LeetCode 155 - Min Stack Problem

Problem Statement

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

  • push(x) -- Push element x onto stack.
  • pop() -- Removes the element on top of the stack.
  • top() -- Get the top element.
  • getMin() -- Retrieve 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:

  • Methods pop, top and getMin operations will always be called on non-empty stacks.

Solution:

class MinStack: def __init__(self): """ initialize your data structure here. """ self.stack = [] def push(self, x: int) -> None: self.stack.append(x) def pop(self) -> None: self.stack.pop() def top(self) -> int: return self.stack[-1] def getMin(self) -> int: return min(self.stack) # 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()