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 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 |
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!