Leetcode - Min Stack Solution
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Leetcode - Min Stack Solution

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(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.

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 in python

class Node:
    def __init__(self, value):
        self.value = value
        self.min = None
    
class MinStack:

    def __init__(self):
        self.stack = []
        self.min = float('inf')

    def push(self, x: int) -> None:
        node = Node(x) 
        node.min = self.min
        self.stack.append(node)
        self.min = min(self.min,x)

    def pop(self) -> None:
        node = self.stack.pop()
        self.min = node.min

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

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


# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()

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