Min Stack

Problem

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

Implement the MinStack class:

• MinStack() initializes the stack object.
• void push(int val) pushes the element val onto the stack.
• void pop() removes the element on the top of the stack.
• int top() gets the top element of the stack.
• int getMin() retrieves the minimum element in the stack.

You must implement a solution with O(1) time complexity for each function.

Example 1:

Input [“MinStack”,”push”,”push”,”push”,”getMin”,”pop”,”top”,”getMin”] [[],[-2],[0],[-3],[],[],[],[]]

Output [null,null,null,null,-3,null,0,-2]

Explanation MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); // return -3 minStack.pop(); minStack.top(); // return 0 minStack.getMin(); // return -2

Constraints:

• -231 <= val <= 231 - 1
• Methods pop, top and getMin operations will always be called on non-empty stacks.
• At most 3 * 104 calls will be made to push, pop, top, and getMin.

Solution

• Stack
• Time: O(1)
• Space: O(n)

class MinStack(object):

    def __init__(self):
        self.stack = []
        self.minStack = []


    def push(self, val):
        """
        :type val: int
        :rtype: None
        """
        self.stack.append(val)
        val = min(val, self.minStack[-1] if self.minStack else val)
        self.minStack.append(val)


    def pop(self):
        """
        :rtype: None
        """
        self.stack.pop()
        self.minStack.pop()


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


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



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

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