{"id":637,"date":"2020-06-10T08:09:30","date_gmt":"2020-06-10T08:09:30","guid":{"rendered":"https:\/\/blog.prepbytes.com\/?p=637"},"modified":"2022-03-28T00:40:31","modified_gmt":"2022-03-28T00:40:31","slug":"array-rotation","status":"publish","type":"post","link":"https:\/\/prepbytes.com\/blog\/array-rotation\/","title":{"rendered":"Array Rotation"},"content":{"rendered":"

\"\"<\/p>\n

CONCEPTS USED:<\/h3>\n
\n

Basic Mathematics<\/p>\n<\/blockquote>\n

DIFFICULTY LEVEL:<\/h3>\n
\n

Medium<\/p>\n<\/blockquote>\n

PROBLEM STATEMENT(<\/code>SIMPLIFIED)<\/code>:<\/h3>\n
\n

With a given array of size N<\/code> and steps K<\/code>, we have to print the array after K<\/code> rotations to the right.<\/p>\n<\/blockquote>\n

<\/strong><\/u><\/a><\/p>\n

For Example:<\/h4>\n
Input: \n\nN = 5, K = 2\nA[] = [1, 2, 3, 4, 5]\n\nOutput: \n\n[4, 5, 1, 2, 3]\n\nExplanation: \n\nWe are supposed to right rotate the array by 2 steps, so each element is shifted to the right by 2 places and elements at the end of the array will be shifted to the starting indexes of the array in a fashion similar to circular arrays.  <\/code><\/pre>\n
\n
NOTE<\/em><\/strong> : Don\u2019t forget to K<\/code> = K<\/code> % N<\/code>, as K<\/code> can be greater than N<\/code>.<\/p>\n<\/blockquote>\n
SOLVING APPROACH:<\/h3>\n
BRUTE FORCE METHOD<\/h4>\n
\n
    \n
  1. Naive solution would be to shift the array K<\/code> times one by one.<\/li>\n
  2. Write a function that shifts all the elements by one place to the right.  <\/li>\n
  3. Run a loop K<\/code> times and execute this function.<\/li>\n
  4. Time Complexity<\/code> of this solution will be O(K*N)<\/code>.<\/li>\n<\/ol>\n<\/blockquote>\n
    SOLUTIONS:<\/h3>\n\t\t\t\t\t\t