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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.
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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.
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Hint: 2
To check if a number is even all you have to check is if number % 2 == 0
.
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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.
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Spoiler Alert!
Solution ahead!
Basic Code Solution:
function sumFibs(num) {
var prevNumber = 0;
var currNumber = 1;
var 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 even, add it to the result.
- Complete the Fibonacci circle by rotating getting the next number and swapping values after.
- Return the result.
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Intermediate Code Solution:
function sumFibs(num) {
// create an array of fib numbers till num
var arrFib = [1];
for (var i = 1; i <=num;) {
arrFib.push(i);
i = arrFib[arrFib.length - 1] + arrFib[arrFib.length - 2];
}
// return the sum of odd numbers from the array
var res = arrFib.reduce(function(prev, curr) {
if (curr%2 !== 0) return prev + curr;
else return prev;
});
return res;
}
// test here
sumFibs(4);
Code Explanation:
- Create an array of fibonacci numbers till num.
- Use
reduce()
method to find the sum of odd members of array. - Return the sum.
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See
Wiki Challenge Solution Template
for reference.