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

Question

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.

Python Solution

class MinStack:

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

    def push(self, x: int) -> None:
        cur_min = self.getMin()
        if cur_min==None or cur_min > x:
            cur_min = x
        self.stack.append([x,cur_min])

    def pop(self) -> None:
        self.stack.pop()

    def top(self) -> int:
        if len(self.stack)==0:
            return None
        return self.stack[-1][0]

    def getMin(self) -> int:
        if len(self.stack)==0:
            return None
        return self.stack[-1][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()