# LeetCode Question - 509. Fibonacci Number π€ͺ

## 6th July 2022 | π Daily LeetCode Challenge - #6

### About the Series

Problem-solving is a key skill set for any tech-related stuff you might be working on.

When it comes to developers it's one of the most crucial skills which is needed in almost any day-to-day code you might be writing.

So, this series of blogs is all about practicing Daily LeetCode Challenges & Problem-solving. π

### Problem Statement

The Fibonacci numbers, commonly denoted

`F(n)`

form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is,

`F(0) = 0, F(1) = 1`

`F(n) = F(n - 1) + F(n - 2), for n > 1.`

Given n, calculate F(n).

**Example 1:**

```
Input: n = 2
Output: 1
Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.
```

**Example 2:**

```
Input: n = 3
Output: 2
Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.
```

**Example 3:**

```
Input: n = 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.
```

### Video Explanation

### Solution 1: Recursive Approach

**Algorithm**

- This is the recursive solution where If
`n`

is`0`

it will return`0`

or if`n`

is`1`

it will return`1`

. This is the base case for recursion - Recursively call the
`f(n-1)`

and`f(n-2)`

to return the final value of`f(n)`

- Finally we get out
`f(n)`

.

**Code in JS π§βπ» **

```
/**
* @param {number} n
* @return {number}
*/
var fib = function (n) {
if (n < 2) return n;
return fib(n - 1) + fib(n - 2);
};
```

**Time Complexity : O(n) **

**Space Complexity: O(n)**

### Solution 2: Iterative approach with an array

**Algorithm**

- First of all,
`n = 0`

return`0`

and for`n = 1`

return`1`

- Create an array and store the first two values of the Fibonacci sequence in the array
- Now update the
`ith`

value of the array with the`i-1`

value + the`i-2`

value of the array. - Repeat the same for n iterations and then the array will have the
`nth`

value too. - Return the
`nth`

value from the array

**Code in JS π§βπ» **

```
/**
* @param {number} n
* @return {number}
*/
var fib = function (n) {
if (n < 2) return n;
var fibArray = [];
fibArray[0] = 0;
fibArray[1] = 1;
for (var i = 2; i <= n; i++) {
fibArray[i] = fibArray[i - 1] + fibArray[i - 2];
}
return fibArray[n];
};
```

**Time Complexity : O(n) **

**Space Complexity: O(n)**

### Solution 3: Iterative approach without an array

**Algorithm**

The approach is similar to the 2nd solution but it is done without using a new array.

- First of all,
`n = 0`

return`0`

and for`n = 1`

return`1`

- Store the last two values of the Fibonacci series in
`f1`

and`f2`

. - Update
`f1`

with`f2`

- Update
`f2`

with the current sum of`f1`

and`f2`

. - Steps 2 and 3 will be repeated till we get the final value after n iterations.
- Return
`f2`

.

**Code in JS π§βπ» **

```
/**
* @param {number} n
* @return {number}
*/
var fib = function (n) {
if (n < 2) return n;
var f1 = 0;
var f2 = 1;
for (var i = 2; i <=n; i++) {
var currentSum = f1 + f2;
f1 = f2;
f2 = currentSum;
}
return f2;
};
```

**Time Complexity : O(n) **

**Space Complexity: O(1)**

### Similar Questions for practice

Now it is time to try some similar questions

- Climbing Stairs
- Split Array into Fibonacci Sequence
- Length of Longest Fibonacci Subsequence
- N-th Tribonacci Number

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