# HackerRank ‘Maximum Element’ Solution

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##### 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:
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 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                   if command == '1':             cmd, value = map(int, unknown.split())             cs.push(value)         elif command == '2':             cs.pop()         else:             cs.printMax()           if __name__ == '__main__':     main() 