How to find nth Fibonacci number using Javascript with O(n) complexity

Question

Trying really hard to figure out how to solve this problem. The problem being finding nth number of Fibonacci with O(n) complexity using javascript.

I found a lot of great articles how to solve this using C++ or Python, but every time I try to implement the same logic I end up in a Maximum call stack size exceeded.

Example code in Python

MAX = 1000

# Create an array for memoization
f = [0] * MAX

# Returns n'th fuibonacci number using table f[]
def fib(n) :
        # Base cases
        if (n == 0) :
                return 0
        if (n == 1 or n == 2) :
                f[n] = 1
                return (f[n])

        # If fib(n) is already computed
        if (f[n]) :
                return f[n]

        if( n & 1) :
                k = (n + 1) // 2
        else : 
                k = n // 2

        # Applyting above formula [Note value n&1 is 1
        # if n is odd, else 0.
        if((n & 1) ) :
                f[n] = (fib(k) * fib(k) + fib(k-1) * fib(k-1))
        else :
                f[n] = (2*fib(k-1) + fib(k))*fib(k)

        return f[n]


// # Driver code
// n = 9
// print(fib(n))

Then trying to port this to Javascript

const MAX = 1000;
let f = Array(MAX).fill(0);
let k;

const fib = (n) => {

    if (n == 0) {
        return 0;
    }

    if (n == 1 || n == 2) {
        f[n] = 1;

        return f[n]
    }

    if (f[n]) {
        return f[n]
    }

    if (n & 1) {
        k = Math.floor(((n + 1) / 2))
    } else {
        k = Math.floor(n / 2)
    }

    if ((n & 1)) {
        f[n] = (fib(k) * fib(k) + fib(k-1) * fib(k-1))
    } else {
        f[n] = (2*fib(k-1) + fib(k))*fib(k)
    }

    return f[n]
}

console.log(fib(9))

That obviously doesn't work. In Javascript this ends up in an infinite loops. So how would you solve this using Javascript?

Thanks in advance

The same way you would do it by hand. Calculate the next number from the prvious two. No recursion needed. No storing of more than the prvious 2 values needed. — MrSmith42, Jan 08 '18 at 12:39
@iremlopsum Do you know that you can post answer to your own question? — user202729, Jan 08 '18 at 12:45
@iremlopsum: Your question isn't asking for the bug in your code, but your comment below suggests that's what you actually wanted. If you didn't want different solutions, why ask for them? — , Jan 08 '18 at 12:53

OmG · Answer 1 · 2018-01-08T12:54:49.377

3

you can iterate from bottom to top (like tail recursion):

var fib_tail = function(n){
        if(n == 0)
             return 0;
        if(n == 1 || n == 2)
             return 1;
        var prev_1 = 1, prev_2 = 1, current; 
        // O(n)
        for(var i = 3; i <= n; i++)
        {
            current = prev_1 + prev_2;
            prev_1 = prev_2;
            prev_2 = current;
        }
        return current;
 }
 console.log(fib_tail(1000))

edited Jan 08 '18 at 12:54

answered Jan 08 '18 at 12:38

OmG

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@rockstar I mean iterating from bottom to top. Anyhow, removed it. – OmG Jan 08 '18 at 12:50

Engineer · Accepted Answer · 2018-01-08T12:53:39.183

1

The problem is related to scope of the k variable. It must be inside of the function:

const fib = (n) => {
    let k;

You can find far more good implementations here list

DEMO

edited Jan 08 '18 at 12:53

answered Jan 08 '18 at 12:43

Engineer

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This the correct answer, thanks, I solved it in a different way initially that works. But you pointed out the correct problem with my code :) – iremlopsum Jan 08 '18 at 12:47
I have edited my answer to add a link, that can be helpfull – Engineer Jan 08 '18 at 12:54

Alok Deshwal · Answer 3 · 2020-01-19T07:37:57.990

1

fibonacci number in O(n) time and O(1) space complexity:

function fib(n) {
    let prev = 0, next =1;
    if(n < 0)
        throw 'not a valid value';
    if(n === prev || n === next)
        return n;
    while(n >= 2) {
        [prev, next] = [next, prev+next];
        n--;
    }
    return next;
}

edited Jan 19 '20 at 07:37

answered Jan 19 '20 at 07:29

Alok Deshwal

1,128
9
20

score 0 · Answer 4 · answered Jan 08 '18 at 12:43

Just use two variables and a loop that counts down the number provided.

function fib(n){
    let [a, b] = [0, 1];
    while (--n > 0) {
      [a, b] = [b, a+b];
    }
    return b;
}

console.log(fib(10));

WesleyAC · Answer 5 · 2021-04-30T03:47:28.843

Here's a simpler way to go about it, using either iterative or recursive methods:

function FibSmartRecursive(n, a = 0, b = 1) {
    return n > 1 ? FibSmartRecursive(n-1, b, a+b) : a;
}

function FibIterative(n) {
    if (n < 2) 
        return n;

    var a = 0, b = 1, c = 1;

    while (--n > 1) {
        a = b;
        b = c;
        c = a + b;
    }

    return c;
}

function FibMemoization(n, seenIt = {}) {//could use [] as well here
    if (n < 2) 
        return n;
    if (seenIt[n]) 
        return seenIt[n];

    return seenIt[n] = FibMemoization(n-1, seenIt) + FibMemoization(n-2, seenIt); 
}

console.log(FibMemoization(25)); //75025
console.log(FibIterative(25)); //75025
console.log(FibSmartRecursive(25)); //75025

PhiAgent · Answer 6 · 2021-08-05T18:48:43.733

0

You can solve this problem without recursion using loops, runtime O(n):

function nthFibo(n) {
    // Return the n-th number in the Fibonacci Sequence
    const fibSeq = [0, 1]
    if (n < 3) return seq[n - 1]
    let i = 1
    while (i < n - 1) {
        seq.push(seq[i - 1] + seq[i])
        i += 1
    }
    return seq.slice(-1)[0]
}

edited Aug 05 '21 at 18:48

answered Aug 04 '21 at 23:01

PhiAgent

158
6

score 0 · Answer 7 · answered Jun 11 '22 at 13:11

0

// Using Recursion
const fib = (n) => {
  if (n <= 2) return 1;
  return fib(n - 1) + fib(n - 2);
}

console.log(fib(4)) // 3
console.log(fib(10)) // 55
console.log(fib(28)) // 317811
console.log(fib(35)) // 9227465

answered Jun 11 '22 at 13:11

dota2pro

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How to find nth Fibonacci number using Javascript with O(n) complexity

7 Answers7