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(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.

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`.

Idea:

Use two stacks to solve this problem

Solution:

``````/**
* initialize your data structure here.
*/
var MinStack = function() {
this.min = [];
this.stack = [];
};

/**
* @param {number} x
* @return {void}
*/
MinStack.prototype.push = function(x) {
this.stack.push(x);
let min = this.getMin();
if (min === undefined || min >= x) {
this.min.push(x);
}
};

/**
* @return {void}
*/
MinStack.prototype.pop = function() {
let val = this.stack.pop();
let min = this.getMin();
if (val === min) this.min.pop();
};

/**
* @return {number}
*/
MinStack.prototype.top = function() {
return this.stack[this.stack.length - 1];
};

/**
* @return {number}
*/
MinStack.prototype.getMin = function() {
return this.min[this.min.length - 1];
};

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