Tags: "leetcode", "stack", "ood", access_time 1-min read

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Min Stack

Created: October 14, 2018 by [lek-tin]

Last updated: October 14, 2018

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:

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []
        self.min_stack = []

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

    def pop(self):
        """
        :rtype: void
        """
        pop = self.stack.pop()
        if pop == self.min_stack[-1]:
            self.min_stack.pop()

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

    def getMin(self):
        """
        :rtype: int
        """
        return self.min_stack[-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()