stack containing min function (js)

Define the data structure of the stack, please implement a min function in this type that can get the smallest element of the stack. In this stack, the time complexity of calling min, push and pop are all O(1).

Example:

MinStack minStack = new MinStack();minStack.push(-2);minStack.push(0);minStack.push(-3);minStack.min(); --> return -3.minStack.pop();minStack.top(); --> return 0.minStack.min(); --> return -2.

Thinking Analysis:

In order to get the smallest element in the stack, it is easy to think of using a variable to store this element, but the question is, after this element goes out, what about the next smallest element?To solve this problem, you have to use an array to store the elements of the first, second, third, and so on. Define two arrays. The main array arr1 is used to pop and push the stack, and the secondary array arr2Used to store small elements.Here comes the point,

When pushing into the stack, if the element is less than or equal to the current minimum element, then the element should also be placed in arr2;

When popping the stack, if the element is the current smallest element, the element at the end of arr2 will also be popped out together;

As a result, the elements stored in arr2 get smaller toward the end, and the last element at the end of each read is the smallest element.

Code implementation:

var MinStack = function() {this.arr1 = []this.arr2 = []};/*** @param {number} x* @return {void}*/MinStack.prototype.push = function(x) {this.arr1.push(x)if (!this.arr2.length || (this.arr2.length && x <= this.arr2[this.arr2.length - 1])) {this.arr2.push(x)}};/*** @return {void}*/MinStack.prototype.pop = function() {const e = this.arr1.pop()if (this.arr2[this.arr2.length - 1] === e) {this.arr2.pop()}return e};/*** @return {number}*/MinStack.prototype.top = function() {return this.arr1[this.arr1.length - 1]};/*** @return {number}*/MinStack.prototype.min = function() {return this.arr2[this.arr2.length - 1]};