Algo #114: Rotate Array

Problem Statement:
         Rotate array by K with constraints
                           Time Complexity: O(n)
                           Space Complexity : O(1)

For K=3

Notice that 1 is poped out and pushed at the back of 10 , similar thing happened for 2 and 3 as well.

Solution: 

Our approach will consist of three steps.
1. Reverse (0 to K )
2. Reverse (K+1 to len-1)
3. Reverse Entire array.

Lets See this by fig.

Fig : Step wise transformation of array.

Implementation of the above approach:

?

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/**   * Rotate array by k   */   void rotate2(int[] arr, int k) {   if(k<=0 || arr==null || arr.length==0)    return ;   int[] old = Arrays.copyOf(arr, arr.length);   k--;       int i=0,j=k,len=arr.length;   for (; i <=k/2 ; i++,j--)    swap(arr, i, j);       int temp=  i + (len-i)/2;   for (i=k+1,j= arr.length-1; i <= temp ; i++,j--)    swap(arr, i, j);           for (i=0,j= len-1; i <len arr="" i="" j--="" j="" old="" rrays.tostring="" swap="" system.out.println="">"+ Arrays.toString(arr));  } </len>

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Happy Coding !! :)