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

@@ -12,7 +12,6 @@
 ->  [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.
-
 """
 from __future__ import print_function
 
@@ -38,7 +37,7 @@ def identity(n):
 def nth_fibonacci(n):
     """
     >>> nth_fibonacci(100)
-    354224848179261915075L
+    354224848179261915075
     >>> nth_fibonacci(-100)
     -100
     """
@@ -90,11 +89,13 @@ def main():
         "100th fibonacci number using matrix exponentiation is %s and using bruteforce is %s \n"
         % (nth_fibonacci(100), nth_fibonacci_test(100))
     )
-    print(
-        "1000th fibonacci number using matrix exponentiation is %s and using bruteforce is %s \n"
-        % (nth_fibonacci(1000), nth_fibonacci_test(1000))
-    )
+
 
 
 if __name__ == "__main__":
     main()
+from timeit import timeit
+print(timeit("nth_fibonacci(1000000)", "from __main__ import nth_fibonacci", number=5))
+print(timeit("nth_fibonacci_test(1000000)", "from __main__ import nth_fibonacci_test", number=5))
+2.3342058970001744
+57.256506615000035