Updated the matrix exponentiation approach of finding nth fibonacci.

- Removed some extra spaces
- Added the complexity of bruteforce algorithm
- Removed unused function called zerro()
- Added some docktest based on request

Author: mahbubcseju

matrix/nth_fibonacci_using_matrix_exponentiation.py

@@ -10,9 +10,8 @@
     ...
     ...
 ->  [f(n),f(n-1)] = [[1,1],[1,0]]^(n-1) * [f(1),f(0)]
-So we just need the n times multiplication of  the matrix [1,1],[1,0]].
-We can decrease the n times multiplication by following  the divide and conquer approach.
-
+So we just need the n times multiplication of the matrix [1,1],[1,0]].
+We can decrease the n times multiplication by following the divide and conquer approach.
 """
 from __future__ import print_function
 
@@ -56,6 +55,12 @@ def nth_fibonacci(n):
 
 
 def nth_fibonacci_test(n):
+    """
+    >>> nth_fibonacci_test(100)
+    354224848179261915075L
+    >>> nth_fibonacci_test(-100)
+    -100
+    """
     if n <= 1:
         return n
     fib0 = 0