leetcode - 155. Min Stack

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.
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:

-2^31 <= val <= 2^31 - 1
Methods pop, top and getMin operations will always be called on non-empty stacks.
At most 3 * 10^4 calls will be made to push, pop, top, and getMin.

Solution

Hint: Consider each node in the stack having a minimum value.

So we keep track of the minimum number of the whole stack, if we popped the minimum element, then we replace the minimum number by the minimum number at the stack top.

Code

class MinStack:
    import heapq

    def __init__(self):
        self.stack = []
        self.min_val = 2**31

    def push(self, val: int) -> None:
        self.min_val = min(self.min_val, val)
        self.stack.append((val, self.min_val))

    def pop(self) -> None:
        self.stack.pop()
        self.min_val = self.stack[-1][1] if self.stack else 2**31

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

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

# 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()