We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Practice
  2. Data Structures
  3. Arrays
  4. Left Rotation
  5. Discussions

Left Rotation

  • atulb7777 + 2 comments

    int main() {

     int n,d,f,j,i,*ar;
cin>>n>>d;
ar=new int[n];
for(i=0;i<n;i++){
    cin>>ar[i];
}
for(i=1;i<=d;i++){
    f=ar[0];
    for(j=0;j<n-1;j++){
        ar[j]=ar[j+1];
    }
    ar[n-1]=f;
}
for(i=0;i<n;i++){
    cout<<ar[i];
    if(i<(n-1)){
        cout<<" ";
    }
}
return 0;

    } test case#8 is showing "terminated due to timeout"! Any suggestions? i am new to this!

    • rohan565singh + 2 comments

      I have the exactly same problem. Did anyone figure out why?

      • Alexbrc + 0 comments

        I don't particularly know what test case 8 is exactly, but I would assume that this test case has set the input parameters of n and d to be very large. If you look at the constraints on n, then you will see that n can be as large as 10,000 and d also can be as large as n. With an approach that manually shifts the array one by one a certain amount of times can lead to a lot of work. Possibly shifting all 10,000 elements to the left 10,000 times racks up to a lot of work. You would have to shift something up to 10,000,000,000 times in this scenario. That is a lot of work and would most likely be the reason the test case times out before finishing.

      • atulb7777 + 3 comments

        I did solve it! It was proably beacuse of the nested loops! they take more time to run!

        • roGue_warRior + 0 comments

          which is known as complexity

        • noahlbw + 1 comment

          how did you solve it ?

          • roGue_warRior + 2 comments

            Its easy man i cant figure out the last test case though !!

            1.Input an array of size n 2.Input number of rotations as r 3. r=r%n -> it reduces the number of repeatative rotations thus decreasing the complexity of our code . suppose the size of the array is 3 and number of rotations is also 3 then our loop wont run it will simply print the same elements .

            1. take a temperory variable and store the leftmost element in it i.e a[o]
            2. Now run the loop from 0 to second last element of the array 6.a[i]=a[i+1] -> shifts the element towards left for eg. for i=0 a[0]=a[1]; it shifts all the elements towards left 7.now simply store the first element in temp variable to the last position.
            3. steps 4 to 7 for first rotation now while loop will run as per the given number of rotations these steps will be repeated until r becomes 0.
            4. At last display the elements in the array.

            :)

            • pratik9044536615 + 1 comment

              Man i have better version of solution. See my code.

              vector array_left_rotation(vector a, int n, int k) { rotate(a.begin(),a.begin()+k,a.end()); return a; }

              int main(){ int n; int k; cin >> n >> k; vector a(n); for(int a_i = 0;a_i < n;a_i++){ cin >> a[a_i]; } vector output = array_left_rotation(a, n, k); for(int i = 0; i < n;i++) cout << output[i] << " "; cout << endl; return 0; }

              • shubh_aagman24 + 0 comments

                chup.. space dedeta kuch to

            • guleriasahil86 + 0 comments

              Doesn't work for 8th test case

        • shockwave97 + 1 comment

          how did u solve??

          • shafagh + 1 comment

            Keep in mind a left rotation of x is equal to (len-x) of righ rotations. Hence, it would be more efficient to the rotation with least amount of "moves".

            • crystalcode + 1 comment

              Thats true. I got accepted for test case 8 using right rotation. We need to choose :

              If #rotations < (size /2) Use Left Rotation If #rotations > (size /2) Use Right Rotation

              • RA16110080543 + 0 comments

                can you elaborate @crystalcode

    • sankeyy + 1 comment

      just type,
      for i=0 to n-1
      print a [ ( i + d ) %n ].

      thats all..