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155. Min Stack

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

class MinStack:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.stack = []
        self.min_stack = []

    def push(self, val: int) -> None:
        """
        Push element val onto stack.
        """
        self.stack.append(val)
        # If min_stack is empty or val is less than or equal to the current minimum, push it onto min_stack
        if not self.min_stack or val <= self.min_stack[-1]:
            self.min_stack.append(val)

    def pop(self) -> None:
        """
        Removes the element on top of the stack.
        """
        val = self.stack.pop()
        # If the popped value is the current minimum, pop it from min_stack as well
        if val == self.min_stack[-1]:
            self.min_stack.pop()

    def top(self) -> int:
        """
        Get the top element.
        """
        return self.stack[-1]

    def getMin(self) -> int:
        """
        Retrieve the minimum element in the stack.
        """
        return self.min_stack[-1]

Thoughts

The problem seemed difficult at first. Had to look at the hint to use two stacks ( one for storing the min value). Time complexity is specified in the question and the space complexity is O(n) for the stack.