Write a function rotate(ar, d, n) that rotates arr of size n by d elements.
Rotation of the above array by 2 will make array
Initialize A = arr[0..d-1] and B = arr[d..n-1] 1) Do following until size of A is equal to size of B a) If A is shorter, divide B into Bl and Br such that Br is of same length as A. Swap A and Br to change ABlBr into BrBlA. Now A is at its final place, so recur on pieces of B. b) If A is longer, divide A into Al and Ar such that Al is of same length as B Swap Al and B to change AlArB into BArAl. Now B is at its final place, so recur on pieces of A. 2) Finally when A and B are of equal size, block swap them.
Here is iterative implementation of the same algorithm. Same utility function swap() is used here.
Time Complexity: O(n)
Please see following posts for other methods of array rotation:
Please write comments if you find any bug in the above programs/algorithms or want to share any additional information about the block swap algorithm.
- Reversal algorithm for right rotation of an array
- Reversal algorithm for array rotation
- C Program for Reversal algorithm for array rotation
- Java Program for Reversal algorithm for array rotation
- Program for array rotation
- Find the Rotation Count in Rotated Sorted array
- Print left rotation of array in O(n) time and O(1) space
- Left Rotation and Right Rotation of a String
- Check if array can be sorted with one swap
- Rotate Linked List block wise
- Sorting possible using size 3 subarray rotation
- Lexicographically minimum string rotation | Set 1
- Find a rotation with maximum hamming distance
- Maximum number of fixed points using atmost 1 swap
- Swap Kth node from beginning with Kth node from end in a Linked List