2620. Counter

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).

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My solution:

var createCounter = function(n) {
    n -= 1
    return function() {
        return n += 1
    };
};

In my solution, I return a function that returns n += 1 initially based off the captured value of n. With how I return the function, it was always 1 off each call, so I needed to add n -= 1 to initialize n at 1 lower.

With an input of: n = 10, ["call","call","call"]

The expected output would be: [10,11,12]

But I was getting [11, 12, 13] before adding the n -= 1. I thought there must be a better way so I looked to other solutions to this problem and discovered something I somehow never knew before about a very familiar operator.

The ++ Operator

When using ++ following the variable instead of +=1, the initial returned value is the base number prior to interation. This would allow us to iterate correctly while returning n initially.

Example of how ++ works.

let a = 1
let b = 1
let c = 1

a++ // 1
a++ // 2

b+=1 // 2
b+=1 // 3

// putting the ++ before the variable iterates first, acting similar to += 1
++c // 2
++c // 3

With this in mind, we can clean up our function a little bit. Taking into account how ++ operates, we can now submit this as our solution:

var createCounter = function(n) {
    return function() {
        return n++
    };
};