Min stack

class MinStack:

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

    def push(self, val: int) -> None:
        if len(self.stack) == 0:
            self.stack.append((val, val))
        else:
            last = self.stack[-1]
            curr_min = last[1]
            if val < curr_min:
                self.stack.append((val, val))
            else:
                self.stack.append((val, curr_min))

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

    def top(self) -> int:
        if self.stack:
            return self.stack[-1][0]
        else:
            return -1

    def getMin(self) -> int:
        if self.stack:
            return self.stack[-1][1]
        else:
            return -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()

Min Stack

Difficulty: Medium

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.