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You want to be able to access the largest element in a stack.
**

Use the built-in Stack class to **implement a new class MaxStack with a function getMax that returns the largest element in the stack.** getMax should not remove the item.

Your stacks will contain only integers.

What if we push several items in increasing numeric order (like 1, 2, 3, 4...), so that there is a *new max* after each push? What if we then pop each of these items off, so that there is a *new max* after each pop? Your algorithm shouldn't pay a steep cost in these edge cases.

You should be able to get a runtime of for push, pop, and getMax.

One lazy approach is to have getMax simply walk through the stack and find the max element. This takes time for each call to getMax. But we can do better.

To get time for getMax, we could store the max integer as a member variable (call it max). But how would we keep it up to date?

For every push, we can check to see if the item being pushed is larger than the current max, assigning it as our new max if so. But what happens when we pop the current max? We could re-compute the current max by walking through our stack in time. So our worst-case runtime for pop would be . We can do better.

What if when we find a new current max (newMax), instead of overwriting the old one (oldMax) we held onto it, so that once newMax was popped off our stack we would know that our max was back to oldMax?

What data structure should we store our set of maxs in? We want something where the last item we put in is the first item we get out ("last in, first out").

We can store our maxs in another stack!

We define *two* new stacks within our MaxStack class—stack holds all of our integers, and maxesStack holds our "maxima." We use maxesStack to keep our max up to date in constant time as we push and pop:

- Whenever we push a new item, we check to see if it's greater than or equal to the current max, which is at the top of maxesStack. If it is, we also push it onto maxesStack.
- Whenever we pop, we also pop from the top of maxesStack if the item equals the top item in maxesStack.

import java.util.Stack;
public class MaxStack {
private Stack<Integer> stack = new Stack<>();
private Stack<Integer> maxesStack = new Stack<>();
// Add a new item to the top of our stack. If the item is greater
// than or equal to the last item in maxesStack, it's
// the new max! So we'll add it to maxesStack.
public void push(int item) {
stack.push(item);
if (maxesStack.empty() || item >= maxesStack.peek()) {
maxesStack.push(item);
}
}
// Remove and return the top item from our stack. If it equals
// the top item in maxesStack, they must have been pushed in together.
// So we'll pop it out of maxesStack too.
public int pop() {
int item = stack.pop();
if (item == maxesStack.peek()) {
maxesStack.pop();
}
return item;
}
// The last item in maxesStack is the max item in our stack.
public int getMax() {
return maxesStack.peek();
}
}

time for push, pop, and getMax. additional space, where m is the number of operations performed on the stack.

Notice that our time-efficient approach takes some additional space, while a lazy approach (simply walking through the stack to find the max integer whenever getMax is called) took no additional space. We've traded some space efficiency for time efficiency.

Notice how in the solution we're *spending time* on push and pop so we can *save time* on getMax. That's because we chose to optimize for the time cost of calls to getMax.

But we could've chosen to optimize for something else. For example, if we expected we'd be running push and pop frequently and running getMax rarely, we could have optimized for faster push and pop functions.

Sometimes the first step in algorithm design is *deciding what we're optimizing for*. Start by considering the expected characteristics of the input.

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