Fibonacci Sums using Array.filter and Array.reduce

I would like to propose another solution to the Intermediate Algorithm Scripting: Sum All Odd Fibonacci Numbers.

This solution uses Array.push, Array.filter, and Array.reduce to get the needed values.

Solution

const sumFibs = (num) => {
  const fibonacci = [0,1,1];
  while ( fibonacci[fibonacci.length - 1] < num ) {
    fibonacci.push( fibonacci[fibonacci.length - 2] + fibonacci[fibonacci.length - 1] );
  }
  return fibonacci.filter(fibNum => fibNum % 2 === 1 && fibNum <= num).reduce(((sum,curVal) => sum + curVal), 0);
}

Explanation

  1. Create an array which holds the first three values of the Fibonacci sequence: 0, 1, and 1.
  2. Use a while loop to check the condition: is the last number in the Fibonacci array less than the num argument?
  3. If not, then add a new number to the array, which is the sum of the second-to-last and the last number in the array.
  4. Once the while loop is completed, Return the combined result of filtering and reducing the Fibonacci array: filter by odd numbers that are less than or equal to num, then reduce the array to the sum of all remaining values.

Tests

/**
 * Tests for sumFibs
 */
let testArr = [];

testArr.push(
  [1000, 1785],
  [4000000, 4613732],
  [4, 5],
  [75024, 60696],
  [75025, 135721],
  [0, 0],
  [1, 2],
  [2, 2],
  [3, 5],
  [undefined, 0],
)

testArr.forEach(arr => {
  console.log(sumFibs(arr[0]) === arr[1] ? `PASS: The fib sum of ${arr[0]} == ${sumFibs(arr[0])} == ${arr[1]}` : `FAIL; ${arr[0]} got ${sumFibs(arr[0])}, expected ${arr[1]}`);
})

I’ll let others chime in, but I usually discourage solutions that generate a Fibonacci array because they are very inefficient.