Is it the efficeint program to rotate array in left direction? - c

Question

#include<stdio.h>
#include<stdlib.h>
main()
{
int i,j,l,m,n;
j=0;
printf("\nenter 5 element single dimension array\n");
printf("enter shift rate\n");
scanf("%d",&n);
/* Here we take input from user that by what times user wants to rotate the array in left. */
int arr[5],arrb[n];
for(i=0;i<=4;i++){
scanf("%d",&arr[i]);
}
/* Here we have taken another array. */
for(i=0;i<=4;i++){
printf("%d",arr[i]);
}
for(i=0;i<n;i++){
arrb[j]=arr[i];
j++;
// These loop will shift array element to left by position which's entered by user.
}
printf("\n");
for(i=0;i<=3;i++){
arr[i]=arr[i+n];
}
for(i=0;i<=4;i++){
if(n==1 && i==4)
break;
if(n==2 && i==3)
break;
if(n==3 && i==2)
break;
printf("%d",arr[i]);
}
//To combine these two arrays. Make it look like single array instead of two
for(i=0;i<n;i++){
printf("%d",arrb[i]);
}
// Final sorted array will get printed here
}

Answer 1

Is it the efficeint program to rotate array in left direction?
Actually, very complicated, and some problems contained:
for(i = 0; i < n; i++)
{
arrb[j] = arr[i];
j++;
}
Why not simply:
for(i = 0; i < n; i++)
{
arrb[i] = arr[i];
}
There is no need for a second variable. Still, if n is greater than five, you get into trouble, as you will access arr out of its bounts (undefined behaviour!). At least, you should check the user input!
for(i = 0; i <=3 ; i++)
{
arr[i] = arr[i + n];
}
Same problem: last accessible index is 4 (four), so n must not exceed 1, or you again access the array out of bounds...
Those many 'if's within the printing loop for the first array cannot be efficient...
You can have it much, much simpler:
int arr[5], arrb[5];
// ^
for(int i = 0; i < 5; ++i)
arrb[i] = arr[(i + n) % 5];
This does not cover negative values of n, though.
arrb[i] = arr[(((i + n) % 5) + 5) % 5];
would be safe even for negative values... All you need now for the output is:
for(int i = 0; i < 5; ++i)
printf("%d ", arrb[i]);
There would be one last point uncovered, though: if user enters for n a value greater than INT_MAX - 4, you get a signed integer overflow, which again is undefined behaviour!
We can again cover this by changing the index formula:
arrb[i] = arr[(5 + i + (n % 5)) % 5];
n % 5 is invariant, so we can move it out of the loop:
n %= 5;
for(int i = 0; i < 5; ++i)
arrb[i] = arr[(5 + i + n) % 5];
Finally, if we make n positive already outside, we can spare the addition in the for loop.
n = ((n % 5) + 5) % 5;
for(int i = 0; i < 5; ++i)
arrb[i] = arr[(i + n) % 5]; // my original formula again...
Last step is especially worth considering for very long running loops.

Answer 2

I think you want to do something like this (you should check that 0 <= n <= 5, too):
int b[5];
int k = 0;
for(i=0; i<5; i++){
if (i < 5 - n)
b[i] = arr[i+n];
else
{
b[i] = arr[k];
k++;
}
}
Array b is used to save the rotated matrix.