# How To Compute Fibonacci Number F(n)

## Problem statement

The Fibonacci numbers make up a sequence denoted as F(n), known as the Fibonacci sequence. Each number in this sequence is the sum of the two preceding numbers, with the sequence starting from 0 and 1. In other words:

F(0) = 0, F(1) = 1
F(n) = F(n - 1) + F(n - 2), for n > 1.


Your task is to calculate the value of F(n) given an integer 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.


### Constraints

• 0 <= n <= 30.

## Solution 1: Recursive

### Code

#include <iostream>
int fib(int n)
{
if (n <= 1)
{
return n;
}
return fib(n - 1) + fib(n - 2);
}
int main()
{
std::cout << fib(2) << std::endl;
std::cout << fib(3) << std::endl;
std::cout << fib(4) << std::endl;
}

Output:
1
2
3


### Code explanation

This solution computes the nth Fibonacci number using a recursive approach. Here's how the code works:

1. The code checks if n is less than or equal to 1. If n is 0 or 1, it returns n itself because the Fibonacci sequence starts with 0 and 1, and the nth Fibonacci number is equal to n for n less than or equal to 1.

2. If n is greater than 1, it calculates the n-th Fibonacci number by recursively calling the fib function for n - 1 and n - 2 and then adding the results. This recursive approach continues until n becomes 0 or 1, at which point it returns the respective value.

### Complexity

The time complexity of this solution is exponential, specifically O(2^n). This is because it repeatedly makes two recursive calls for each n, resulting in an exponential number of function calls and calculations. As n grows larger, the execution time increases significantly.

The space complexity of the given recursive Fibonacci solution is O(n). This space complexity arises from the function call stack when making recursive calls.

When you call the fib function with a value of n, it generates a call stack with a depth of n, as each call to fib leads to two more recursive calls (one for n - 1 and one for n - 2) until n reaches the base cases (0 or 1). The space required to store the function call stack is proportional to the depth of the recursion, which is n.

Therefore, the space complexity is linearly related to the input value n, making it O(n). This can be a concern for large values of n because it could lead to a stack overflow if n is too large.

• Runtime: O(2^n).
• Extra space: O(n).

## Solution 2: Dynamic programming

### Code

#include <iostream>
#include <vector>
int fib(int n)
{
if (n <= 1)
{
return n;
}
std::vector<int> f(n + 1);
f = 0;
f = 1;
for (int i = 2; i <= n; i++)
{
f[i] = f[i - 1] + f[i - 2];
}
return f[n];
}
int main()
{
std::cout << fib(2) << std::endl;
std::cout << fib(3) << std::endl;
std::cout << fib(4) << std::endl;
}

Output:
1
2
3


### Code explanation

This solution calculates the n-th Fibonacci number using dynamic programming and an array to store intermediate results. Here's a step-by-step explanation:

1. The code first checks if n is less than or equal to 1. If n is 0 or 1, it returns n because these are the base cases of the Fibonacci sequence.

2. For values of n greater than 1, it initializes a vector f of size n + 1 to store the Fibonacci numbers from 0 to n.

3. It sets the initial values of f and f to 0 and 1, respectively, as these are the known base cases of the Fibonacci sequence.

4. Then, it enters a loop starting from i = 2 up to n. In each iteration, it calculates the Fibonacci number at index i as the sum of the two previous Fibonacci numbers, f[i - 1] and f[i - 2], and stores it in f[i].

5. After the loop completes, it returns f[n], which contains the nth Fibonacci number.

### Complexity

This solution uses dynamic programming to avoid redundant calculations by storing and reusing previously computed Fibonacci numbers. It has a time complexity of O(n) because it iterates through the values from 2 to n once, calculating each Fibonacci number once. Additionally, it has a space complexity of O(n) due to the f array used to store intermediate results.

• Runtime: O(n).
• Extra space: O(n).

## Solution 3: Reduce space for dynamic programming

### Code

#include <iostream>
int fib(int n)
{
if (n <= 1)
{
return n;
}
int f0 = 0;
int f1 = 1;
for (int i = 2; i <= n; i++)
{
int f2 = f1 + f0;
f0 = f1;
f1 = f2;
}
return f1;
}
int main()
{
std::cout << fib(2) << std::endl;
std::cout << fib(3) << std::endl;
std::cout << fib(4) << std::endl;
}

Output:
1
2
3


### Code explanation

This solution calculates the nth Fibonacci number iteratively using two variables to keep track of the last two Fibonacci numbers. Here's a step-by-step explanation:

1. The code first checks if n is less than or equal to 1. If n is 0 or 1, it returns n because these are the base cases of the Fibonacci sequence.

2. For values of n greater than 1, it initializes three integers: f0, f1, and f2. f0 is initialized to 0, f1 to 1, and f2 will be used to store the current Fibonacci number.

3. It enters a loop starting from i = 2 up to n. In each iteration:

• It calculates the next Fibonacci number f2 by adding f1 and f0.
• It updates f0 to be the previous f1, which will be used in the next iteration.
• It updates f1 to be the previous f2, which will be used in the next iteration.
4. After the loop completes, it returns f1, which contains the nth Fibonacci number.

### Complexity

This solution effectively calculates the Fibonacci sequence without the need for an array to store intermediate results. It has a time complexity of O(n) because it iterates through the values from 2 to n once, calculating each Fibonacci number once. It also has a space complexity of O(1) because it uses only a constant amount of additional memory to store the variables f0, f1, and f2.

• Runtime: O(n).
• Extra space: O(1).

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