Recursion Assistance

[lizsaunderscoaching]

Hello,

Can someone please explain the -1 in recursion? I understand .length-1 but when -1 is added to these formulas I don’t get it. Thank you!

function rangeOfNumbers(startNum, endNum) {
if (endNum - startNum === 0){
return [startNum];
} else {
var numbers = rangeOfNumbers(startNum, endNum - 1);
numbers.push(endNum);
return numbers;
}
};

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**Challenge:**  Use Recursion to Create a Range of Numbers

**Link to the challenge:**
https://www.freecodecamp.org/learn/javascript-algorithms-and-data-structures/basic-javascript/use-recursion-to-create-a-range-of-numbers

[snowmonkey]

@lizsaunderscoaching

function rangeOfNumbers(startNum, endNum) {
  if (endNum - startNum === 0){
    return [startNum];
  } else {
    var numbers = rangeOfNumbers(startNum, endNum - 1);
    numbers.push(endNum);
    return numbers;
  }
};

Sure, let’s see if we can dive into what this is doing. A few key parts to recursion we need to define, to start:

• What is our exit condition? At what point have we called our function deep enough, and can begin returning back up the chain?
• What is our recursive behavior? What do we do, each time we call the function?
• How are we approaching that exit condition? Somehow, we need to be changing each call to our function, to approach our exit.

In this case, our exit condition is if(endNum-startNum===0). so at each iteration, we need to either increment startNum or decrement endNum. They need to eventually be the same number, otherwise we will be looping forever. So our exit condition sounds like “when startNum === endNum, we can start returning everything.”

Now, what’s our recursive behavior? What do we do each time? We take whatever is stored in the var numbers recursive call, we push a value onto it, and we return that numbers variable.

And how are we approaching that exit condition (remember, we want startNum===endNum eventually)? By reducing endNum on each iteration, each time we call rangeOfNumbers(...), we have reduced the gap between the two by one.
At our innermost call, startNum===endNum, and we begin returning.

So let’s unpack what that’s actually doing.

We call our original function, say rangeOfNumbers(1, 4). That checks if 1===4 and… nope. So it goes to the else side, and says var numbers = rangeOfNumbers(1, 3). Again with the exit check, and again, no exit.

Eventually, we get to var numbers = rangeOfNumbers(1, 1). At this point, we check our exit condition, is 1===1? It is indeed. So THIS time, we take the if branch. We return [1], which is passed to the next function out.

The next function out was the call to rangeOfNumbers(1,2), and our returned [1] is being stored in the var numbers - so now var numbers=[1]. At that point, we numbers.push(2), as endNum===2 here, and then we return [1,2]` to the next function out.

That gets passed to the rangeOfNumbers(1,3)'s else branch, where we were assigning something to var numbers = .... In this case, we’ve set numbers=[1,2], as that was the returned value. And we’re pushing onto that, and returning [1,2,3].

And we do that until we reach our ultimate final call, where we called our function (not where it called itself). And it returns to us [1,2,3,4].