I am trying to find complexity of Fibonacci series using a recursion tree and concluded height of tree = O(n) worst case, cost of each level = cn, hence complexity = n*n=n^2 ... It is theta (F_N) (and so O(2^N)), even if you consider the complexity of adding two n bits numbers to be Polynomial in N, I believe. ... As a followup question, is the time ... More on stackoverflow.com

Solution For What is the expression for time complexity for finding Fibonacci number using recursion? a. T ( n ) = T ( n - 1 ) + O ( n ) b. T ( n ) = More on askfilo.com

Surprised I don't see this mentioned, but there is a nifty way to do it without memoization -- in O(log N). Matrix Exponentiation. Basically, consider your processing step: you have two values, the previous and current numbers. You do a nice clean simple linear transformation, and end up with the current and next. Flip it around a bit: you have an incoming 2-vector, you do a matrix multiply, giving a new 2-vector. Doing it several times to get new numbers just stacks up a sequence of matrix multiplies, which you can re-associate to be a matrix to the Nth power times the starting vector. More thinking. If I wanted a matrix to the 8th power, I'd not multiply it eight times. I'd square it, then square that, then square that. So O(log N) matrix multiplies (this is called Russian Peasant Exponentiation, or The Egyptian Method). Up to some N, the O(N) method using fast steps will be faster than the O(log N) method which does Matrix Multiplies, but at some N, the O(log N) will pull ahead. OK, trying some Bad Text Graphics ... One FIB step, as matrix multiply: ⎡ F₂ ⎤ ⎡ 0 1 ⎤ ⎡ F₁ ⎤ ⎢ ⎥ := ⎢ ⎥ × ⎢ ⎥ ⎣ F₃ ⎦ ⎣ 1 1 ⎦ ⎣ F₂ ⎦ Eight FIB steps: ⎡ F₈ ⎤ ⎡ 0 1 ⎤⁸ ⎡ F₁ ⎤ ⎢ ⎥ := ⎢ ⎥ × ⎢ ⎥ ⎣ F₉ ⎦ ⎣ 1 1 ⎦ ⎣ F₂ ⎦ So for N steps, compute the Nth power of that matrix, which can be done in O(log N). You can see a bunch of different "takes" on this method here: https://rosettacode.org/wiki/Fibonacci_matrix-exponentiation [Edit to add: looking for even faster ways to do the matrix exponentiation will lead you to looking at very cool stuff that totally does not apply to this problem. You know, just in case you are math-curious ;] More on reddit.com

Merge sort is a true divide-and-conquer algorithm: the two recursive calls are each responsible for half of the list. A naive recursive Fibonacci implementation would be more accurately described as "duplicate-and-conquer". When calculating fib(n) and fib(n+1) for an addition, the latter call duplicates all the work of the former call. But you can have a linear recursive Fibonacci implementation by passing additional parameters: fib'(0,x,y) = x fib'(1+n,x,y) = fib'(n, x+y, x) fib(n) = fib'(n,1,0) Think of it as a dynamic programming solution filling in an array of Fibonacci numbers, but keeping only the bits of the array you need for the next step. More on reddit.com