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 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]
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.
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 left Rotate arr[] of size n by 1*/ void leftRotatebyOne( int arr[], int n) { int temp = arr[0], i; for (i = 0; i < n - 1; i++) arr[i] = arr[i + 1]; arr[n-1] = temp; } /*Function to left rotate arr[] of size n by d*/ void leftRotate( int arr[], int d, int n) { for ( int i = 0; i < d; i++) leftRotatebyOne(arr, n); } /* utility function to print an array */ void printArray( int arr[], int n) { for ( int i = 0; i < n; 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 left Rotate arr[] of size n by 1*/ void leftRotatebyOne( int arr[], int n); /*Function to left rotate arr[] of size n by d*/ 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 temp = arr[0], i; for (i = 0; i < n - 1; i++) arr[i] = arr[i + 1]; arr[n-1] = temp; } /* utility function to print an array */ void printArray( int arr[], int n) { int i; for (i = 0; i < n; i++) printf ( "%d " , arr[i]); } /* Driver program to test above functions */ int main() { int arr[] = { 1, 2, 3, 4, 5, 6, 7 }; leftRotate(arr, 2, 7); printArray(arr, 7); return 0; } |
Java
// Java program to rotate an array by // d elements class RotateArray { /*Function to left rotate arr[] of size n by d*/ void leftRotate( int arr[], int d, int n) { for ( int 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[n- 1 ] = temp; } /* utility function to print an array */ void printArray( int arr[], int n) { for ( int i = 0 ; i < n; i++) System.out.print(arr[i] + " " ); } // 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): for i in range (d): leftRotatebyOne(arr, n) # Function to left Rotate arr[] of size n by 1*/ def leftRotatebyOne(arr, n): temp = arr[ 0 ] for i in range (n - 1 ): arr[i] = arr[i + 1 ] arr[n - 1 ] = temp # utility function to print an array */ def printArray(arr, size): for i in range (size): print ( "% d" % arr[i], end = " " ) # Driver program to test above functions */ arr = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 ] leftRotate(arr, 2 , 7 ) printArray(arr, 7 ) # 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) { for ( int i = 0; i < d; i++) leftRotatebyOne(arr, n); } static void leftRotatebyOne( int [] arr, int n) { int i, temp = arr[0]; for (i = 0; i < n - 1; i++) arr[i] = arr[i + 1]; arr[n-1] = temp; } /* utility function to print an array */ static void printArray( int [] arr, int size) { for ( int i = 0; i < size; i++) Console.Write(arr[i] + " " ); } // 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 |
PHP
<?php // PHP program to rotate an array // by d elements /*Function to left Rotate arr[] of size n by 1*/ function leftRotatebyOne(& $arr , $n ) { $temp = $arr [0]; for ( $i = 0; $i < $n - 1; $i ++) $arr [ $i ] = $arr [ $i + 1]; $arr [ $n -1] = $temp ; } /*Function to left rotate arr[] of size n by d*/ function leftRotate(& $arr , $d , $n ) { for ( $i = 0; $i < $d ; $i ++) leftRotatebyOne( $arr , $n ); } /* utility function to print an array */ function printArray(& $arr , $n ) { for ( $i = 0; $i < $n ; $i ++) echo $arr [ $i ] . " " ; } // Driver Code $arr = array ( 1, 2, 3, 4, 5, 6, 7 ); $n = sizeof( $arr ); // Function calling leftRotate( $arr , 2, $n ); printArray( $arr , $n ); // This code is contributed // by ChitraNayal ?> |
Javascript
<script> // JavaScript program to rotate an array by // d elements /* Function to left rotate arr of size n by d */ function leftRotate(arr , d , n) { for (i = 0; i < d; i++) leftRotatebyOne(arr, n); } function leftRotatebyOne(arr , n) { var i, temp; temp = arr[0]; for (i = 0; i < n - 1; i++) arr[i] = arr[i + 1]; arr[n - 1] = temp; } /* utility function to print an array */ function printArray(arr , n) { for (i = 0; i < n; i++) document.write(arr[i] + " " ); } // Driver program to test above functions var arr = [ 1, 2, 3, 4, 5, 6, 7 ]; leftRotate(arr, 2, 7); printArray(arr, 7); // This code is contributed by todaysgaurav </script> |
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}
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; /*Fuction 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); /*Fuction 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]); } /*Fuction 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] + " " ); } /*Fuction 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 = " " ) # Fuction 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] + " " ); } /* Fuction 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 |
Output :
3 4 5 6 7 1 2
Time complexity : O(n)
Auxiliary Space : O(1)
Please see 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 bug in above programs/algorithms.
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