Anzhari Purnomo

Posted on

# Problem Statement

Write a MinMaxStack class for a Min-Max Stack with these features:

• Push & pop values from the stack.
• Peek value at the top of the stack.
• Get the current min and max value of the stack in constant time.

# Sample Input & Output

``````stack = MinMaxStack()
stack.push(5) # min: 5, max: 5, current value: 5
stack.push(7) # min: 5, max: 7, current value: 7
stack.push(2) # min: 2, max: 7, current value: 2
stack.pop(2)
stack.pop(7) # min: 5, max: 5, current value: 5
``````

# Code #1

``````class MinMaxStack:
def __init__(self):
self.stack = []

def peek(self):
if len(self.stack) == 0:
return None
else:
return self.stack[len(self.stack) - 1]['value']

def pop(self):
if len(self.stack) == 0:
return None
else:
return self.stack.pop()['value']

def push(self, number):
self.stack.append(
{
'value': number,
'min': self._calc_min(number),
'max': self._calc_max(number)
}
)

def get_min(self):
if len(self.stack) == 0:
return None
else:
return self.stack[len(self.stack) - 1]['min']

def get_max(self):
if len(self.stack) == 0:
return None
else:
return self.stack[len(self.stack) - 1]['max']

def _calc_min(self, number):
if len(self.stack) == 0:
return number
elif number < self.stack[len(self.stack) - 1]['min']:
return number
else:
return self.stack[len(self.stack) - 1]['min']

def _calc_max(self, number):
if len(self.stack) == 0:
return number
elif number > self.stack[len(self.stack) - 1]['max']:
return number
else:
return self.stack[len(self.stack) - 1]['max']

``````

# Notes

• Store current, min, and max value state during pop and push.