---
title: '155. Min Stack'
description: Design a stack that supports push, pop, top, and retrieving the minimum element in constant time
sidebar:
label: 'Min Stack'
badge: 'Medium'
---
Stack
::::warning
You must implement a solution with O(1) time complexity for each function.
::::
### Example 1:
- Input: ``
- Output: ``
- Explanation: 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.
## Approach
```mermaid
flowchart TD
subgraph state["two parallel stacks — same depth, always"]
A["stack — the values"]
B["minStack — the minimum as of that depth"]
end
state --> P
subgraph P["push(value)"]
P1["stack.append(value)"] --> P2["minStack.append(min(value, minStack[-1]))"]
end
state --> O
subgraph O["pop()"]
O1["pop both stacks together"]
end
state --> R
subgraph R["top() / getMin() — O(1)"]
R1(["stack[-1] / minStack[-1]"])
end
```
## Solution
```py
class MinStack:
def __init__(self):
self.stack = []
self.minStack = []
def push(self, value: int) -> None:
self.stack.append(value)
value = min(value, self.minStack[-1] if self.minStack else value)
self.minStack.append(value)
def pop(self) -> None:
self.stack.pop()
self.minStack.pop()
def top(self) -> int:
return self.stack[-1]
def getMin(self) -> int:
return self.minStack[-1]
# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(value)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()
```
## Explanation