Min Stack medium

Problem Statement

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 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

Example 2:

Input
["MinStack","push","push","getMin","pop","getMin"]
[[], [2],[0],[],[],[]]
Output
[null,null,null,0,null,2]
Explanation
MinStack minStack = new MinStack();
minStack.push(2);
minStack.push(0);
minStack.getMin(); // return 0
minStack.pop();
minStack.getMin(); // return 2

Steps:

1. Data Structure: Use two stacks: stack to store the elements and min_stack to store the minimum element at each stage.
2. push(val): Push val onto stack. If min_stack is empty or val is less than or equal to the top of min_stack, push val onto min_stack as well.
3. pop(): Pop the top element from stack. If the popped element is equal to the top of min_stack, pop from min_stack too.
4. top(): Return the top element of stack.
5. getMin(): Return the top element of min_stack.

Explanation:

The key idea is to maintain a second stack (min_stack) that always keeps track of the minimum element encountered so far. Whenever we push an element, we also check if it's the new minimum and push it onto min_stack. When we pop, we check if the popped element is the current minimum; if it is, we pop it from min_stack as well. This ensures min_stack always contains the current minimum element at its top.

Code:

class MinStack:
    def __init__(self):
        self.stack = []
        self.min_stack = []

    def push(self, val: int) -> None:
        self.stack.append(val)
        if not self.min_stack or val <= self.min_stack[-1]:
            self.min_stack.append(val)

    def pop(self) -> None:
        if self.stack:
            if self.stack[-1] == self.min_stack[-1]:
                self.min_stack.pop()
            self.stack.pop()

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

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

Complexity:

• Time Complexity:
  • push(), pop(), top(), getMin(): O(1) All operations involve accessing the top of the stack, which is a constant-time operation.
• Space Complexity: O(N), where N is the number of elements in the stack. In the worst case, both stack and min_stack can store up to N elements.