Python3 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:
Python3
def minMovesToSort(arr, N) :
count = 0
index = 0
for i in range (N - 1 ):
if (arr[i] < arr[i + 1 ]) :
count + = 1
index = i
if (count = = 0 ) :
print ( "0" )
elif (count = = N - 1 ) :
print ( N - 1 )
elif (count = = 1
and arr[ 0 ] < = arr[N - 1 ]) :
print (index + 1 )
else :
print ( "-1" )
arr = [ 2 , 1 , 5 , 4 , 2 ]
N = len (arr)
minMovesToSort(arr, N)
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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!
Last Updated :
04 Feb, 2022
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