diff --git a/guide/english/certifications/javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-odd-fibonacci-numbers/index.md b/guide/english/certifications/javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-odd-fibonacci-numbers/index.md
index b53a361a56..e8962cacbf 100644
--- a/guide/english/certifications/javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-odd-fibonacci-numbers/index.md
+++ b/guide/english/certifications/javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-odd-fibonacci-numbers/index.md
@@ -80,7 +80,7 @@ As you get the next odd one, don't forget to add it to a global variable that ca
// 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
@@ -88,11 +88,8 @@ As you get the next odd one, don't forget to add it to a global variable that ca
arrFib.unshift(nextFib);
}
- // Sum only the odd numbers and return the value
- // First, reverse the array to avoid starting acc with the first/greater number when it's even
- return arrFib.reverse().reduce((acc, curr) => {
- return acc + curr * (curr % 2);
- }, 0);
+ // 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
@@ -103,15 +100,19 @@ As you get the next odd one, don't forget to add it to a global variable that ca
### Code Explanation:
* Create an array of fibonacci numbers till **num**.
-* Use `reduce()` method to find the sum of odd members of array.
+* Use `filter()` method to filter out even numbers.
+* Use `reduce()` method to sum the remaining (odd) values.
* Return the sum.
#### Relevant Links
* JS Array Prototype Push
* JS For Loops Explained
+* JS Array Prototype Filter
* JS Array Prototype Reduce
+
+
##  NOTES FOR CONTRIBUTIONS:
*  **DO NOT** add solutions that are similar to any existing solutions. If you think it is **_similar but better_**, then try to merge (or replace) the existing similar solution.