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1472.DesignBrowserHistory(Dictionary).py
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1472.DesignBrowserHistory(Dictionary).py
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"""
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.
Example:
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.
"""
#Difficulty: Medium
#71 / 71 test cases passed.
#Runtime: 224 ms
#Memory Usage: 16 MB
#Runtime: 224 ms, faster than 99.17% of Python3 online submissions for Design Browser History.
#Memory Usage: 16 MB, less than 100.00% of Python3 online submissions for Design Browser History.
class BrowserHistory:
def __init__(self, homepage: str):
self.position = 0
self.db = {self.position : homepage}
self.length = 1
def visit(self, url: str) -> None:
self.position += 1
self.length = self.position + 1
self.db[self.position] = url
def back(self, steps: int) -> str:
self.position = max(0, self.position - steps)
return self.db[self.position]
def forward(self, steps: int) -> str:
self.position = min(self.length - 1, self.position + steps)
return self.db[self.position]
# Your BrowserHistory object will be instantiated and called as such:
# obj = BrowserHistory(homepage)
# obj.visit(url)
# param_2 = obj.back(steps)
# param_3 = obj.forward(steps)