Tags: "leetcode", "stack", access_time 2-min read

# Max Stack

Design a max stack that supports push, pop, top, peekMax and popMax.
1. `push(x`) – Push element x onto stack. 2. `pop()` – Remove the element on top of the stack and return it. 3. `top()` – Get the element on the top. 4. `peekMax()` – Retrieve the maximum element in the stack. 5. `popMax()` – Retrieve the maximum element in the stack, and remove it. If you find more than one maximum elements, only remove the top-most one.

### Example 1:

``````MaxStack stack = new MaxStack();
stack.push(5);
stack.push(1);
stack.push(5);
stack.top(); -> 5
stack.popMax(); -> 5
stack.top(); -> 1
stack.peekMax(); -> 5
stack.pop(); -> 1
stack.top(); -> 5
``````

### Note:

1. `-1e7 <= x <= 1e7`
2. Number of operations won’t exceed `10000`.
3. The last four operations won’t be called when stack is empty.

### Solution:

``````class MaxStack {

Stack<Integer> stack;
Stack<Integer> maxStack;
/** initialize your data structure here. */
public MaxStack() {
stack = new Stack<>();
maxStack = new Stack<>();
}

public void push(int x) {
int max = maxStack.isEmpty() ? x : maxStack.peek();
// Assign a max to every newcomer
maxStack.push(max > x ? max : x);
stack.push(x);
}

public int pop() {
System.out.println(maxStack.pop());
return stack.pop();
}

public int top() {
return stack.peek();
}

public int peekMax() {
return maxStack.peek();
}

public int popMax() {
int val = peekMax();
Stack<Integer> buffer = new Stack();
while (top() != val) { buffer.push(pop()); }
pop();
while (!buffer.isEmpty()) { push(buffer.pop()); }
return val;
}
}

/**
* Your MaxStack object will be instantiated and called as such:
* MaxStack obj = new MaxStack();
* obj.push(x);
* int param_2 = obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.peekMax();
* int param_5 = obj.popMax();
*/
``````

Time Complexity: `O(N)` for the popMax operation, and `O(1)` for the other operations, where `N` is the number of operations performed.
Space Complexity: `O(N)`, the maximum size of the stack.