Design Browser History

Design Browser History - Problem

Array Linked List Stack Design Doubly-Linked List Data Stream Medium

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

Implement the BrowserHistory class:

BrowserHistory(string homepage): Initializes the object with the homepage of the browser.
void visit(string url): Visits url from the current page. It clears up all the forward history.
string back(int steps): Move steps back in history. If you can only return x steps in the history and steps > x, you will return only x steps. Return the current url after moving back in history at most steps.
string forward(int steps): Move steps forward in history. If you can only forward x steps in the history and steps > x, you will forward only x steps. Return the current url after forwarding in history at most steps.

Input & Output

Example 1 — Basic Browser Navigation

$ Input: operations = ["BrowserHistory","visit","visit","visit","back","back","forward"], parameters = [["leetcode.com"],["google.com"],["facebook.com"],["youtube.com"],[1],[1],[2]]

› Output: [null,null,null,null,"facebook.com","leetcode.com","facebook.com"]

💡 Note: Initialize with leetcode.com, visit google.com, facebook.com, youtube.com. Back 1 step to facebook.com, back 1 more to leetcode.com, forward 2 steps to facebook.com

Example 2 — Forward History Cleared

$ Input: operations = ["BrowserHistory","visit","back","visit","forward"], parameters = [["home.com"],["page1.com"],[1],["page2.com"],[1]]

› Output: [null,null,"home.com",null,"page2.com"]

💡 Note: Start at home.com, visit page1.com, back to home.com, visit page2.com (clears page1.com), forward stays at page2.com

Example 3 — Boundary Navigation

$ Input: operations = ["BrowserHistory","back","forward"], parameters = [["start.com"],[5],[5]]

› Output: [null,"start.com","start.com"]

💡 Note: Only one page in history, so both back 5 steps and forward 5 steps return start.com

Constraints

1 ≤ homepage.length ≤ 20
1 ≤ url.length ≤ 20
1 ≤ steps ≤ 100
homepage and url consist of '.' or lower case English letters.
At most 5000 calls will be made to visit, back, and forward.

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8 Apple 6

The key insight is to maintain a dynamic array of URLs with a current index pointer. When visiting new pages, clear forward history by truncating the array. Best approach uses array with index tracking: Time: O(1), Space: O(n)

Common Approaches

✓ Array with Current Index

⏱️ Time: O(1) Space: O(n)

Use a dynamic array to store all visited URLs and maintain a current index. When visiting a new URL, clear forward history by truncating the array.

Dynamic Array with Capacity Management

⏱️ Time: O(1) Space: O(n)

Similar to basic approach but with better memory management. Pre-allocate reasonable capacity and only resize when necessary to reduce allocation overhead.

Array with Current Index — Algorithm Steps

Step 1: Initialize array with homepage and set current index to 0
Step 2: For visit(), add URL after current position and clear forward history
Step 3: For back(), move index backward within bounds
Step 4: For forward(), move index forward within bounds

Visualization

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Step-by-Step Walkthrough

Initialize

Create array with homepage at index 0

Visit URLs

Add new URLs and update current index

Navigate

Move current index for back/forward operations

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct {
    char history[1000][101];  // Max 1000 URLs, each up to 100 chars
    int current;
    int size;
} BrowserHistory;

BrowserHistory* browserHistoryCreate(char* homepage) {
    BrowserHistory* obj = (BrowserHistory*)malloc(sizeof(BrowserHistory));
    strcpy(obj->history[0], homepage);
    obj->current = 0;
    obj->size = 1;
    return obj;
}

void browserHistoryVisit(BrowserHistory* obj, char* url) {
    // Clear forward history and add new URL
    obj->current++;
    obj->size = obj->current + 1;
    strcpy(obj->history[obj->current], url);
}

char* browserHistoryBack(BrowserHistory* obj, int steps) {
    obj->current = (obj->current - steps > 0) ? obj->current - steps : 0;
    return obj->history[obj->current];
}

char* browserHistoryForward(BrowserHistory* obj, int steps) {
    int maxPos = obj->size - 1;
    obj->current = (obj->current + steps < maxPos) ? obj->current + steps : maxPos;
    return obj->history[obj->current];
}

void browserHistoryFree(BrowserHistory* obj) {
    free(obj);
}

int main() {
    // For this implementation, we'll handle a basic test case
    BrowserHistory* browser = browserHistoryCreate("leetcode.com");
    
    browserHistoryVisit(browser, "google.com");
    browserHistoryVisit(browser, "facebook.com");
    browserHistoryVisit(browser, "youtube.com");
    
    printf("[null,null,null,null,\"%s\",\"%s\",\"%s\"]\n", 
           browserHistoryBack(browser, 1),
           browserHistoryBack(browser, 1), 
           browserHistoryForward(browser, 2));
    
    browserHistoryFree(browser);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

All operations are constant time - array access and index updates

✓ Linear Growth

Space Complexity

O(n)

Store up to n URLs in the history array where n is number of unique visits

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Array with Current Index — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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