Find (1^n + 2^n + 3^n + 4^n) mod 5 | Set 2

Given a very large number N. The task is to find (1ⁿ + 2ⁿ + 3ⁿ + 4ⁿ) mod 5.
Examples: 
 

Input: N = 4 
Output: 4 
(1 + 16 + 81 + 256) % 5 = 354 % 5 = 4
Input: N = 7823462937826332873467731 
Output: 0 
 

Approach: (1ⁿ + 2ⁿ + 3ⁿ + 4ⁿ) mod 5 = (1^{n mod ?(5)} + 2^{n mod ?(5)} + 3^{n mod ?(5)} + 4^{n mod ?(5)}) mod 5. 
This formula is correct because 5 is a prime number and it is coprime with 1, 2, 3, 4. 
Know about ?(n) and modulo of large number 
?(5) = 4, hence (1ⁿ + 2ⁿ + 3ⁿ + 4ⁿ) mod 5 = (1^{n mod 4} + 2^{n mod 4} + 3^{n mod 4} + 4^{n mod 4}) mod 5
Below is the implementation of the above approach: 
 

4

Time Complexity: O(|N|), where |N| is the length of the string.
Auxiliary Space: O(1), since no extra space has been taken.