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# 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 ≤ 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 ≤ 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!

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