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

nth fibonacci number javascript

fibonacci series in javascript

program to find nth fibonacci number in python

write a program to fibonacci series in javascript hackerrank

fibonacci series in c using function

fibonacci series sum javascript

11 write ac function to generate the n th fibonacci number

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 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

**Program to find the nth Fibonacci in javascript,** Learn how to find the nth fibonacci number in javascript. We will We are iterating till the given number, so Time complexity is O(n). We are� Time complexity: O(n). Space complexity: O(n). Time and Space complexity. We are using memoization to optimize the recursive algorithm by storing the value of the already computed functions with given number i.e we are just calling the functions with distinct numbers, so Time complexity is O(n).

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))

**JavaScript: Compute the nth Fibonacci Number,** The simplest answer is to do it recursively. This has a O(2^n) time complexity but if you memoize the function, this comes down to O(n). // Recursive, O(2^n) const fib = (n) => n < 2 ? n : fib(n-1) + fib(n-2); } // Memoize O(n) // use underscore _. Question: Write a function to calculate the Nth fibonacci number.. There are many possible approaches to this problem. The simplest answer is to do it recursively.This has a O(2^n) time complexity but if you memoize the function, this comes down to O(n).

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));

**Fibonacci sequence algorithm in Javascript,** Given a number N return the index value of the Fibonacci sequence, where the sequence is: The important part here is that we calculate the next number by adding And the good news is that has a O(n) time complexity. Output: Fibonacci(5): 3 Fibonacci(8): 13. By using recursion: As we know that the nth Fibonacci number is the summation of n-1 and n-2 term and the n-1 term is the summation of n-2 and n-3 term.

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; }

**Program for Fibonacci numbers,** HTML � CSS � JavaScript � PHP � JQuery Polynomial Time Approximation Scheme � --- A Time Complexity Question In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence Write a function int fib(int n) that returns Fn. Following are different methods to get the nth Fibonacci number. There are many other ways to find the n-th Fibonacci number, even better than Dynamic Programming with respect to time complexity also space complexity, I will also introduce to you one of those by using a formula and it just takes a constant time O(1) to find the value: F n = {[(√5 + 1)/2] ^ n} / √5

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

function FibSmartRecursive(n, a = 0, b = 1) { return n > 0 ? 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

**Time complexity of recursive Fibonacci program,** Program to print first n Fibonacci Numbers | Set 1 � Program to find last two digits of Nth Fibonacci number. Article Tags : Analysis � Fibonacci. Practice Tags� The runtime of O(log n) ignores the work required to multiply together the numbers, which isn't trivial because the Fibonacci numbers grow exponentially. Only O(log n) multiplies are required, but those multiplies can take a long time. – templatetypedef Aug 29 '13 at 0:06

**Program to find n'th Fibonacci number,** Write a program to calculate n'th Fibonacci number where n is a given positive number. The sequence Fn of Fibonacci numbers is defined by the recurrence relation: Function to find nth Fibonacci number The time complexity of above iterative solution is O(n) since it contains a loop that repeats n-1 times, but it only � If we abstract from those and count all our calculations as one operation then the time complexity of our fibonacci() function is O(1). That is it. I think this is all I had about sequencing Fibonacci numbers with JavaScript. If I will find some way of calculation that is not described in this article, I will definitely let you know.

**Fibonacci Number,** Use recursion to compute the Fibonacci number of a given integer. space complexity by only storing the value of the two previous numbers and on the matrix equation for the Fibonacci sequence, to find the Nth Fibonacci number: Try watching this video on www.youtube.com, or enable JavaScript if it� The important part here is that we calculate the next number by adding the current number to the old number. And the good news is that has a O(n) time complexity. Fair enough for the first try right?

**Time complexity for all Fibonacci numbers from 0 to n,** Fibonacci search has an average- and worst-case complexity of O(log n) (see Big O JavaScript, C++ , , the integer sequence in which every number after the first two Here is the recursive solution for calculating the Nth Fibonacci number in� Below is one more interesting recurrence formula that can be used to find n’th Fibonacci Number in O(Log n) time. If n is even then k = n/2: F(n) = [2*F(k-1) + F(k)]*F(k) If n is odd then k = (n + 1)/2 F(n) = F(k)*F(k) + F(k-1)*F(k-1)

##### Comments

- Using tail recursion or dynamic programming.
- 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.
- @OmG thanks, solved it
`In Javascript this ends up in an infinite loops`

...- @iremlopsum Do you know that you can post answer to your own question?
- 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 :)
- I have edited my answer to add a link, that can be helpfull
- @rockstar I mean iterating from bottom to top. Anyhow, removed it.