Rotate an array by k steps in c++ - c++

Question

We have given an array of n size and we to have to rotate it by k times where k can be greater than n in some cases .
My try -
#include<iostream>
using namespace std;
void inputarray(int arr[],int size){
for(int i=0;i<size;i++){
cin>>arr[i];
}
}
int main(){
int n;
cin>>n;
int arr[100];
inputarray(arr,n);
int ansarr[n];
int k;
cin>>k;
k = k%n;
int j=0;
for(int i =n-k;i<n;i++){
ansarr[j++] = arr[i];
}
for(int i=0;i<=k;i++){
ansarr[j++]=arr[i];
}
for(int i=0;i<n;i++){
cout<<ansarr[i];
}
}
My output is coming correct for k>2 like my output for k=3 is -
6
1 2 3 4 5 6
3
456123
Which is correct but for k <2 like k = 2 , 1 and 0 my output is not coming correct . Like for K=2 my output is -
6
1 2 3 4 5 6
2
561234199699
So where am I doing wrong can anyone please tell .

Answer 1

You can perform the rotation with a single loop, and this is completely bulletproof, using
for (int i= 0; i < n; i++)
{
ans[i]= arr[(i + k) % n];
}
Now you can eliminate the modulo inside the loop by noting that i + k remains in [0, n) for i in [- k, n - k), thus in [0, n - k), and i + k - n remains in [0, n) for i in [n - k, 2n - k), thus in [n - k, n). So you can rewrite with two successive loops
k= k % n;
int i;
for (i= 0; i < n - k; i++)
{
ans[i]= arr[i + k];
}
for ( ; i < n ; i++)
{
ans[i]= arr[i + k - n];
}
Caution: due to the behavior of the operator %, the above code will fail for k < 0.

Answer 2

For starters pay attention to that variable length arrays like this
int ansarr[n];
are not a standard C++ feature. You should declare the array similarly to the array arr
int ansarr[100];
Or instead of arrays you could use standard container std::vector.
In any case your code is wrong at least because this for loop
for(int i =n-k;i<n;i++){
ansarr[j++] = arr[i];
}
should start from i equal to k instead of i = n - k
for(int i = k;i<n;i++){
ansarr[j++] = arr[i];
}
and correspondingly the second for loop should look like
for(int i=0;i < k;i++){
ansarr[j++]=arr[i];
}
That is the condition of the loop should look like i < k.
Your program works for k equal to 3 because the expression n - k when n is equal to 6 yields the same value 3.
Bear in mind that there is standard algorithm std::reverse_copy declared in the header <algorithm> that you could use.
Here is a demonstration program
#include <iostream>
#include <algorithm>
int main()
{
const size_t N = 6;
for (size_t k = 0; k < N; k++)
{
int a[N] = { 1, 2, 3, 4, 5, 6 };
int b[N];
std::rotate_copy( a, a + k, a + N, b );
std::cout << k << ":";
for (size_t i = 0; i < N; i++)
{
std::cout << ' ' << b[i];
}
std::cout << '\n';
}
}
The program output is
0: 1 2 3 4 5 6
1: 2 3 4 5 6 1
2: 3 4 5 6 1 2
3: 4 5 6 1 2 3
4: 5 6 1 2 3 4
5: 6 1 2 3 4 5
As for your approach with for loop then it is enough to use only one for loop. For example
#include <iostream>
int main()
{
const size_t N = 6;
for (size_t k = 0; k < N; k++)
{
int a[N] = { 1, 2, 3, 4, 5, 6 };
int b[N];
for (size_t j = 0, i = k; j < N; i = ( i + 1 ) % N)
{
b[j++] = a[i];
}
std::cout << k << ":";
for (size_t i = 0; i < N; i++)
{
std::cout << ' ' << b[i];
}
std::cout << '\n';
}
}
The program output is the same as shown above
0: 1 2 3 4 5 6
1: 2 3 4 5 6 1
2: 3 4 5 6 1 2
3: 4 5 6 1 2 3
4: 5 6 1 2 3 4
5: 6 1 2 3 4 5

Answer 3

I'm not sure how either the question description or the current answers are "rotate it by k times," where "it" is the give array. They all seem to create a new array.
To rotate an array by k, first take k modulo the array length, then perform three steps (for example, {1, 2, 3, 4, 5}, k = 3):
Reverse the part up to index k:
{3, 2, 1, 4, 5}
Reverse the part from index k to the end:
{3, 2, 1, 5, 4}
Reverse the whole array:
{4, 5, 1, 2, 3}