> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Design Browser History Python Solution

> Tested Python solution for LeetCode 1472 with 18 pytest cases. Generate a practice environment with lcpy.

LeetCode 1472, [Medium](/catalog/medium). Topics: [Array](/catalog/topics/array), [Linked List](/catalog/topics/linked-list), [Stack](/catalog/topics/stack), [Design](/catalog/topics/design), Doubly-Linked List, [Data Stream](/catalog/topics/data-stream). [View on LeetCode](https://leetcode.com/problems/design-browser-history/description/).

Generate this problem as a practice environment: tested reference solution, 18 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 1472   # by problem number
lcpy gen -s design_browser_history   # by problem name
```

## Problem

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

### Examples

```
Input
["BrowserHistory","visit","visit","visit","back","back","forward","visit","forward","back","back"]
[["leetcode.com"],["google.com"],["facebook.com"],["youtube.com"],[1],[1],[1],["linkedin.com"],[2],[2],[7]]
Output
[null,null,null,null,"facebook.com","google.com","facebook.com",null,"linkedin.com","google.com","leetcode.com"]
```

**Explanation:**

```
BrowserHistory browserHistory = new BrowserHistory("leetcode.com");
browserHistory.visit("google.com");   // You are in "leetcode.com". Visit "google.com"
browserHistory.visit("facebook.com"); // You are in "google.com". Visit "facebook.com"
browserHistory.visit("youtube.com");  // You are in "facebook.com". Visit "youtube.com"
browserHistory.back(1);               // You are in "youtube.com", move back to "facebook.com" return "facebook.com"
browserHistory.back(1);               // You are in "facebook.com", move back to "google.com" return "google.com"
browserHistory.forward(1);            // You are in "google.com", move forward to "facebook.com" return "facebook.com"
browserHistory.visit("linkedin.com"); // You are in "facebook.com". Visit "linkedin.com"
browserHistory.forward(2);            // You are in "linkedin.com", you cannot move forward any steps.
browserHistory.back(2);               // You are in "linkedin.com", move back two steps to "facebook.com" then to "google.com". return "google.com"
browserHistory.back(7);               // You are in "google.com", you can move back only one step to "leetcode.com". return "leetcode.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`.

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/design_browser_history/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/design_browser_history/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class BrowserHistory:
    # Time: __init__ O(1), visit O(n), back O(steps), forward O(steps)
    # Space: O(n)
    def __init__(self, homepage: str) -> None:
        self.history: list[str] = [homepage]
        self.cur = 0

    def visit(self, url: str) -> None:
        del self.history[self.cur + 1 :]
        self.history.append(url)
        self.cur += 1

    def back(self, steps: int) -> str:
        self.cur = max(0, self.cur - steps)
        return self.history[self.cur]

    def forward(self, steps: int) -> str:
        self.cur = min(len(self.history) - 1, self.cur + steps)
        return self.history[self.cur]
```

## Complexity

| Time | Space |
| - | - |
| **init** O(1), visit O(n), back O(steps), forward O(steps) | O(n) |

## Tags

[NeetCode All](/catalog/neetcode).


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