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:
Javascript
<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!
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