##### 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 isĀ 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()

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