3
06/11/2024 7:40 am
Onderwerp starter
You have an empty sequence, and you will be given N 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, N. The next N lines each contain an above mentioned query. (It is guaranteed that each query is valid.)
Output Format: For each type 3 query, print the maximum element in the stack on a new line.
Constraints:
1 ≤ N ≤ 105
1 ≤ x ≤ 109
1 ≤ type ≤ 3
Here is an examples:
1 antwoord
2
06/11/2024 7:41 am
Here is the answer my friend:
import java.util.Scanner; import java.util.Stack; public class Pr_03_MaximumElement { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int commandCount = sc.nextInt();//GETTING THE FIRST NUMBER - HOW MANY NUMBERS WILL BE Stack<Integer> stack = new Stack<>(); Stack<Integer> maxStack = new Stack<>(); int max = Integer.MIN_VALUE; for (int i = 0; i < commandCount; i++) { int command = sc.nextInt();//THE COMMAND NUMBER - WHICH IS IN FRONT OF THE NUMBER if (command == 1) { int numToPush = sc.nextInt();//THE NUMBER AFTER THE COMMAND ONE stack.push(numToPush); if (max <= numToPush) { max = numToPush; maxStack.push(max); } } else if (command == 2) { int poppedItem = stack.pop();//IN THIS CASE WE ARE REMOVING THE ELEMENT PRESENT AT THE TOP OF THE STACK if (poppedItem == max) { maxStack.pop(); if (maxStack.size() > 0) { max = maxStack.peek(); } else { max = Integer.MIN_VALUE; } } } else {//FOR THE COMMAND WITH NUMBER 3 FOR PRINTING THE MAXIMUM ELEMENT IN THE STACK System.out.println(max); } } } }
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