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

Circular Array Rotation

  • karthik1301 + 1 comment

    Failed for one test case.

    • anuj_pancholi_1 + 2 comments

      Test case 4? Segmentation fault?

      • karthik1301 + 1 comment

        Yes

        • anuj_pancholi_1 + 3 comments

          Here's what worked for me in C++. After taking values of n and k:

          while(k greater than n)k=k-n;

          The value of k can be several times greater than n, so by putting this can lead to referencing an index in the array greater than n, which causes segmentation fault. And the output will be correct because performing 5 rotations in an array of 4 elements is same as performing 1 rotation.

          • astromahi + 2 comments

            you could use k=k%n, what if the k is doulble or trible the size of n. subtraction always reduce one rotation.

            Consider if k = 3n => k = 3n -n => k = 2n. but in case of modular it reduces to n rotation. k = 3n => k = 3n%n => k = 0.

            • anuj_pancholi_1 + 1 comment

              If K=2n then the loop will run twice and k will be set to 0. If k is a multiple of n then the array after k rotations will be same as initial array. k=k%n could also do the trick.

              • jarvian + 1 comment

                if k is a multiple of n then won't the remainder be 0?

                • jcacooper + 0 comments

                  Yes. Which means, in effect, that no rotation should be performed upon the array, because rotating each element by k will have each of them end up right back where they started. Using 0 for k will have that effect.

            • luqmanalfarisi + 0 comments

              thank you sir.

          • jonathan_cheung1 + 1 comment

            This helped me out thanks

            • nijwm21 + 2 comments

              how it worked please can u explain me??

              • jonathan_cheung1 + 1 comment

                You can determine where in the array each element will end up with a simple calculation using the number of elements (n), the number of times you perform the operation (k), and an incrementing variable (use a for loop). Try to figure out the equation and use that answer as the array assignment. You need to use the while loop right before the array assignment because sometimes the calculation you perform will leave you with an array assignment value that is greater (hint) than the amount of array values so you just repeat the same equation until you get a valid value (something < n).

                • AhmadAliSabir + 0 comments

                  (Y)

              • Dinesh1306 + 0 comments

                Here i have explained it in detail. Circular Array Rotation

          • saintlyzero + 0 comments

            Thanks, It worked in Python. In fact, it would work in any language.

      • 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());