Counter | #2620 | LeetCode Solution

Problem : Counter | #2620 | LeetCode

Given an integer n, return a counter function. This counter function initially returns n and then returns 1 more than the previous value every subsequent time it is called (n, n + 1, n + 2, etc).

Example 1:

Input: 

n = 10 

["call","call","call"]

Output: [10,11,12]

Explanation: 

counter() = 10 // The first time counter() is called, it returns n.

counter() = 11 // Returns 1 more than the previous time.

counter() = 12 // Returns 1 more than the previous time.

Example 2:

Input: 

n = -2

["call","call","call","call","call"]

Output: [-2,-1,0,1,2]

Explanation: counter() initially returns -2. Then increases after each sebsequent call.

Solution:

    /**

     * @param {number} n

     * @return {Function} counter

     */

    var createCounter = function(n) {

        return function() {

            return n++

        };

    };

    /**

     * const counter = createCounter(10)

     * counter() // 10

     * counter() // 11

     * counter() // 12

     */

Explanation:

The provided code defines a function `createCounter` that takes an integer `n` as an argument and returns another function. This returned function acts as a counter that starts at `n` and increments by 1 each time it's called. 

Let's break down the code and provide an explanation:

    var createCounter = function(n) {

        return function() {

            return n++;

        };

    };

1. `createCounter` is a function that takes an integer `n` as its parameter.

2. Inside `createCounter`, it returns another function (an anonymous function) that captures the value of `n`.

3. This inner function has no parameters and when called, it returns the current value of `n` and then increments `n` by 1 using the `n++` expression.