Given an array arr[] of size N (consisting of duplicates), the task is to check if the given array can be converted to a non-decreasing array by rotating it. If it’s not possible to do so, then print “No“. Otherwise, print “Yes“.

Examples:

Input: arr[] = {3, 4, 5, 1, 2}
Output: Yes
Explanation: After 2 right rotations, the array arr[] modifies to {1, 2, 3, 4, 5}

Input: arr[] = {1, 2, 4, 3}
Output: No

Approach: The idea is based on the fact that a maximum of N distinct arrays can be obtained by rotating the given array and check for each individual rotated array, whether it is non-decreasing or not. Follow the steps below to solve the problem:

• Initialize a vector, say v, and copy all the elements of the original array into it.
• Sort the vector v.
• Traverse the original array and perform the following steps:
  • Rotate by 1 in each iteration.
  • If the array becomes equal to vector v, print “Yes“. Otherwise, print “No“.

Below is the implementation of the above approach:

# Python 3 program for the above approach


# Function to check if a
# non-decreasing array can be obtained
# by rotating the original array
def rotateArray(arr, N):
  
    # Stores copy of original array
    v = arr

    # Sort the given vector
    v.sort(reverse = False)

    # Traverse the array
    for i in range(1, N + 1, 1):
      
        # Rotate the array by 1
        x = arr[N - 1]
        i = N - 1
        while(i > 0):
            arr[i] = arr[i - 1]
            arr[0] = x
            i -= 1
            
        # If array is sorted
        if (arr == v):
            print("YES")
            return

    # If it is not possible to
    # sort the array
    print("NO")

# Driver Code
if __name__ == '__main__':
  
    # Given array
    arr =  [3, 4, 5, 1, 2]

    # Size of the array
    N = len(arr)

    # Function call to check if it is possible
    # to make array non-decreasing by rotating
    rotateArray(arr, N)
    
    # This code is contributed by ipg2016107.

YES

Time Complexity: O(N²)
Auxiliary Space: O(N)

Method 2: Identify the index of the smallest element in the array using the index method, and then creates a rotated version of the array where the smallest element is the first element. It then checks if the rotated array is non-decreasing by traversing through it and comparing each element with the next element.

• First, we identify the index of the smallest element in the array using the index method.
• We then create a rotated version of the array where the smallest element is the first element. We do this by slicing the original array into two parts and concatenating them in reverse order.
• Finally, we traverse through the rotated array and check if it is non-decreasing. If we find an element that is greater than the next element, we return “NO”. Otherwise, we return “YES”.

Implementation:

def check_rotation(arr):
    n = len(arr)
    
    # Identify the index of the smallest element in the array
    idx = arr.index(min(arr))
    
    # Rotate the array to make the smallest element the first element
    rotated_arr = arr[idx:] + arr[:idx]
    
    # Check if the rotated array is non-decreasing
    for i in range(n-1):
        if rotated_arr[i] > rotated_arr[i+1]:
            return "NO"
    
    return "YES"

# Example usage
arr = [3, 4, 5, 1, 2]
result = check_rotation(arr)
print(result) # Output: NO

YES

Time complexity: O(n)
Auxiliary Space: O(n)

Please refer complete article on Modify given array to a non-decreasing array by rotation for more details!

kartik

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Article Tags :

• Arrays
• DSA
• Mathematical
• Python
• Python Programs
• Sorting
• array-rearrange
• rotation

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Practice Tags :

• Arrays
• Mathematical
• python
• Sorting