Java Program to Count of rotations required to generate a sorted array

Given an array arr[], the task is to find the number of rotations required to convert the given array to sorted form.
Examples:

Input: arr[] = {4, 5, 1, 2, 3} 
Output: 2 
Explanation:
Sorted array {1, 2, 3, 4, 5} after 2 anti-clockwise rotations.

Input: arr[] = {2, 1, 2, 2, 2} 
Output: 1 
Explanation:
Sorted array {1, 2, 2, 2, 2} after 1 anti-clockwise rotations.

Naive Approach:
To solve the problem mentioned above the first observation is if we have n elements in the array then after sorting, the largest element is at (n – 1)^th position. After k number of anti-clockwise rotations, the largest element will be at index (k – 1) (k^th element from start). Another thing to note here is that, after rotation, the next element of the largest element will always be the smallest element, (unless the largest element is at last index, possible if there was no rotation). 
Hence,

Number of rotations (k) = index of smallest element (k) in the array

Below is the implementation of the above approach:

Java

// Java Program to find the
// count of rotations
public class GFG {

    // Function to return the count of
    // rotations
    public static int countRotation(int[] arr,
                                    int n)
    {
        for (int i = 1; i < n; i++) {
            // Find the smallest element
            if (arr[i] < arr[i - 1]) {
                // Return its index
                return i;
            }
        }
        // If array is not
        // rotated at all
        return 0;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int[] arr1 = { 4, 5, 1, 2, 3 };

        System.out.println(
            countRotation(
                arr1,
                arr1.length));
    }
}

Output:

2

Time Complexity: O(N) 
Auxiliary Space: O(1)

Efficient Approach:
To optimize the above approach, we will use Binary Search. We can notice that, after being sorted and rotated, the given array is divided into two halves with non-decreasing elements, which is the only pre-requisite for binary search. Perform a recursive binary search in the array to find the index of the smallest element.

Below is the implementation of the above approach:

Java

// Java Program to implement
// the above approach

public class GFG {

    // Function to return the
    // count of rotations
    public static int countRotation(int[] arr,
                                    int low,
                                    int high)
    {
        // If array is not rotated
        if (low > high) {
            return 0;
        }

        int mid = low + (high - low) / 2;

        // Check if current element is
        // greater than the next
        // element

        if (mid < high
            && arr[mid] > arr[mid + 1]) {
            // the next element is
            // the smallest
            return mid + 1;
        }

        // Check if current element is
        // smaller than it's previous
        // element
        if (mid > low
            && arr[mid] < arr[mid - 1]) {
            // Current element is
            // the smallest
            return mid;
        }

        // Check if current element is
        // greater than lower bound

        if (arr[mid] > arr[low]) {
            // The sequence is increasing
            // so far
            // Search for smallest
            // element on the right
            // subarray
            return countRotation(arr,
                                mid + 1,
                                high);
        }

        if (arr[mid] < arr[high]) {
            // Smallest element lies on the
            // left subarray
            return countRotation(arr,
                                low,
                                mid - 1);
        }

        else {
            // Search for the smallest
            // element on both subarrays
            int rightIndex = countRotation(arr,
                                        mid + 1,
                                        high);
            int leftIndex = countRotation(arr,
                                        low,
                                        mid - 1);

            if (rightIndex == 0) {
                return leftIndex;
            }

            return rightIndex;
        }
    }

    // Driver Program
    public static void main(String[] args)
    {
        int[] arr1 = { 4, 5, 1, 2, 3 };

        System.out.println(
            countRotation(
                arr1,
                0, arr1.length
                    - 1));
    }
}

Output:

2

Time Complexity: O(N) 
The complexity will be O(logN) for an array without duplicates. But if the array contains duplicates, then it will recursively call the search for both halves. So the worst-case complexity will be O(N).

Auxiliary Space:O(N) 
At worst case, the recursion call stack will have N/2 recursion calls at a time.

Please refer complete article on Count of rotations required to generate a sorted array for more details!