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] 

Array

Rotation of the above array by 2 will make array

ArrayRotation1

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 C implementation of the above approach :

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// C/C++ program for reversal algorithm of array rotation
#include <stdio.h>
  
/*Utility function to print an array */
void printArray(int arr[], int size);
  
/* Utility function to reverse arr[] from start to end */
void rvereseArray(int arr[], int start, int end);
  
/* Function to left rotate arr[] of size n by d */
void leftRotate(int arr[], int d, int n)
{
    rvereseArray(arr, 0, d - 1);
    rvereseArray(arr, d, n - 1);
    rvereseArray(arr, 0, n - 1);
}
  
/*UTILITY FUNCTIONS*/
/* function to print an array */
void printArray(int arr[], int size)
{
    int i;
    for (i = 0; i < size; i++)
        printf("%d ", arr[i]);
}
  
/*Function to reverse arr[] from index start to end*/
void rvereseArray(int arr[], int start, int end)
{
    int temp;
    while (start < end) {
        temp = arr[start];
        arr[start] = arr[end];
        arr[end] = temp;
        start++;
        end--;
    }
}
  
/* Driver program to test above functions */
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int d = 2;
    leftRotate(arr, d, n);
    printArray(arr, n);
    return 0;
}

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Output:

3 4 5 6 7 1 2

Time Complexity : O(n)

Please refer complete article on Reversal algorithm for array rotation for more details!



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