C# Program for Reversal algorithm for array rotation
Write a function rotate(arr[], d, n) that rotates arr[] of size n by d elements.
Example :
Input : arr[] = [1, 2, 3, 4, 5, 6, 7]
d = 2
Output : arr[] = [3, 4, 5, 6, 7, 1, 2]
Rotation of the above array by 2 will make array
The first 3 methods to rotate an array by d elements has been discussed in this post.
Method 4 (The Reversal Algorithm) :
Algorithm :
rotate(arr[], d, n)
reverse(arr[], 1, d) ;
reverse(arr[], d + 1, n);
reverse(arr[], 1, n);
Let AB are the two parts of the input array where A = arr[0..d-1] and B = arr[d..n-1]. The idea of the algorithm is :
- Reverse A to get ArB, where Ar is reverse of A.
- Reverse B to get ArBr, where Br is reverse of B.
- Reverse all to get (ArBr) r = BA.
Example :
Let the array be arr[] = [1, 2, 3, 4, 5, 6, 7], d =2 and n = 7
A = [1, 2] and B = [3, 4, 5, 6, 7]
- Reverse A, we get ArB = [2, 1, 3, 4, 5, 6, 7]
- Reverse B, we get ArBr = [2, 1, 7, 6, 5, 4, 3]
- Reverse all, we get (ArBr)r = [3, 4, 5, 6, 7, 1, 2]
Below is the implementation of the above approach :
C#
using System;
class GFG {
static void leftRotate( int [] arr, int d)
{
if (d == 0)
return ;
int n = arr.Length;
d = d % n;
reverseArray(arr, 0, d - 1);
reverseArray(arr, d, n - 1);
reverseArray(arr, 0, n - 1);
}
static void reverseArray( int [] arr, int start,
int end)
{
int temp;
while (start < end) {
temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
static void printArray( int [] arr)
{
for ( int i = 0; i < arr.Length; i++)
Console.Write(arr[i] + " " );
}
public static void Main()
{
int [] arr = { 1, 2, 3, 4, 5, 6, 7 };
int n = arr.Length;
int d = 2;
leftRotate(arr, d);
printArray(arr);
}
}
|
Output :
3 4 5 6 7 1 2
Time Complexity : O(n), where n represents the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Reversal algorithm for array rotation for more details!
Last Updated :
28 May, 2022
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