155. Min Stack

• python3

📎 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.

Plan Solution

Min 값을 스택에서 O(1) 시간으로 검색하는 법은 가장 윗 스택에 Min값을 함께 저장하는 것이다. 그러므로 값을 넣으면서 최소값과 현재값을 비교해 현재값, 최소값을 쌍으로 저장한다.

Code

class MinStack:

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

    def push(self, val: int) -> None:
        if not self.stack:
            self.stack.append((val,val))
        else:
            self.stack.append((val, min(val, self.stack[-1][1])))


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

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

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


# 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()

Result

• Time Complexity : O(1) each func
• Space Complexity : O(n) for stack