Refactor a Loop in JavaScript to Use Recursion

Thank you for this useful course. I really would like to know that which IDE you've used in this course materials?

Hi Tyler, thanks for the course.

Just a FYI: I've been playing around with the final code of this lesson and figured out that final function doesn't work for all arrays. e.g. findSix for const items =[[1,2,3],[4,5,[6],[9]]] will return no because hasSix variable will be overridden in consecutive iteration. Don't know if it's the most efficient way, but I merely put if (hasSix === 'yes') return; in the beginning of forEach loop to prevent it from overriding.

@Dr.Emmett it's VS Code. The extension that shows results of console.log right in the editor window is Quokka.js.

''' const somewhere = [6,6,undefined,131212312, 123,123 , 132, '', null, [[8]]];

function findNumber(num, arr) { if(arr === num) return true; if(!arr || !Array.isArray(arr)) return false; for(let i=0; i < arr.length; i++) { if(arr[i] === num) return true; if(Array.isArray(arr[i])) isFound = findNumber(num, arr[i]); } return false; } console.time('test1'); console.log(findNumber(6, somewhere)); console.timeEnd('test1'); '''

Can someone please delete this and the above comment ?!. Thank you.

Good catch @Stanislav, that is a good way of doing it!

Thank you for this useful course. I really would like to know that which IDE you've used in this course materials?

I am using the pro version of https://quokkajs.com/

Thank you Tyler, it is wonderful course, I learn a lot

Thank you Tyler, it is wonderful course, I learn a lot

Hi,Tyler, if I have const items = [[1, 2, 3],[4, 5,[6]],7,[8,9],[10,6]], how can I find all of 6 ?

Hi,Tyler, if I have const items = [[1, 2, 3],[4, 5,[6]],7,[8,9],[10,6]], how can I find all of 6 ?

Adding a 6 in the first array wouldn't work (I think the recursive case wasn't properly handled). So I had to modify the code to this:

const items = [[1, 2, 3, [6]], [4, 5, 5]];

function findSix(i) {
  let hasSix = 'no!';
  i.forEach(a => {
    if (hasSix === 'yes!') return;

    if (a === 6) {
      hasSix = 'yes!';
    }

    if (Array.isArray(a)) {
      hasSix = findSix(a);
    }
  });
  return hasSix;
}

console.log(findSix(items)); //=> true

https://codesandbox.io/s/recursive-findsix-lo9tb

In a functional style (while returning a boolean), I would do this:

//recursive flatten deep
function flatten(array) {
  let flattend = [];
  !(function flat(array) {
    array.forEach(function(el) {
      if (Array.isArray(el)) flat(el);
      else flattend.push(el);
    });
  })(array);
  return flattend;
}

const findSix = arr => flatten(arr).some(equals(6));

The poinfree version with ramda and crocks would look like this:

const findSix = composeB(any(equals(6)), flatten);

The video example is missing the base case of hasSix === yes!. Otherwise great tutorial