Question

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

• push(x) – Push element x onto stack.
• pop() – Removes the element on top of the stack.
• top() – Get the top element.
• getMin() – Retrieve the minimum element in the stack.

Example:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> Returns -3.
minStack.pop();
minStack.top();      --> Returns 0.
minStack.getMin();   --> Returns -2.

Solution

class MinStack(object):

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = list()
        self.min_vals = list()


    def push(self, x):
        """
        :type x: int
        :rtype: void
        """
        if len(self.min_vals) > 0:
            self.min_vals.append(min(x, self.min_vals[-1]))
        else:
            self.min_vals.append(x)
        self.stack.append(x)

    def pop(self):
        """
        :rtype: void
        """
        self.min_vals.pop()
        self.stack.pop()


    def top(self):
        """
        :rtype: int
        """
        return self.stack[-1]

    def getMin(self):
        """
        :rtype: int
        """
        return self.min_vals[-1]



# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()