Problem with Intermediate Algorithm Scripting: Sum All Odd Fibonacci Numbers

JavaScript
1

“Given a positive integer num, return the sum of all odd Fibonacci numbers that are less than or equal to num.
The first two numbers in the Fibonacci sequence are 1 and 1. Every additional number in the sequence is the sum of the two previous numbers. The first six numbers of the Fibonacci sequence are 1, 1, 2, 3, 5 and 8.
For example, sumFibs(10) should return 10 because all odd Fibonacci numbers less than or equal to 10 are 1, 1, 3, and 5.
Remember to use Read-Search-Ask if you get stuck. Try to pair program. Write your own code.”

So, I came up with a rather long solution for 5 out of the 6 tests but I cannot make it work when the test runs with a value of 4.000.000.
I could just look it up and use somebody else’s code but now I wonder why mine doesn’t work for 4.000.000 in particular!

Here is my code:

function sumFibs(num) {
  let fullArray = [];
  let fibonacciArray = [0, 1, 1];
  let oddFibonacci = [];
  let fCounter = 2;
  let summed;

  for (let i = 1; i <= num; i++) {
    fullArray.push(i);
  }

  for (let j = 1; j <= fullArray.length; j++) {
    if (fullArray[j] === fibonacciArray[fCounter] + fibonacciArray[fCounter - 1]) {
      fibonacciArray.push(fullArray[j])
      fCounter ++;
    }     
  }

  oddFibonacci = fibonacciArray.filter((el) => {
    return el % 2 != 0;
  });

  summed = oddFibonacci.reduce((accumulator, currentValue) => accumulator + currentValue);
  
  console.log(summed)
}

sumFibs(4000000);

Thank you in advance.

2

without looking at your code, I will guess (as others here have reported) that your solution is time-intensive and takes too long to solve the last test-case which is why it hits a ‘timeout’ that FCC sets up to catch infinite-loops. (even though your code is not infinitely running)

there are workarounds that may help you but you’ll have to search the forum for them as I don’t recall the exact method. (some of the posts may also have suggestions on changing the algorithm to make it run faster)

3

Very interesting, thanks for your answer.
I planned to put the remainder operator in the conditional statement instead of using the filter afterwards, which would have made the code shorter and the runtime faster, but for some reason it wasn’t working. So since we’re here, let’s see if someone can tell me why:

for (let j = 1; j <= fullArray.length; j++) {
    if (fullArray[j] === fibonacciArray[fCounter] + fibonacciArray[fCounter - 1] && fullArray[j] % 2 != 0) {
      fibonacciArray.push(fullArray[j])
      fCounter ++;
    }     
  }
4

In fact, that would have also saved me the fullArray and I would have pushed the odd fibonaccies directly to the fibonacciArray.