Java Program to Count rotations required to sort given array in non-increasing order

Given an array arr[] consisting of N integers, the task is to sort the array in non-increasing order by minimum number of anti-clockwise rotations. If it is not possible to sort the array, then print “-1”. Otherwise, print the count of rotations.

Examples:

Input: arr[] = {2, 1, 5, 4, 3}
Output: 2
Explanation: Two anti-clockwise rotations are required to sort the array in decreasing order, i.e. {5, 4, 3, 2, 1}

Input: arr[] = {2, 3, 1}
Output: -1

Approach: The idea is to traverse the given array arr[] and count the number of indices satisfying arr[i + 1] > arr[i]. Follow the steps below to solve the problem:

• Store the count of arr[i + 1] > arr[i] in a variable and also store the index when arr[i+1] > arr[i].
• If the value of count is N – 1, then the array is sorted in non-decreasing order. The required steps are exactly (N – 1).
• If the value of count is 0, then the array is already sorted in non-increasing order.
• If the value of count is 1 and arr[0] ≤ arr[N – 1], then the required number of rotations is equal to (index + 1), by performing shifting of all the numbers upto that index. Also, check if arr[0] ≤ arr[N – 1] to ensure if the sequence is non-increasing.
• Otherwise, it is not possible to sort the array in non-increasing order.

Below is the implementation of the above approach:

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to count minimum anti-
// clockwise rotations required to
// sort the array in non-increasing order
static void minMovesToSort(int arr[], int N)
{

    // Stores count of arr[i + 1] > arr[i]
    int count = 0;

    // Store last index of arr[i+1] > arr[i]
    int index = 0;

    // Traverse the given array
    for(int i = 0; i < N - 1; i++)
    {

        // If the adjacent elements are
        // in increasing order
        if (arr[i] < arr[i + 1])
        {

            // Increment count
            count++;

            // Update index
            index = i;
        }
    }

    // Print the result according
    // to the following conditions
    if (count == 0)
    {
        System.out.print("0");
    }
    else if (count == N - 1)
    {
        System.out.print(N - 1);
    }
    else if (count == 1 &&
            arr[0] <= arr[N - 1])
    {
        System.out.print(index + 1);
    }

    // Otherwise, it is not
    // possible to sort the array
    else
    {
        System.out.print("-1");
    }
}

// Driver Code
public static void main(String[] args)
{

    // Given array
    int[] arr = { 2, 1, 5, 4, 2 };
    int N = arr.length;

    // Function Call
    minMovesToSort(arr, N);
}
}

// This code is contributed by susmitakundugoaldanga

Output:

2

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

Please refer complete article on Count rotations required to sort given array in non-increasing order for more details!