Javascript 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:

<script>

// JavaScript program for the above approach

   
// Function to count minimum anti-
// clockwise rotations required to
// sort the array in non-increasing order
function minMovesToSort(arr, N)
{
    
    // Stores count of arr[i + 1] > arr[i]
    let count = 0;
  
    // Store last index of arr[i+1] > arr[i]
    let index = 0;
  
    // Traverse the given array
    for(let 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 result according
    // to the following conditions
    if (count == 0) 
    {
        document.write("0");
    }
    else if (count == N - 1)
    {
        document.write(N - 1);
    }
    else if (count == 1 && 
             arr[0] <= arr[N - 1]) 
    {
        document.write(index + 1);
    }
  
    // Otherwise, it is not
    // possible to sort the array
    else
    {
        document.write("-1");
    }
}
   
// Driver Code

// Given array
let arr = [2, 1, 5, 4, 2];
let N = arr.length;
 
// Function Call
minMovesToSort(arr, N);


</script>

2

Time Complexity: O(N)
Auxiliary Space: O(1)

kartik

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Article Tags :

• Searching
• Sorting
• JavaScript
• Web Technologies
• DSA
• Arrays
• rotation

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Practice Tags :

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• Searching
• Sorting