Find minimum of the current stack in O(1) time.

You have to give a data structure to implement push & pop (as like a simple satck ) in O(1) time &
find minimum of the current stack also in O(1) time.
Solution :

push(x)
{
if(stack.isEmpty())
{
stack.push(x);
stack.push(x);
}
else
{
if(x > stack.top)
{
min=stack.top;
stack.push(x);
stack.push(min);
}
else
{
stack.push(x);
stack.push(x);
}
}
}

int pop()
{
stack.pop();
return (stack.pop());

}


int min()
{
return stack.top;
}