Program for array rotation
Write a function rotate(arr[], d, n) that rotates arr[] of size n by d elements.
Rotation of the above array by 2 will make an array
Method 1 (Using temp array) :
Input arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2, n =7 1) Store the first 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]
Below is the implementation of the above approach :
Python3
def rotate(L,d,n): k = L.index(d) new_lis = [] new_lis = L[k + 1 :] + L[ 0 :k + 1 ] return new_lis d = 2 n = 7 print (rotate([ 1 , 2 , 3 , 4 , 5 , 6 , 7 ],d,n)) |
[3, 4, 5, 6, 7, 1, 2]
Time complexity : O(n)
Auxiliary Space : O(n)
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.
Below is the implementation of the above approach:
Java
/*package whatever //do not write package name here */ import java.io.*; class GFG { public static void main(String[] args) { int a[] = { 1 , 2 , 3 , 4 , 5 , 6 , 7 }; int n = a.length; // Rotate 2 times int d = 2 ; int p = 1 ; while (p <= d) { int last = a[ 0 ]; for ( int i = 0 ; i < n - 1 ; i++) { a[i] = a[i + 1 ]; } a[n - 1 ] = last; p++; } System.out.println( "Array after Rotation :" ); for ( int i = 0 ; i < n; i++) { System.out.print(a[i] + " " ); } } } // contributed by keerthikarathan123 |
Array after Rotation : 3 4 5 6 7 1 2
Time Complexity: O(N*d), Where N is the length of the given array and d is the rotation number.
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 into different sets
where the number of sets is equal to the GCD of n and d and moves 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}
Below is the implementation of the above approach :
C++
// C++ program to rotate an array by // d elements #include <bits/stdc++.h> using namespace std; /*Function to get gcd of a and b*/ int gcd( int a, int b) { if (b == 0) return a; else return gcd(b, a % b); } /*Function to left rotate arr[] of siz n by d*/ void leftRotate( int arr[], int d, int n) { /* To handle if d >= n */ d = d % n; int g_c_d = gcd(d, n); for ( int i = 0; i < g_c_d; i++) { /* move i-th values of blocks */ int temp = arr[i]; int j = i; while (1) { int k = j + d; if (k >= n) k = k - n; if (k == i) break ; arr[j] = arr[k]; j = k; } arr[j] = temp; } } // Function to print an array void printArray( int arr[], int size) { for ( int i = 0; i < size; i++) cout << arr[i] << " " ; } /* Driver program to test above functions */ int main() { int arr[] = { 1, 2, 3, 4, 5, 6, 7 }; int n = sizeof (arr) / sizeof (arr[0]); // Function calling leftRotate(arr, 2, n); printArray(arr, n); return 0; } |
C
// C program to rotate an array by // d elements #include <stdio.h> /* function to print an array */ void printArray( int arr[], int size); /*Function to get gcd of a and b*/ int gcd( int a, int b); /*Function to left rotate arr[] of siz n by d*/ void leftRotate( int arr[], int d, int n) { int i, j, k, temp; /* To handle if d >= n */ d = d % n; int g_c_d = gcd(d, n); for (i = 0; i < g_c_d; i++) { /* move i-th values of blocks */ temp = arr[i]; j = i; while (1) { k = j + d; if (k >= n) k = k - n; if (k == i) break ; arr[j] = arr[k]; j = k; } arr[j] = temp; } } /*UTILITY FUNCTIONS*/ /* function to print an array */ void printArray( int arr[], int n) { int i; for (i = 0; i < n; i++) printf ( "%d " , arr[i]); } /*Function to get gcd of a and b*/ int gcd( int a, int b) { if (b == 0) return a; else return gcd(b, a % b); } /* Driver program to test above functions */ int main() { int arr[] = { 1, 2, 3, 4, 5, 6, 7 }; leftRotate(arr, 2, 7); printArray(arr, 7); getchar (); return 0; } |
Java
// Java program to rotate an array by // d elements class RotateArray { /*Function to left rotate arr[] of siz n by d*/ void leftRotate( int arr[], int d, int n) { /* To handle if d >= n */ d = d % n; int i, j, k, temp; int g_c_d = gcd(d, n); for (i = 0 ; i < g_c_d; i++) { /* move i-th values of blocks */ 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; } } /*UTILITY FUNCTIONS*/ /* function to print an array */ void printArray( int arr[], int size) { int i; for (i = 0 ; i < size; i++) System.out.print(arr[i] + " " ); } /*Function to get gcd of a and b*/ int gcd( int a, int b) { if (b == 0 ) return a; else return gcd(b, a % b); } // Driver program to test above functions 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 ); } } // This code has been contributed by Mayank Jaiswal |
Python3
# Python3 program to rotate an array by # d elements # Function to left rotate arr[] of size n by d def leftRotate(arr, d, n): d = d % n g_c_d = gcd(d, n) for i in range (g_c_d): # move i-th values of blocks temp = arr[i] j = i while 1 : k = j + d if k > = n: k = k - n if k = = i: break arr[j] = arr[k] j = k arr[j] = temp # UTILITY FUNCTIONS # function to print an array def printArray(arr, size): for i in range (size): print ( "% d" % arr[i], end = " " ) # Function to get gcd of a and b def gcd(a, b): if b = = 0 : return a; else : return gcd(b, a % b) # Driver program to test above functions arr = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 ] n = len (arr) d = 2 leftRotate(arr, d, n) printArray(arr, n) # This code is contributed by Shreyanshi Arun |
C#
// C# program for array rotation using System; class GFG { /* Function to left rotate arr[] of size n by d*/ static void leftRotate( int [] arr, int d, int n) { int i, j, k, temp; /* To handle if d >= n */ d = d % n; int g_c_d = gcd(d, n); for (i = 0; i < g_c_d; i++) { /* move i-th values of blocks */ 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; } } /*UTILITY FUNCTIONS*/ /* Function to print an array */ static void printArray( int [] arr, int size) { for ( int i = 0; i < size; i++) Console.Write(arr[i] + " " ); } /* Function to get gcd of a and b*/ static int gcd( int a, int b) { if (b == 0) return a; else return gcd(b, a % b); } // Driver code public static void Main() { int [] arr = { 1, 2, 3, 4, 5, 6, 7 }; leftRotate(arr, 2, 7); printArray(arr, 7); } } // This code is contributed by Sam007 |
Javascript
<script> // JavaScript program to rotate an array by // d elements /*Function to get gcd of a and b*/ function gcd( a, b){ if (b == 0) return a; else return gcd(b, a % b); } /*Function to left rotate arr[] of siz n by d*/ function leftRotate(arr, d, n){ /* To handle if d >= n */ d = d % n; let g_c_d = gcd(d, n); for (let i = 0; i < g_c_d; i++) { /* move i-th values of blocks */ let temp = arr[i]; let j = i; while (1) { let k = j + d; if (k >= n) k = k - n; if (k == i) break ; arr[j] = arr[k]; j = k; } arr[j] = temp; } } // Function to print an array function printArray(arr, size){ for (let i = 0; i < size; i++) document.write(arr[i] + " " ); } /* Driver program to test above functions */ let arr = [ 1, 2, 3, 4, 5, 6, 7 ]; let n = arr.length; // Function calling leftRotate(arr, 2, n); printArray(arr, n); </script> |
3 4 5 6 7 1 2
Time complexity : O(n)
Auxiliary Space : O(1)
Please see the following posts for other methods of array rotation:
Block swap algorithm for array rotation
Reversal algorithm for array rotation
Please write comments if you find any bugs in the above programs/algorithms.