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  1. Prepare
  2. Data Structures
  3. Arrays
  4. Left Rotation
  5. Discussions

Left Rotation

  • 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.

    :)