird little data structure I’d seen in a textbook — the monotonic stack — and realized it might be the magic wand I needed. The Revelation (The Insight) Here’s the thing: a monotonic stack isn’t just a stack with a funny name; it’s a way to capture relationships between elements without ever looking backward more than once . Imagine you’re walking through a line of people sorted by height, and you want to know, for each person, who is the first taller person standing ahead of them. If you keep a stack of people whose heights are strictly decreasing as you move from left to right, then whenever you see a new person taller than the one on top of the stack, you’ve just found the answer for that stacked person: the current person is their “next greater.” You pop them off, record the distance, and keep going. Because each index is pushed once and popped at most once , the total work is linear. The same idea works for “next smaller,” “previous greater,” or any problem where you need the nearest element that satisfies a monotonic condition. The stack does the heavy lifting of remembering candidates that could still be relevant, discarding the ones that are alrea

[While preparing for interviews, I realized I had a strange habit. I would solve a problem, get stuck, open the solution, understand it, and move on feeling productive. A few days later, I couldn’t solve a similar problem on my own. The issue wasn’t lack of practice. The issue was that I was consuming solutions faster than I was developing problem-solving skills. So I changed my approach. Instead of looking for answers, I started forcing myself to think longer, write down my ideas, identify where I was stuck, and only then seek guidance. That worked much better. But I couldn’t find a tool that supported this style of learning. Most platforms either: Give you the answer. Give you the editorial. Give you AI that writes the code for you. So I started building my own. The goal was simple: An AI coach that guides the thought process instead of generating the solution. Over time I added: DSA practice System Design preparation Low-Level Design preparation Company-wise interview questions Topic-wise strength and weakness analysis Personalized revision lists The interesting part wasn’t building it. The interesting part was realizing that interview preparation is less about collecting solutions and more about training how you think. What has helped you improve more during interview prep? Reading solutions? Or struggling with the problem first? Sde vault - https://sdevaultweb.onrender.com/

A linked list is an ordered linear data structure where elements are not stored in sequential memory locations, instead they are stored in nodes that are linked together by a pointers. Linked list are used in data intensive application because linked list offer specific benefits for high frequency data manipulation, this benefits include: Efficient insertion and deletion Adding and removing elements from a linked list is highly efficient, unlike arrays which requires shifting all subsequent elements to maintain indexing. A linked list only requires updating the pointer. Dynamic sizing: linked list can grow or shrink during runtime without needing to pre-allocate memory. Memory management: Nodes in a linked list are only allocated when needed which prevents memory wastage. Flexible Traversal: Doubly and circular list allow you to move forward or backward, which makes them helpful for complex navigation The first node in a linked list is called the head which signifies the start of the list, while the last node is called the tail and has a pointer of null except in a circular linked list. Each node in a linked list has two things which are: the actual data the pointer or reference There are three main types of linked list: Singly linked list Doubly linked list Circular linked list Singly Linked List: Singly linked list are lists where each node has a next pointer that points to the next node. Doubly Linked List: Doubly linked list are list where each node has a next and previous pointer that points to the previous and next node. Circular Linked List: Circular linked list are list where the last node points back to the first node, forming a circle. Table of Contents create node class create linked list class isEmpty and getSize Methods prepend and append Method removeHead and removeTail Methods insert and search Methods getIndex and removeIndex Methods clear and print Methods create node class First let's open our code editor and create a new file called singlyLinkedLi