I did something similar, but I used mod in case

K > arr.lenth + idx.

Ah. I was doing some weird modular arithmetic, which led to funky outcomes for the unknown test cases.

Can you please explain your solution ? why are you taking mod here ?

Failed for one test case.

What's the space and time efficiency for this code?

n, k, m = map(int, input().strip().split())
arr = list(map(int, input().strip().split()))
k %= n
arr = arr[-k:] + arr[:-k]
for i in range(m):
    print(arr[int(input().strip())])

Wow! Lovely Solution.. But what if the direction of rotation is left?

I have given it a try, please let me know if I am wrong:

if(idx+rot <= arr.length) System.out.println(arr[idx+rot]); else System.out.println(arr[arr.length-(idx+rot)]);

no Rotation but still does the job :v CHECK THIS out c++ lovers

im noob in coding so i wrote it so big :(

 int b,n,k,q,ans=0;
cin>>n>>k>>q;
int a[n];
for(int i=0;i<n;i++)
    cin>>a[i];
    if(k==n){
        for(int i=0;i<q;i++)
    {
     cin>>b;
     cout<<a[b]<<endl;
    }
    }
    else if(k>n)
    { while(k>n)
        k=k-n;
        for(int i=0;i<q;i++)
        {
            ans=n-k;
            cin>>b;
            ans+=b;
            if(ans>n-1)
                ans-=n;
            cout<<a[ans]<<endl;
        }
    } else if(k<n)
        {

           for(int i=0;i<q;i++)
        {
            ans=n-k;
            cin>>b;
            ans+=b;
            if(ans> n-1)
                ans-=n;
            cout<<a[ans]<<endl;
        }
        }
return 0;

}

Very elegant solution!

(\(\
(=':')
(,(")(")

You can further reduce the time. No need to rotate the array just apply simple logic and you can do it in O(1) complexity per query. Check this solution Circular Array Rotation

Thanks buddy!

How is that O(1) space? You're using an n element array.

Actually, you can do it in O(1) and O(1) space by simply using modulo.

n,k,q=raw_input().split()
n,k,q=[int(n),int(k),int(q)]
s=raw_input()
s=map(int,s.split())
while(q!=0):
    m=input()
    print s[(m-k)%n]
    q=q-1

bitch plz python rules

n,k,q=raw_input().split()
n,k,q=[int(n),int(k),int(q)]
s=raw_input()
s=map(int,s.split())
while(q!=0):
    m=input()
    print s[(m-k)%n]
    q=q-1

bitch plz python rules

so good I could cry

Wow!It is so effective compared to mine

awsome

@piyush121 I would like to understand what you did with the modulo operation and how does rot help us in solving thiis code. I am beginner so please be patient and help me.

I was faithfully doing a for loop for 1 rotation inside a dummy for loop to have k rotations. Then another for loop to output the m'th position. It worked, but O(n) became n*k; too long for 6 test cases. This comment inspired me to use modulus by n and only print the mth position numbers. I tried to understand it in my own way and shorten your code to:

System.out.println(a[(m-k+n)%n]);

Could you please explain why time complexiety is O(N) instead of O(Q)?

Hi piyush can you explain the math behind your solution that would be great

This can be easily done with a one liner O(1) time and O(1) space

cout << a[(n+(m-k)%n)%n] << endl;

You don't need to use extra variable "rot". One can overwrite K with K%N if K is larger than N. Also, there is no need to call arr.length because you already have N.

This is pretty much more elegant than my solution of using a queue, but a question that I'm having is why use k%n? Can someone please explain this logic?

Tx

good catch on the k which can be bigger than the number of elements!!

Can you guys explain to me this line

if(idx-rot>=0)
              System.out.println(arr[idx-rot]);
           else
            System.out.println(arr[idx-rot+arr.length]);

I don't understand why it can refer to the roation element of array?

arr.unshift(arr.pop())

Can anyone explain this easily from start ?

Wow, its crazy to see how yours is so similar yet so different from my code. I instead placed the integer at the index it was going to be at from the very beginning

public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int cycles = in.nextInt();
        int queries = in.nextInt();
        int[] a = new int[n];
        for(int a_i=0; a_i < n; a_i++){
            a[(a_i+cycles) % n] = in.nextInt();
        }
        for(int a0 = 0; a0 < queries; a0++){
            int m = in.nextInt();
            System.out.println(a[m]);
        }
    }

does it work in *4*th case?

Can Someone Explain me Logic Behind it pls

Thank you!!

thanks man, its awesome. :)

i got "Segmentation Fault". Why?

if (idx - rot >= 0) System.out.println(arr[idx - rot]); else System.out.println(arr[idx - rot + arr.length]); } will u plz make me understand ur logic?? how did it come to ur mind?

It's O(1) time(for each query) and O(n) space, not the other way around.

Can somebody please explain the logic ? I did it with loop for shifting array elements but it shows timeout error for large testcases.

Please explain for me why use a[(n-(k%n) +m) %n]?

can you please describe this to me:(a[(n - (k % n)+ m) % n])?

Just wanted to ask how do you came up with the one line solution to the problem.

System.out.println(a[(n - (k % n)+ m) % n]);

why do we have to use (k%n) instead of just k i.e. (n-k+m)? All TCs except for TC#4 succeed with k.

