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)
Please refer complete article on Count rotations required to sort given array in non-increasing order for more details!