ARR_ROT -Editorial

editorial
,
#1

PROBLEM LINK:

Practice

Author: Kunal Demla
Editorialist: Kunal Demla

DIFFICULTY:

CAKEWALK

PREREQUISITES:

Array Indexing

PROBLEM:

Given an Array Arr and an Integer K, print the array after K steps of rotation.

QUICK EXPLANATION:

Traverse the array with j=(i+(size(Arr) -K)) \bmod size(Arr).

EXPLANATION:

It is observable that for any K>N, the apparent K will be K \bmod N.
We can also conclude for any Integer K<N, the element at index N-K comes to 1st index.
Hence we start traversing the array from same.
Using a Loop with, i=0 and i<N, we can conclude the output to be ((N-K)+i), but since the array is still unaltered and non repeating with last index at N, the output shall be made to ((N-K)+i) \bmod N.

ALTERNATE EXPLANATION:

Using another array Brr of size N, we declare Brr[(i+K) \bmod N]= Arr[i] and output Brr.

Actually rotating the whole array is also a method but its non-essential.

SOLUTIONS:

Setter's Solution 
#include <iostream>
using namespace std;
#define ll long long int
void solve()
{
    int i,j,n,k;
    cin>>n>>k;
    ll a[n];
    for(i=0;i<n;i++)
    cin>>a[i];
    k=k%n;
    for(i=0;i<n;i++)
    {
        cout<<a[((n-k+i)%(n))]<<" ";
    }
    cout<<endl;
    
}

int main() {
    int t;
    cin>>t;
    while(t--)
    {
        solve();
    }
    return 0;
}