# 155-min-stack


Try it on <a href='https://leetcode.com/problems/155-min-stack'>leetcode</a>

## Description
<div class="description">
<div><p>Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.</p>

<p>Implement the <code>MinStack</code> class:</p>

<ul>
	<li><code>MinStack()</code> initializes the stack object.</li>
	<li><code>void push(int val)</code> pushes the element <code>val</code> onto the stack.</li>
	<li><code>void pop()</code> removes the element on the top of the stack.</li>
	<li><code>int top()</code> gets the top element of the stack.</li>
	<li><code>int getMin()</code> retrieves the minimum element in the stack.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre><strong>Input</strong>
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

<strong>Output</strong>
[null,null,null,null,-3,null,0,-2]

<strong>Explanation</strong>
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
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>-2<sup>31</sup> &lt;= val &lt;= 2<sup>31</sup> - 1</code></li>
	<li>Methods <code>pop</code>, <code>top</code> and <code>getMin</code> operations will always be called on <strong>non-empty</strong> stacks.</li>
	<li>At most <code>3 * 10<sup>4</sup></code> calls will be made to <code>push</code>, <code>pop</code>, <code>top</code>, and <code>getMin</code>.</li>
</ul>
</div>
</div>

## Solution(Python)
```Python
class MinStack:
    def __init__(self):
        self.stack = []
        self.min = None

    def push(self, val: int) -> None:
        if not self.stack:
            self.min = val
            self.stack.append(val)
        elif val < self.min:
            y = (2 * val) - self.min
            self.min = val
            self.stack.append(y)
        else:
            self.stack.append(val)

    def pop(self) -> None:

        if self.stack:
            y = self.stack[-1]
            if y < self.min:
                self.min = (2 * self.min) - y
            self.stack.pop()

    def top(self) -> int:
        if self.stack[-1] < self.min:
            return self.min
        else:
            return self.stack[-1]

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


# 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()

```