### LeetCode Challenge #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`.
##### Video Solution
###### Java Solution
```				```
class MinStack {
private Stack<Integer> st;
private Stack<Integer> min;
public MinStack() {
st = new Stack<>();
min = new Stack<>();
}

public void push(int val) {
if(st.size()==0 || min.peek()>=val){
min.push(val);
}

st.push(val);
}

public void pop() {
int ele1 = st.pop();
int ele2 = min.peek();

if(ele1==ele2){
min.pop();
}

}

public int top() {
return st.peek();
}

public int getMin() {
return min.peek();
}
}

/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(val);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
```
```