I have been learning basic javascript on freecodecamp.com, and this is the second problem that I encountered a bit difficult of understanding. It wanted me to modify function multiplyAll so that it multiplies the product variable by each number in the sub-arrays of arr.

function multiplyAll(arr) {
  var product = 1;
  // Only change code below this line

  // Only change code above this line
  return product;
}

// Modify values below to test your code
multiplyAll([[1,2],[3,4],[5,6,7]]);

The challenge is:

Modify function multiplyAll so that it multiplies the product variable by each number in the sub-arrays of arr

• multiplyAll([[1],[2],[3]]); should return 6
• multiplyAll([[1,2],[3,4],[5,6,7]]); should return 5040
• multiplyAll([[5,1],[0.2, 4, 0.5],[3, 9]]); should return 54

I wasn’t sure how to manipulate elements inside arrays within an array. I decided to research a bit about array. There are few different types of arrays, and are exemplified below:

homogeneous arrays: stores single type of data (i.e. string, or int, or boolean) in an array

var array = ["Matthew", "John", "Luke"];
var array = [27, 30, 35];
var array = [true, false, true];

heterogeneous array: stores mixed data types in an array

var array = ["Matthew", 27, true];

multidimensional array: an array of arrays, or array-ception like inception

var array = [
    ["Matthew", "27"],
    ["Simon", "24"],
    ["Luke", "30"]
];

jagged array like multidimensional array but not in uniformed set

var array = [
    ["Matthew", "27", "Developer"],
    ["Simon", "24"],
    ["Luke"]
];

The problem I try to solve is multidimensional array, or 2D array since there is array in an array (2 arrays). Meaning two loops needed. I decided to play with the freecodecamp’s provided example using repl.it a better console to practice with.

var arr = [
  [1,2], [3,4], [5,6]
];
for (var i=0; i < arr.length; i++) {
  for (var j=0; j < arr[i].length; j++) {
    console.log(arr[i][j]);
  }
}

// 1, 2, 3, 4, 5, 6
// => undefined

What happens if I used console.log(arr[i]); without j? It will print out:

[ 1, 2 ]
[ 1, 2 ]
[ 3, 4 ]
[ 3, 4 ]
[ 5, 6 ]
[ 5, 6 ]

The invisible action is that j loop went through each element in the arrays, but arr[i] prints out its “elements” which are arrays therefore its element was printed twice because of j loop. If I take out for (var j=0; j < arr[i].length; j++), and then it will print arr[i]’s elements:

[1, 2]
[2, 4]
[5, 6]

Now what happens if I used console.log(arr[j]); instead? It comes out:

[ 1, 2 ]
[ 3, 4 ]
[ 1, 2 ]
[ 3, 4 ]
[ 1, 2 ]
[ 3, 4 ]

Didn’t even touch third array within primary array. Let’s find out why by this console.log(arr[i], i); :

[ 1, 2 ] 0
[ 1, 2 ] 0
[ 3, 4 ] 1
[ 3, 4 ] 1
[ 5, 6 ] 2
[ 5, 6 ] 2

Can deduce that there is a difference between [i] and i. [i] prints elements (secondary arrays in this case) and i prints its elements’ indexes. Again, test with console.log(arr[i], j); :

[ 1, 2 ] 0
[ 1, 2 ] 1
[ 3, 4 ] 0
[ 3, 4 ] 1
[ 5, 6 ] 0
[ 5, 6 ] 1

[i] prints all secondary arrays. We can see that j prints index of each element inside secondary arrays. [i] loop went every secondary arrays and j loop went through elements in each secondary array separately. Let’s try console.log(arr[j], i); :

[ 1, 2 ] 0
[ 3, 4 ] 0
[ 1, 2 ] 1
[ 3, 4 ] 1
[ 1, 2 ] 2
[ 3, 4 ] 2

[j] didn’t cover third array’s elements and i has a total of three indexes even though there isn’t a third array. Why? Test this one console.log(arr[j], j); :

[ 1, 2 ] 0
[ 3, 4 ] 1
[ 1, 2 ] 0
[ 3, 4 ] 1
[ 1, 2 ] 0
[ 3, 4 ] 1

Almost similar result and from these results, it might be obvious because of this j < arr[i].length; which allowed j loop to go only twice since i is the total of three indexes. i loop goes through indexes of arrays and j loop goes through indexes of elements.

Lastly… test console.log(arr[j], [i]);:

1
2
3
4
undefined
undefined

arr[j] only has two indexes, couldn’t go through third array. [i] fetches the elements and print them out. Letters are not really the culprits. It is the matter of loops. First [] loop goes through secondary arrays, and second [] loop prints secondary arrays’ elements out.

So what happened with arr[i][j]?

1) [] is an array symbol and focuses on elements rather than indexes.
2) [i] loop goes through primary array’s elements which are secondary arrays.
3) [j] loop goes through and print secondary arrays’ elements.

Back to the challenge and the solution lies below:

function multiplyAll(arr) {
  var product = 1;

  // Only change code below this line
  for (var i = 0; i < arr.length; i++) {
    for (var j = 0; j < arr[i].length; j++) {
      product = product * arr[i][j];
    }
  }
  // Only change code above this line
  return product;
}

// Modify values below to test your code
multiplyAll([[1,2],[3,4],[5,6,7]]);

Now we know that arr[i][j] fetches all secondary arrays’ elements by using two loops.

• Create product that equals to 1.
• product follows j loop in i loop, and arr[i][j] fetches an element at a time.
• product multiples with arr[i][j] element then becomes the “new” product.
• j loops again and “new” product multiples with the next fetched element.
• j loop repeats until there is not any element left and leaving the final product.