Rotate Array by K times && Quick Left Rotation, GCD of two or more numbers, LCD

Rotate Array by K times:

Lets says array [1,2,3,4,5,6,7,8], K = 2
                result = [7,8,1,2,3,4,5,6]
if K>N where N = length of array and K = numer of rotations
let's say k=20, then k%n gives us the number of rotations

Can do it in 3 ways:
Approach 1: (Naive or Brute Force Approach)
     [1,2,3,4,5,6,7,8] -> [8,1,2,3,4,5,6,7] -> [7,8,1,2,3,4,5,6]

If k=5, then it will be more run time
   Time Complexity: O(N*K) where N = length of array and K = numer of rotations
   Space Complexity: O(1) as no space is used

Approach 2:(Using Extra Space)
 Create a new array of same size, and then arrange elements accordingly
 

   Time Complexity: O(N) where N = length of array and K = numer of rotations

   Space Complexity: O(N) as elements of N space is used

Approach 3:(Optimized)

Three main points:

Create your own Reverse Function which takes the array, start and end indexes

          1) Reverse all Elements

          2) Reverse first K elements

          3) Reverse N-K Elements
or
similar to above algorithm
1) Make two parts of array as (N-K) and K {A,B}
2) Reverse A and B
3) Reverse whole A and B that we got from 2

   Time Complexity: O(N) where N = length of array and K = numer of rotations

   Space Complexity: O(1) as No extra space is used

Quickly find multiple left rotations of an array:

Coming soon...

References:
https://www.geeksforgeeks.org/quickly-find-multiple-left-rotations-of-an-array/

Land Right Rotation of a String:
coming soon...

References:
https://www.geeksforgeeks.org/left-rotation-right-rotation-string-2/

Array Rotation in Place:
Accomplish above by using the Array Rotation in place:
Example:  [1,2,3,4,5,6,7,8,9]
output: [4,5,6,7,8,9,1,2,3]

Steps:
     1) Divide array into sets & rotate each element in set k postitions to left
      2) outer loop: number of sets, inner loop: rotate elements of a set k postions to the left

number of sets?? (Use GCD to calculate the number of sets)

References:
 https://www.youtube.com/watch?v=utE_1ppU5DY

GCD of 1 or more numbers:
-> GCD between two: GCD(a,b)
-> GCD between 3: GCD(a,GCD(b,c))
gcd(int a,int b){
  if(b==0){
     return a;
   }else{
      return gcd(b,a%b);
   }
}
References:
https://www.geeksforgeeks.org/gcd-two-array-numbers/
LCD between a and b = (a/GCD) * b;