# 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:

• -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.
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 class MinStack: def __init__(self): self.data = [] self.min_stack = [] def push(self, val: int) -> None: self.data.append(val) if not self.min_stack or val < self.min_stack[-1][0]: self.min_stack.append([val, 1]) elif val == self.min_stack[-1][0]: self.min_stack[-1][1] += 1 def pop(self) -> None: if self.min_stack[-1][0] == self.data[-1]: self.min_stack[-1][1] -= 1 if self.min_stack[-1][1] == 0: self.min_stack.pop() self.data.pop() def top(self) -> int: return self.data[-1] def getMin(self) -> int: return self.min_stack[-1][0] # Your MinStack object will be instantiated and called as such: # obj = MinStack() # obj.push(val) # obj.pop() # param_3 = obj.top() # param_4 = obj.getMin()