Sum of Fibonacci Numbers in a range

Given a range [l, r], the task is to find the sum fib(l) + fib(l + 1) + fib(l + 2) + ..... + fib(r) where fib(n) is the n^th Fibonacci number.

Examples: 

Input: l = 2, r = 5 
Output: 11 
fib(2) + fib(3) + fib(4) + fib(5) = 1 + 2 + 3 + 5 = 11

Input: l = 4, r = 8 
Output: 50  

Naive approach: Simply calculate fib(l) + fib(l + 1) + fib(l + 2) + ..... + fib(r) in O(r - l) time complexity. 
In order to find fib(n) in O(1) we will take the help of the Golden Ratio. 

Fibonacci's calculation using Binet's Formula  

fib(n) = phiⁿ - psiⁿ) / ?5 
Where, 
phi = (1 + sqrt(5)) / 2 which is roughly equal to 1.61803398875 
psi = 1 - phi = (1 - sqrt(5)) / 2 which is roughly equal to 0.61803398875 
 

Below is the implementation of the above approach:  

50

Efficient approach: The idea is to find the relationship between the sum of Fibonacci numbers and n^th Fibonacci number and use Binet's Formula to calculate its value.

Relationship Deduction 

1. F(i) refers to the i^th Fibonacci number.
2. S(i) refers to the sum of Fibonacci numbers till F(i). 

We can rewrite the relation F(n + 1) = F(n) + F(n - 1) as below: 
F(n - 1) = F(n + 1) - F(n) 
Similarly, 
F(n - 2) = F(n) - F(n - 1) 
... 
... 
... 
F(0) = F(2) - F(1)
Adding all the equations, on left side, we have 
F(0) + F(1) + ... + F(n - 1) which is S(n - 1) 
 

Therefore, 
S(n - 1) = F(n + 1) – F(1) 
S(n - 1) = F(n + 1) – 1 
S(n) = F(n + 2) – 1 

In order to find S(n), simply calculate the (n + 2)^th Fibonacci number and subtract 1 from the result. 
Therefore, 
S(l, r) = S(r) - S(l - 1) 
S(l, r) = F(r + 2) - 1 - (F(l + 1) - 1) 
S(l, r) = F(r + 2) - F(l + 1) 

50

Time Complexity: O(log r)
Auxiliary Space: O(1)