##### Short Problem Definition:

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

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

##### Link

##### Complexity:

time complexity is `O(N)`

space complexity is `O(N)`

##### Execution:

I really enjoyed this problem. I did not see the solution at first, but after it popped up, it was really simple.

Keep two stacks. One for the actual values and one (non-strictly) increasing stack for keeping the maxima.

##### Solution:

```
class CustomStack:
def __init__(self):
self.stack = []
self.maxima = []
def push(self, value):
self.stack.append(value)
if not self.maxima or value >= self.maxima[-1]:
self.maxima.append(value)
def printMax(self):
print self.maxima[-1]
def pop(self):
value = self.stack.pop()
if value == self.maxima[-1]:
self.maxima.pop()
def main():
cs = CustomStack()
N = int(raw_input())
for _ in xrange(N):
unknown = raw_input()
command = unknown[0]
if command == '1':
cmd, value = map(int, unknown.split())
cs.push(value)
elif command == '2':
cs.pop()
else:
cs.printMax()
if __name__ == '__main__':
main()
```