Modular exponentiation (Recursive)

Given three numbers a, b and c, we need to find (a^b) % c
Now why do “% c” after exponentiation, because a^b will be really large even for relatively small values of a, b and that is a problem because the data type of the language that we try to code the problem, will most probably not let us store such a large number.
Examples: 
 

Input : a = 2312 b = 3434 c = 6789
Output : 6343

Input : a = -3 b = 5 c = 89 
Output : 24

Auxiliary Space: O(1)
 

The idea is based on below properties.
Property 1: 
(m * n) % p has a very interesting property: 
(m * n) % p =((m % p) * (n % p)) % p
Property 2: 
if b is even: 
(a ^ b) % c = ((a ^ b/2) * (a ^ b/2))%c ? this suggests divide and conquer 
if b is odd: 
(a ^ b) % c = (a * (a ^( b-1))%c
Property 3: 
If we have to return the mod of a negative number x whose absolute value is less than y: 
then (x + y) % y will do the trick
Note: 
Also as the product of (a ^ b/2) * (a ^ b/2) and a * (a ^( b-1) may cause overflow, hence we must be careful about those scenarios 
 

Power is 6

Time Complexity : O(logn)

Auxiliary Space: O(logn)

Iterative modular exponentiation.