# HackerRank 'Maximum Element' Solution

Martin Kysel · January 29, 2016

##### Short Problem Definition:

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

1. 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.

Maximum Element

##### 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()
``````