Maximum Element HackerRank Solution

Maximum Element HackerRank Solution
Source : https://www.hackerrank.com/challenges/maximum-element

You have an empty sequence, and you will be given queries. Each query is one of these three types:

1 x  -Push the element x into the stack.2    -Delete the element present at the top of the stack.3    -Print the maximum element in the stack.

Input Format

The first line of input contains an integer, . The next lines each contain an above mentioned query. (It is guaranteed that each query is valid.)

Constraints

Output Format

For each type query, print the maximum element in the stack on a new line.

Sample Input

101 9721 2021 261 20231 913

Sample Output

2691

Source : https://www.hackerrank.com/challenges/maximum-element

Solution

	// Karthikalapati.blogspot.com
	import java.util.Scanner;
	import java.util.Stack;
	// All 3 queries (1:push, 2:delete, 3:print max) are all O(1) runtime
	public class Solution {
	public static void main(String[] args) {
	Stack<Integer> stack = new Stack<Integer>();
	Stack<Integer> maxStack = new Stack<Integer>(); // keeps track of maximums
	Scanner scan = new Scanner(System.in);
	int N = scan.nextInt();
	for (int i = 0; i < N; i++) {
	int query = scan.nextInt();
	switch (query) {
	case 1:
	int x = scan.nextInt();
	stack.push(x);
	if (maxStack.isEmpty() || x >= maxStack.peek()) {
	maxStack.push(x);
	}
	break;
	case 2:
	int poppedValue = stack.pop();
	if (poppedValue == maxStack.peek()) {
	maxStack.pop();
	}
	break;
	case 3:
	System.out.println(maxStack.peek());
	break;
	default:
	break;
	}
	}
	scan.close();
	}
	}