C PROGRAM TO ROTATE THE SECOND HALF OF THE ARRAY TO THE LEFT FOR K NUMBER OF TIMES

  

Consider the following I/O samples.

SAMPLE INPUT: 5

                            1 2 3 4 5

                             5

SAMPLE OUTPUT: 1 2 5 3 4

The algorithm used here is :

1. Input the values
2. Store the half of the size of the array to a temporary variable.
3. Swap the second half of the array for K times
4. Print the output

C PROGRAM:

#include<stdio.h>

int main()

{

int arr[100],n,i,temp,rot;

scanf("%d",&n);

for(i=0;i<n;i++)

    scanf("%d",&a[i]);

scanf("%d",&rot);

while(rot --)

{

    temp = arr[n/];

    for(i=n/2;i<n-1;i++)

        arr[i]=arr[i+1];

    arr[i] = temp;

}

for(i=0;i<n;i++)

    printf("%d ",arr[i]);

return 0;

}

OUTPUT:

The given source code in C program "to rotate the second half of the array to the left for K number of times" is short and simple to understand. The source code is well tested in DEV-C++ and is completely error free. 

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