sorry, typo while(top>p)

I was asked of this question as well. I not sure how to implment a second stack in a way that the max number is always on top.

I was asked of this question as well. I not sure how to implment a second stack in a way that the max number is always on top.

Create a separate stack for the maximums.

just saw this, i think u can just map your number into an object, that has both the maximum number, and the last number that you've pushed in just peek the stack before you push, compare the peek'd value 'vs' your new pushed number, then replace or update the max number as you see fit? if you're only allowed to push numbers into your stack, just push your number in the stack more than once (at-least three times) indicating that every value that you put in is a sequence of 2 numbers, and so, push another one (the 3rd, which will always be your maximum number) so on every 3 pushes, you've stored the maximum value, then everytime you push something in, just peek the last 3 values in the stack, knowing that the 2nd and 3rd value was probably the number you pushed in, and the first 1 was the max value it's funny

Sort the array so that the numbers are in descending order this way you know for sure the first element is the maximum number.

Maintain your Stack ADT like this: class Node { T data; T max; Node next; } class Stack { Node top; push(T value) { } }

Maintain your Stack ADT like this: class Node { T data; T max; Node next; } class Stack { Node top; void push(T value) { Node node = new Node(); node.value = value; if(top == null) { node.max = value; } else { node.max = value.compareTo(top.max) > 1 ? value : top.max; } node.next = top; top = node; } T pop() { if(top == null) { return null; } T val = top.val; top = top.next; return val; } T max() { return top == null ? null : top.max; } } public static void main(String[] args) { Stack stack = new Stack(); stack.push(3); stack.push(2); stack.push(5); System.out.println(stack.max()); // 5 System.out.println(stack.pop()); // 5 System.out.println(stack.max()); // 3 System.out.println(stack.pop()); // 2 System.out.println(stack.max()); // 3 System.out.println(stack.pop()); // 3 }

maintain a sorted stack, the insert would look like insert(int p,stack s){ stack large; int top = s.peek(); while(top>s){ large.push(s.pop()); top = s.peek(); } s.push(p); while(!large.empty()){ s.push(large.pop()); } }