155. Min Stack — LeetCode(Python)

I got you!

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.

Solution:

class MinStack:
def __init__(self):
        self.stack = []
        self.min_val = []
        
    def push(self, val: int) -> None:
        self.stack += [val]
        if self.min_val:
            self.min_val += [min(val, self.min_val[-1])]
        else:
            self.min_val += [val]
def pop(self) -> None:
        self.stack = self.stack[:-1]
        self.min_val = self.min_val[:-1]
def top(self) -> int:
        return self.stack[-1]
def getMin(self) -> int:
        return self.min_val[-1]

Explanation —

This is a simple implementation of the Stack Abstract Data Type (ADT).

While implementing basic functionalities of the stack, we also keep a min_val stack that stores the minimum value in the stack at a given point in time.

Everytime we push a new element onto the stack, we update our min_val.

And everytime we pop an element from the stack, we do the same for our min_val stack as well.

Time and Space Complexity —

All member functions in our class perform constant time operations.

And, since we are using a different stack to keep track of the minimum value in our stack, our code requires a linear amount of extra space.

Thus, if n elements are going to be pushed onto the stack,

Time Complexity: O(1)

Space Complexity: O(n)

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