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:
Python3
# Python program for the above approach # Function to count minimum anti-# clockwise rotations required to# sort the array in non-increasing orderdef minMovesToSort(arr, N) : # Stores count of arr[i + 1] > arr[i] count = 0 # Store last index of arr[i+1] > arr[i] index = 0 # Traverse the given array for i in range(N-1): # If the adjacent elements are # in increasing order if (arr[i] < arr[i + 1]) : # Increment count count += 1 # Update index index = i # Print result according # to the following conditions if (count == 0) : print("0") elif (count == N - 1) : print( N - 1) elif (count == 1 and arr[0] <= arr[N - 1]) : print(index + 1) # Otherwise, it is not # possible to sort the array else : print("-1") # Driver Code# Given arrayarr = [ 2, 1, 5, 4, 2 ]N = len(arr) # Function CallminMovesToSort(arr, N)# This code i contributed by sanjoy_62. |
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!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
