Java Rotate and Translate Example

Write a function rotate(ar[], d, n) that rotates arr[] of size n by d elements.

Rotation of the above array by 2 will make array

METHOD 1 (Using temp array)

Input arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2, n =7 1) Store d elements in a temp array    temp[] = [1, 2] 2) Shift rest of the arr[]    arr[] = [3, 4, 5, 6, 7, 6, 7] 3) Store back the d elements    arr[] = [3, 4, 5, 6, 7, 1, 2]

Time complexity : O(n)
Auxiliary Space : O(d)

METHOD 2 (Rotate one by one)

leftRotate(arr[], d, n) start   For i = 0 to i < d     Left rotate all elements of arr[] by one end

To rotate by one, store arr[0] in a temporary variable temp, move arr[1] to arr[0], arr[2] to arr[1] …and finally temp to arr[n-1]

Let us take the same example arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2
Rotate arr[] by one 2 times
We get [2, 3, 4, 5, 6, 7, 1] after first rotation and [ 3, 4, 5, 6, 7, 1, 2] after second rotation.

Java

class RotateArray

{

void leftRotate( int arr[], int d, int n)

{

int i;

for (i = 0 ; i < d; i++)

leftRotatebyOne(arr, n);

}

void leftRotatebyOne( int arr[], int n)

{

int i, temp;

temp = arr[ 0 ];

for (i = 0 ; i < n - 1 ; i++)

arr[i] = arr[i + 1 ];

arr[i] = temp;

}

void printArray( int arr[], int size)

{

int i;

for (i = 0 ; i < size; i++)

System.out.print(arr[i] + " " );

}

public static void main(String[] args)

{

RotateArray rotate = new RotateArray();

int arr[] = { 1 , 2 , 3 , 4 , 5 , 6 , 7 };

rotate.leftRotate(arr, 2 , 7 );

rotate.printArray(arr, 7 );

}

}

Output :

3 4 5 6 7 1 2            

Time complexity : O(n * d)
Auxiliary Space : O(1)
METHOD 3 (A Juggling Algorithm)
This is an extension of method 2. Instead of moving one by one, divide the array in different sets
where number of sets is equal to GCD of n and d and move the elements within sets.
If GCD is 1 as is for the above example array (n = 7 and d =2), then elements will be moved within one set only, we just start with temp = arr[0] and keep moving arr[I+d] to arr[I] and finally store temp at the right place.

Here is an example for n =12 and d = 3. GCD is 3 and

Let arr[] be {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}  a)    Elements are first moved in first set – (See below diagram for this movement)                            arr[] after this step --> {                4              2 3                              7              5 6                              10              8 9                              1              11 12}  b)    Then in second set.           arr[] after this step --> {4                              5                            3 7                              8                            6 10                              11                            9 1                              2                            12}  c)    Finally in third set.           arr[] after this step --> {4 5                              6              7 8                              9              10 11                              12              1 2                              3              }            

Java

class RotateArray

{

void leftRotate( int arr[], int d, int n)

{

int i, j, k, temp;

for (i = 0 ; i < gcd(d, n); i++)

{

temp = arr[i];

j = i;

while ( true )

{

k = j + d;

if (k >= n)

k = k - n;

if (k == i)

break ;

arr[j] = arr[k];

j = k;

}

arr[j] = temp;

}

}

void printArray( int arr[], int size)

{

int i;

for (i = 0 ; i < size; i++)

System.out.print(arr[i] + " " );

}

int gcd( int a, int b)

{

if (b == 0 )

return a;

else

return gcd(b, a % b);

}

public static void main(String[] args) {

RotateArray rotate = new RotateArray();

int arr[] = { 1 , 2 , 3 , 4 , 5 , 6 , 7 };

rotate.leftRotate(arr, 2 , 7 );

rotate.printArray(arr, 7 );

}

}

Output :

3 4 5 6 7 1 2            

Time complexity : O(n)
Auxiliary Space : O(1)

Please refer complete article on Program for array rotation for more details!

binionandeed.blogspot.com

Source: https://www.geeksforgeeks.org/java-program-for-array-rotation/

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