You have never declared fib to be an array. Use var fib = []; to solve this.
Also, you're never modifying the y variable, neither using it.
The code below makes more sense, plus, it doesn't create unused variables:
var i;
var fib = [0, 1]; // Initialize array!
for (i = 2; i <= 10; i++) {
// Next fibonacci number = previous + one before previous
// Translated to JavaScript:
fib[i] = fib[i - 2] + fib[i - 1];
console.log(fib[i]);
}
Answer from Rob W on Stack OverflowYou have never declared fib to be an array. Use var fib = []; to solve this.
Also, you're never modifying the y variable, neither using it.
The code below makes more sense, plus, it doesn't create unused variables:
var i;
var fib = [0, 1]; // Initialize array!
for (i = 2; i <= 10; i++) {
// Next fibonacci number = previous + one before previous
// Translated to JavaScript:
fib[i] = fib[i - 2] + fib[i - 1];
console.log(fib[i]);
}
According to the Interview Cake question, the sequence goes 0,1,1,2,3,5,8,13,21. If this is the case, this solution works and is recursive without the use of arrays.
function fibonacci(n) {
return n < 1 ? 0
: n <= 2 ? 1
: fibonacci(n - 1) + fibonacci(n - 2)
}
console.log(fibonacci(4))
Think of it like this.
fibonacci(4) .--------> 2 + 1 = 3
| / |
'--> fibonacci(3) + fibonacci(2)
| ^
| '----------- 2 = 1 + 1 <----------.
1st step -> | ^ |
| | |
'----> fibonacci(2) -' + fibonacci(1)-'
Take note, this solution is not very efficient though.
You were not using the previous two numbers that are already in the array to > generate the new fibonacci number to be inserted into the array.
https://www.mathsisfun.com/numbers/fibonacci-sequence.html
Here I have used the sum of result[i-2] and result[i-1] to generate the new fibonacci number and pushed it into the array.
Also to generate n number of terms you need the condition to be i < n and not i <= n.
function fib(n) {
const result = [0, 1];
for (var i = 2; i < n; i++) {
result.push(result[i-2] + result[i-1]);
}
return result; // or result[n-1] if you want to get the nth term
}
console.log(fib(8));
Return result[n-1] if you want to get the nth term.
My solution for Fibonacci series:
const fibonacci = n =>
[...Array(n)].reduce(
(acc, val, i) => acc.concat(i > 1 ? acc[i - 1] + acc[i - 2] : i),
[]
)
You have never declared fib to be an array. Use var fib = []; to solve this.
Also, you're never modifying the y variable, neither using it.
The code below makes more sense, plus, it doesn't create unused variables:
var i;
var fib = [0, 1]; // Initialize array!
for (i = 2; i <= 10; i++) {
// Next fibonacci number = previous + one before previous
// Translated to JavaScript:
fib[i] = fib[i - 2] + fib[i - 1];
console.log(fib[i]);
}
According to the Interview Cake question, the sequence goes 0,1,1,2,3,5,8,13,21. If this is the case, this solution works and is recursive without the use of arrays.
function fibonacci(n) {
return n < 1 ? 0
: n <= 2 ? 1
: fibonacci(n - 1) + fibonacci(n - 2)
}
console.log(fibonacci(4))
Think of it like this.
fibonacci(4) .--------> 2 + 1 = 3
| / |
'--> fibonacci(3) + fibonacci(2)
| ^
| '----------- 2 = 1 + 1 <----------.
1st step -> | ^ |
| | |
'----> fibonacci(2) -' + fibonacci(1)-'
Take note, this solution is not very efficient though.