CodingGyan

Browser History (Back and Forward)

medium Java Collections

Problem

A browser starts on the page home. Commands: visit url goes to a new page and forgets all forward pages; back goes to the previous page (nothing happens if there is none); forward goes to the page you came back from (nothing happens if there is none). After every command print the current page. Use two stacks: one for the pages behind you and one for the pages ahead.

Constraints

  • 1 <= commands <= 10^5

Examples

Input

visit a
visit b
back
forward

Output

a
b
a
b

Solutions

Best approachTime O(n)Space O(n)

Two stacks model any undo-redo style history: moving between them keeps the order right. A new visit clears the 'ahead' stack, because the old future no longer exists. Java's ArrayDeque is the recommended stack; the old Stack class is slow and out of date.

Same code works in Java810172125Basic features only, so nothing changes between versions.
/*
 * Problem : Process lines.
 * Input   : Several lines.
 * Output  : The current page after each command.
 * Approach: Two stacks.
 * Time    : O(n)
 * Space   : O(n)
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
    // This method solves the task. It receives the input and returns the answer.
    static String[] solve(String[] lines) {
        // Step 1: "behind" holds the pages we can go back to. "ahead" holds the pages we can go forward to.
        java.util.ArrayDeque<String> behind = new java.util.ArrayDeque<>();
        java.util.ArrayDeque<String> ahead = new java.util.ArrayDeque<>();
        String current = "home";
        String[] output = new String[lines.length];
        for (int i = 0; i < lines.length; i++) {
            String[] part = lines[i].split(" ");
            if (part[0].equals("visit")) {
                // Step 2: A new page clears the forward history, like in a real browser.
                behind.push(current);
                current = part[1];
                ahead.clear();
            } else if (part[0].equals("back") && !behind.isEmpty()) {
                // Step 3: Going back moves the current page to the "ahead" stack.
                ahead.push(current);
                current = behind.pop();
            } else if (part[0].equals("forward") && !ahead.isEmpty()) {
                behind.push(current);
                current = ahead.pop();
            }
            output[i] = current;
        }
        return output;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        java.util.List<String> all = new java.util.ArrayList<>();
        while (scanner.hasNextLine()) {
            String line = scanner.nextLine().trim();
            if (!line.isEmpty()) {
                all.add(line);
            }
        }
        String[] lines = all.toArray(new String[0]);
        System.out.println(String.join("\n", solve(lines)));
    }
}
Hints (1)
  1. Why must visit clear the forward stack?