155-min-stack

Try it on leetcode

Description

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.

 

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(Python)

class MinStack:
    def __init__(self):
        self.stack = []
        self.min = None

    def push(self, val: int) -> None:
        if not self.stack:
            self.min = val
            self.stack.append(val)
        elif val < self.min:
            y = (2 * val) - self.min
            self.min = val
            self.stack.append(y)
        else:
            self.stack.append(val)

    def pop(self) -> None:

        if self.stack:
            y = self.stack[-1]
            if y < self.min:
                self.min = (2 * self.min) - y
            self.stack.pop()

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

    def getMin(self) -> int:
        return self.min


# 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()