# Sum All Odd Fibonacci Numbers

## Problem Explanation

You will need to gather all the **Fibonacci** numbers and then check for the odd ones. Once you get the odd ones then you will add them all. The last number should be the number given as a parameter if it actually happens to be an off Fibonacci number.

#### Relevant Links

## Hints

### Hint 1

To get the next number of the series, you need to add the current one to the previous and that will give you the next one.

### Hint 2

To check if a number is even all you have to check is if `number % 2 == 0`

.

### Hint 3

As you get the next odd one, don’t forget to add it to a global variable that can be returned at the end. `result += currNumber;`

will do the trick.

## Solutions

## Solution 1 (Click to Show/Hide)

```
function sumFibs(num) {
let prevNumber = 0;
let currNumber = 1;
let result = 0;
while (currNumber <= num) {
if (currNumber % 2 !== 0) {
result += currNumber;
}
currNumber += prevNumber;
prevNumber = currNumber - prevNumber;
}
return result;
}
// test here
sumFibs(4);
```

#### Code Explanation

- Create a variable to keep record of the current and previous numbers along with the result that will be returned.
- Use a while loop to make sure we do not go over the number given as parameter.
- We use the modulo operand to check if the current number is odd or even. If it is odd, add it to the result.
- Complete the Fibonacci circle by rotating getting the next number and swapping values after.
- Return the result.

#### Relevant Links

## Solution 2 (Click to Show/Hide)

```
function sumFibs(num) {
// Perform checks for the validity of the input
if (num <= 0) return 0;
// Create an array of fib numbers till num
const arrFib = [1, 1];
let nextFib = 0;
// We put the new Fibonacci numbers to the front so we
// don't need to calculate the length of the array on each
// iteration
while ((nextFib = arrFib[0] + arrFib[1]) <= num) {
arrFib.unshift(nextFib);
}
// We filter the array to get the odd numbers and reduce them to get their sum.
return arrFib.filter(x => x % 2 != 0).reduce((a, b) => a + b);
}
// test here
sumFibs(4);
```

#### Code Explanation

- Create an array of fibonacci numbers till
**num**. - Use
`filter()`

method to filter out even numbers. - Use
`reduce()`

method to sum the remaining (odd) values. - Return the sum.

Note that this solution will be slower than Solution 1, as dynamically creating an array is rather slow, especially in JavaScript.