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}

#include 
    
     #include 
     
      int gcd(int a,int b){    if(b==0)        return a;    else        return gcd(b, a%b);}void leftRo(int *array, int step, int len){    int i, j, k, tmp;    for (i = 0; i < gcd(step, len); i++) {        tmp = array[i];        j = i;        while (1) {            k = j + step;            if (k >= len) {                k -= len;            }            if (k == i) break;            array[j] = array[k];            j = k;        }        array[j] = tmp;    }}void printArr(int *arr, int len) {    int i;    for (i = 0; i < len; i++) {        printf("%d ", arr[i]);    }    printf("\n");}int main(int argc, char *argv[]){    int arr[12] = {1,2,3,4,5,6,7,8,9,10,11,12};    leftRo(arr, 3, 12);    printArr(arr, 12);    return 0;}