1472.design-browser-history.kt

/*
 * @lc app=leetcode id=1472 lang=kotlin
 *
 * [1472] Design Browser History
 *
 * https://leetcode.com/problems/design-browser-history/description/
 *
 * algorithms
 * Medium (58.86%)
 * Likes:    75
 * Dislikes: 10
 * Total Accepted:    7.1K
 * Total Submissions: 12.1K
 * Testcase Example:  '["BrowserHistory","visit","visit","visit","back","back","forward","visit","forward","back","back"]\r\n' +
  '[["leetcode.com"],["google.com"],["facebook.com"],["youtube.com"],[1],[1],[1],["linkedin.com"],[2],[2],[7]]\r'
 *
 * 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.
 * 
 */

// @lc code=start
class BrowserHistory(homepage: String) {
    val st = Array<String>(5005) {""}
    var cnt = 1
    var top = 1

    init {
        st[1] = homepage
    }

    fun visit(url: String) {
        if (cnt == top) {
            st[++cnt] = url;
            ++top;
        } else {
            st[cnt+1] = url;
            top = cnt+1;
            ++cnt;
        }
    }

    fun back(steps: Int): String {
        if (steps >= cnt) {
            cnt = 1;
            return st[cnt];
        } else {
            cnt -= steps;
            return st[cnt];
        }
    }

    fun forward(steps: Int): String {
        if (steps+cnt >= top) {
            cnt = top;
            return st[cnt];
        } else {
            cnt += steps;
            return st[cnt];
        }
    }

}

/**
 * Your BrowserHistory object will be instantiated and called as such:
 * var obj = BrowserHistory(homepage)
 * obj.visit(url)
 * var param_2 = obj.back(steps)
 * var param_3 = obj.forward(steps)
 */
// @lc code=end