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  1. Practice
  2. Algorithms
  3. Implementation
  4. Circular Array Rotation
  5. Discussions

Circular Array Rotation

  • AhmadAliSabir + 7 comments
    int n,k,q,temp,m,z;
cin>>n>>k>>q;

int a[n];

for(int x=0;x<n;x++)
{
    cin>>a[x];
}

k=k%n;
for(int y=0;y<k;y++)
{
   temp = a[n-1];

   for(int q=n-1;q>=1;q--)
   {
       a[q]=a[q-1];
   }

    a[0]=temp;
}
for(int u=0;u<q;u++)
{
   cin>>m;
   cout<<a[m]<<endl;                
}

    "Terminated due to timed out" on Case 5,12,13,14 ... help please .. Thanks !

    • astromahi + 1 comment

      try to solve in O(n).

    • rjdp3 + 1 comment

      I was using the nested loop algo, was getting timed out. Then i used this (JAVA) Collections.rotate(Arrays.asList(array_name), k);

      i am a beginner, can anyone tell me is this a bad way to solve a problem, if yes why?

      • Stephen26 + 0 comments

        The Actual vision of the problem is to solve it by without actually making any rotations.because rotation takes more time and space.thats why you may get timeout errors if you try to solve it by rotations .we gotta have to solve it by doing calculations .Example: write a simple logic to determine what will be the output after k rotations.if K is equal to N then no rotations needed coz the array will be the same as original after rotations.

    • kumarnikhil104 + 0 comments

      hey ,,if u got the solution of your code please inform me also...thanxx;

    • Dinesh1306 + 1 comment

      No need to rotate the array you can do this problem much faster. Take a look here, i have explained it in detail. Circular Array Rotation

      Hope it helped you

      • poonam2992 + 1 comment

        hi @Dinesh1306, I tried your solution, but I get this error when I try for large inputs as given in test cases: Sorry, we can't accept your submission. The custom input size should not exceed 50Kb.

        • Dinesh1306 + 1 comment

          hi @poon12, the solution is working fine, you must have done something wrong. Can you post your code here so that i can check.

          • poonam2992 + 0 comments

            hi @Dinesh1306 your solution got accepted. thanks :) but yes when i try the solution with large inputs, it gives me that custom input size error.

    • ashishgulati21 + 2 comments

      Same for me :/ someone please help

      int main() { int n,k,q,r=0; cin >>n>>k>>q;

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

      }

      • Dinesh1306 + 1 comment

        Check this

        • chikithapaleti99 + 0 comments
          int main() {
   long int n; 
   long  int k; 
    long  int q,i,t; 
    scanf("%li %li %li", &n, &k, &q);
    long int a [n];
    for (long int a_i = 0; a_i < n; a_i++) {
       scanf("%li",&a[a_i]);
    }
    long int *m = malloc(sizeof(long int) * q);
    for (long int m_i = 0; m_i < q; m_i++) {
       scanf("%li",&m[m_i]);
    }
   long int b[n];
    for(i=0;i<n;i++){
        t=i+k;
      t=t%n;
        b[t]=a[i];
        
    }
     for (long  int m_i = 0; m_i < q; m_i++) {
         printf("%ld\n",b[m[m_i]]);
     }
    return 0;
}
      • chikithapaleti99 + 0 comments
        [deleted]
    • gozer_goose + 0 comments

      You can make another array and use modular arithematic approach to insert values in that array. In this way you will solve it in O(n) and there will be no timeout errors :D

    • teodora_vasilas + 0 comments

      i used this code insted of the second for:

      rotate(a.begin(),a.end()-k,a.end());