Issue with Nesting For Loops

JavaScript
#1

Tell us what’s happening:
Hi everyone. English is not my native language , I’m sorry if there are some errors in spelling or grammar.
I got two problem with this lecture.

The first:
I know the code works but I don’t understand why the result is 5040.
Can someone show me the math behind it?

The other problem is the following:
I know " for (var i = 0; i < arr.length; i++) " checks the number of arrays.
I know " for (var j = 0; j < arr[i].length; j++ " checks the numbers in the arrays.
But it only checks the length of the numbers.
The actual array is passed into the function as argument as I understand it.
How does the function calculate with the arguments when I only check the length of an array?
I’m sorry this all might sound weird but I can’t explain it any better.
Maybe some one can explain this whole function to me at a very basic level?

Your code so far


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;
}

multiplyAll([[1,2],[3,4],[5,6,7]]);

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Challenge: Nesting For Loops

Link to the challenge:

#2

Hi and welcome to the forum!

This function takes in an array of arrays and finds the product of all numbers in these arrays.

Your outer loop basically says ‘go through all of the arrays in our big array’.

The inner loop says ‘go through all of the numbers in each sub array’.

With each number in each sub array, we multiply that number by the current value of the product.

The length there means ‘how many subarrays’ and ‘how many numbers’. We aren’t taking the length of any numbers themselves. That would generate an error.

1 Like
#3

The inner loop is accessing the array elements, the math is done here product = product * arr[i][j];

Try logging out product * arr[i][j] in the inner loop (you have to add braces {} to the inner loop)

2 Likes
#4

Hey @Laviniss!

Nested loops can be really confusing at first. So it just takes times to get used to how they work. But you will have plenty of chances to work with them throughout the FCC curriculum.

I think the key thing to remember is that product doesn’t’ just stay at 1. It is constantly being updated.

How the math works

  • when i=0
  • product = 1 * 1(arr[0][0]), the result for product is still 1
  • product = 1 * 2(arr[0][1]), the result for product is now updated to 2
  • when i=1
  • product = 2 * 3(arr[1][0]), product is now updated to 6
  • product = 6 * 4(arr[1][1]), the result for product is now updated to 24
  • when i=2
  • product = 24 * 5(arr[2][0]), product is now updated to 120
  • product = 120 * 6(arr[2][1]), the result for product is now updated to 720
  • product = 720 * 7(arr[2][2]), the result for product is now updated to 5040

Hope that makes sense!

1 Like
#5

Thanks a lot for the great answers!
I really appreciate it.
All answers were helpful and I could finally understand the math behind the multiplication :slight_smile:

1 Like
#6

Hi jwilkins,

I have really tried to make sense of this but just can’t. For example, when i-0, why is the product 2 here?

  • product = 1 * 2(arr[0][1]), the result for product is now updated to 2

Arghhhh

#7

Hi @mtahir2020 !

When i=0 we targeting this first array here [1,2]
If we want to access each number in that array that is where the inner loop j comes in.

So arr[0] refers to this first array [1,2]
and arr[0][1] refers to the number 2.
Remember that arrays are zero based index.

So 1*2 = 2

or

1 * arr[0][1] = 2

Make sense?

1 Like
#8

Got it - thank you!!!