Whats the logic? Could you please explain it..

why my code gives wrong answers ...?

vector <int> circularArrayRotation(vector <int> a, vector <int> m,int n,int k, int q) 
{
        while(k!=0)
        {
            int end=a[n-1];
            for(int i=n-2;i>=0;i--)
            {
                a[i+1]=a[i];
            }
            a[0]=end;
            k--;
        }
        
    return a;
}

so, if you change the whole format of main method then how will it take the auto generated test cases?

what if the rotation was in left direction ?

Can you please explain, how did you find the solution?

shouldnt you circulate the whole array instead of manipulating the indexes to find the answer.

O(m) time and O(1) space.

def circularArrayRotation(a, k, queries):
    return [a[((len(a) - (k % len(a)))+q)%len(a)] for q in queries]

With Love, Python

where are you rotating the array?

u did amazing job...what i do...i cirular shift the whole array...can you suggest me the way...when you see a problem like this.

Nice one!
Here is my code in C.

#include <stdio.h>
#include <stdlib.h>


int main(){
    int n,k,q;
    scanf("%d %d %d",&n,&k,&q);
    int a[100000],m,i = 0;
    
    while(i < n)
        scanf("%d",&a[(i++ + k) % n]); 
    
    while(q--){
        scanf("%d",&m);
        printf("%d\n",a[m]);
    }
    return 0;
}

Position resulting because of rotation is computed first using ((i++ + k) % n).Then the array elements are stored in these positions in the read ing stage itself using scanf() . Suggestions are welcome!

Why am I getting this error when I have not even touched main() ?

Here's my function code

int i,j,arr[queries_count];
*result_count=queries_count;
k = k % a_count;
for(i=1;i<=queries_count;i++)
{
    j=queries[i]-k;
    if(j<0)
        arr[i]=a[a_count+j];
    else
        arr[i]=a[j];
}
return arr;

}

Segmentation Fault Error (stderr)

GDB trace: Reading symbols from solution...done. [New LWP 2592] Core was generated by `solution'. Program terminated with signal SIGSEGV, Segmentation fault.

0 fprintf (__fmt=0x400cc4 "%d", __stream=0x155d010)

at /usr/include/x86_64-linux-gnu/bits/stdio2.h:97

97 return __fprintf_chk (__stream, __USE_FORTIFY_LEVEL - 1, __fmt,

0 fprintf (__fmt=0x400cc4 "%d", __stream=0x155d010)

at /usr/include/x86_64-linux-gnu/bits/stdio2.h:97

1 main () at solution.c:97

Can anyone explain the maths behind this solution?

python 3 https://www.hackerrank.com/challenges/circular-array-rotation/forum/comments/514048

sir can you please help me.Whats wrong with my code ??

import java.io.; import java.math.; import java.security.; import java.text.; import java.util.; import java.util.concurrent.; import java.util.regex.*;

public class Solution {

// Complete the circularArrayRotation function below.
static int[] circularArrayRotation(int[] a, int k, int[] queries) {
    int last=a[a.length-1];
    for(int i=a.length-2;i>=0;i--)
    {
        a[i+1]=a[i];
    }
    a[0]=a[last];
    for(int i=0;i<queries.length;i++)
    {
        int c=queries[i];
        int m=a[c];
        return m;
        break;
    }


}

private static final Scanner scanner = new Scanner(System.in);

public static void main(String[] args) throws IOException {
    BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

    String[] nkq = scanner.nextLine().split(" ");

    int n = Integer.parseInt(nkq[0]);

    int k = Integer.parseInt(nkq[1]);

    int q = Integer.parseInt(nkq[2]);

    int[] a = new int[n];

    String[] aItems = scanner.nextLine().split(" ");
    scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

    for (int i = 0; i < n; i++) {
        int aItem = Integer.parseInt(aItems[i]);
        a[i] = aItem;
    }

    int[] queries = new int[q];

    for (int i = 0; i < q; i++) {
        int queriesItem = scanner.nextInt();
        scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
        queries[i] = queriesItem;
    }

    int[] result = circularArrayRotation(a, k, queries);

    for (int i = 0; i < result.length; i++) {
        bufferedWriter.write(String.valueOf(result[i]));

        if (i != result.length - 1) {
            bufferedWriter.write("\n");
        }
    }

    bufferedWriter.newLine();

    bufferedWriter.close();

    scanner.close();
}

}

It's for C (''Godspeed!")

#include <stdio.h>

int main() {
	
  int n, k, m, mx;
  scanf("%d %d %d", &n, &k, &m);
  int nums[n];
  for (int i = 0; i < n; i++)
    scanf("%d", &nums[i]);
  while (m--) {
    scanf("%d", &mx);
    printf("%d\n", nums[(n + mx - (k % n)) % n]);
  }
  return 0;
}

what this line is doing "arr[i] = in.nextInt();"

Wow. The core of the solution is two lines in Ruby (or one line if you don't mind calling a.length three times).

len = a.length
queries.collect { |i| a[((len - k) % len + i) % len] }

It may be possible to simplify my expression, but I wasn't able to see how to reduce it.

gr8 plz send your all the solutions in the discussion section thnks bro