Home/dsa/Design/Design Browser History

Design Browser History

# System & Data Structure Design Design problems in DSA interviews test your ability to translate requirements into a functional, efficient, and maintainable class structure. Unlike standard algorithmic problems, the focus here is on **State Management** and **API Design**. ### Core Principles 1. **Encapsulation:** Keep data private and expose functionality through well-defined methods. 2. **Trade-offs:** Every design choice has a cost. Is it better to have $O(1)$ read and $O(N)$ write, or vice versa? 3. **State Consistency:** Ensure that your internal data structures (e.g., a Map and a List) stay in sync after every operation. ### Common Design Patterns #### 1. HashMap + Doubly Linked List (DLL) The "Gold Standard" for $O(1)$ caching (LRU/LFU). ```text [Head] <-> [Node A] <-> [Node B] <-> [Node C] <-> [Tail] ^ ^ ^ ^ ^ (MRU) (Data) (Data) (Data) (LRU) ``` - **HashMap:** Provides $O(1)$ lookups for keys to their corresponding nodes. - **DLL:** Provides $O(1)$ addition/removal of nodes at both ends, maintaining the order of access. #### 2. Amortized Analysis (Rebalancing) Commonly used in **Queue using Stacks** or **Dynamic Arrays**. - Instead of doing heavy work on every call, we batch it. Pushing to a stack is $O(1)$, and "flipping" elements to another stack happens only when necessary, averaging $O(1)$ per operation. #### 3. Ring Buffers (Circular Arrays) Used for fixed-size memory management (e.g., **Circular Queue**, **Hit Counter**). ```text [0] [1] [2] [3] [4] [5] ^ ^ ^ Head (Data) Tail (Pops) (Next Push) ``` - Use `(index + 1) % capacity` to wrap around the array. #### 4. Concurrency & Thread Safety For "Hard" design problems (e.g., **Bounded Blocking Queue**). - Use **Mutexes** (Locks) to prevent data races. - Use **Condition Variables** (`wait`/`notify`) to manage producer-consumer logic efficiently without busy-waiting. ### How to Approach a Design Problem 1. **Identify the API:** What methods do you need to implement? (`get`, `put`, `push`, etc.) 2. **Define the State:** What variables represent the current state? (Size, Capacity, Pointers). 3. **Choose the Data Structures:** Select the combination that minimizes time complexity for the most frequent operations. 4. **Dry Run:** Trace the state changes through a sequence of operations based on your chosen structure.

Design Browser History

You have a browser of one tab where you start on a homepage and you can visit another url, get back in the history number of steps or move forward in the history number of steps.

Medium
Approach 1

Level I: Two Stacks

Intuition

Maintain history and forward stacks. When visiting, clear forward stack.

O(1) all ops.💾 O(N).

Detailed Dry Run

Visit(A), Visit(B), Back(1) -> Back is at A.

java
import java.util.*;\nclass BrowserHistory {\n    private Stack<String> history = new Stack<>(), forward = new Stack<>();\n    public BrowserHistory(String homepage) { history.push(homepage); }\n    public void visit(String url) { history.push(url); forward.clear(); }\n    public String back(int steps) {\n        while(steps-- > 0 && history.size() > 1) forward.push(history.pop());\n        return history.peek();\n    }\n    public String forward(int steps) {\n        while(steps-- > 0 && !forward.isEmpty()) history.push(forward.pop());\n        return history.peek();\n    }\n}
Approach 2

Level III: Single Array with Pointer (Optimal)

Intuition

Use a single dynamic array (list) and an index pointer. Overwrite forward history on visit and simply move the index on back / forward.

O(1) average all ops.💾 O(N).

Detailed Dry Run

Index starts at 0. Visit moves it to 1. Back(1) moves it to 0.

java
import java.util.*;\nclass BrowserHistory {\n    private List<String> history = new ArrayList<>();\n    private int curr = 0, last = 0;\n    public BrowserHistory(String homepage) { history.add(homepage); }\n    public void visit(String url) {\n        curr++;\n        if(curr < history.size()) history.set(curr, url);\n        else history.add(url);\n        last = curr;\n    }\n    public String back(int steps) { curr = Math.max(0, curr - steps); return history.get(curr); }\n    public String forward(int steps) { curr = Math.min(last, curr + steps); return history.get(curr); }\n}