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Count of rotations required to generate a sorted array

  • Difficulty Level : Easy
  • Last Updated : 10 Jun, 2021

Given an array arr[], the task is to find the number of rotations required to convert the given array to sorted form.
Examples:

Input: arr[] = {4, 5, 1, 2, 3} 
Output:
Explanation: 
Sorted array {1, 2, 3, 4, 5} after 2 anti-clockwise rotations.

Input: arr[] = {2, 1, 2, 2, 2} 
Output:
Explanation: 
Sorted array {1, 2, 2, 2, 2} after 1 anti-clockwise rotations.

Naive Approach:
To solve the problem mentioned above the first observation is if we have n elements in the array then after sorting, the largest element is at (n – 1)th position. After k number of anti-clockwise rotations, the largest element will be at index (k – 1) (kth element from start). Another thing to note here is that, after rotation, the next element of the largest element will always be the smallest element, (unless the largest element is at last index, possible if there was no rotation). 
Hence,

Number of rotations (k) = index of smallest element (k) in the array



Below is the implementation of the above approach:

C++




// C++ program to find the
// count of rotations
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count
// of rotations
int countRotation(int arr[], int n)
{
    for(int i = 1; i < n; i++)
    {
         
       // Find the smallest element
       if (arr[i] < arr[i - 1])
       {
           // Return its index
           return i;
       }
    }
     
    // If array is not
    // rotated at all
    return 0;
}
 
// Driver Code
int main()
{
    int arr1[] = { 4, 5, 1, 2, 3 };
    int n = sizeof(arr1) / sizeof(int);
     
    cout << countRotation(arr1, n);
}
 
// This code is contributed by jrishabh99

Java




// Java Program to find the
// count of rotations
public class GFG {
 
    // Function to return the count of
    // rotations
    public static int countRotation(int[] arr,
                                    int n)
    {
        for (int i = 1; i < n; i++) {
            // Find the smallest element
            if (arr[i] < arr[i - 1]) {
                // Return its index
                return i;
            }
        }
        // If array is not
        // rotated at all
        return 0;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr1 = { 4, 5, 1, 2, 3 };
 
        System.out.println(
            countRotation(
                arr1,
                arr1.length));
    }
}

Python3




# Python3 program to find the
# count of rotations
 
# Function to return the count
# of rotations
def countRotation(arr, n):
     
    for i in range (1, n):
         
        # Find the smallest element
        if (arr[i] < arr[i - 1]):
             
            # Return its index
            return i
     
    # If array is not
    # rotated at all
    return 0
 
# Driver Code
if __name__ == "__main__":
     
    arr1 = [ 4, 5, 1, 2, 3 ]
    n = len(arr1)
     
    print(countRotation(arr1, n))
 
# This code is contributed by chitranayal

C#




// C# program to find the count of rotations
using System;
class GFG{
 
// Function to return the count of
// rotations
public static int countRotation(int[] arr,
                                int n)
{
    for(int i = 1; i < n; i++)
    {
         
    // Find the smallest element
    if (arr[i] < arr[i - 1])
    {
             
        // Return its index
        return i;
    }
    }
     
    // If array is not
    // rotated at all
    return 0;
}
 
// Driver Code
public static void Main(String[] args)
{
    int[] arr1 = { 4, 5, 1, 2, 3 };
 
    Console.WriteLine(countRotation(arr1,
                                    arr1.Length));
}
}
 
// This code is contributed by gauravrajput1

Javascript




<script>
 
// JavaScript program to find the
// count of rotations
 
// Function to return the count of
// rotations
function countRotation(arr, n)
{
    for(let i = 1; i < n; i++)
    {
         
        // Find the smallest element
        if (arr[i] < arr[i - 1])
        {
             
            // Return its index
            return i;
        }
    }
     
    // If array is not
    // rotated at all
    return 0;
}
 
// Driver Code
let arr1 = [ 4, 5, 1, 2, 3 ];
 
document.write(countRotation(
    arr1, arr1.length));
     
// This code is contributed by sanjoy_62
 
</script>
Output: 
2

 

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

Efficient Approach: 
To optimize the above approach, we will use Binary Search. We can notice that, after being sorted and rotated, the given array is divided into two halves with non-decreasing elements, which is the only pre-requisite for binary search. Perform a recursive binary search in the array to find the index of the smallest element.

Below is the implementation of the above approach:

C++




// C++ program to implement the
// above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the
// count of rotations
int countRotation(int arr[], int low,
                             int high)
{
     
    // If array is not rotated
    if (low > high)
    {
        return 0;
    }
 
    int mid = low + (high - low) / 2;
 
    // Check if current element is
    // greater than the next
    // element
    if (mid < high && arr[mid] > arr[mid + 1])
    {
         
        // The next element is
        // the smallest
        return mid + 1;
    }
 
    // Check if current element is
    // smaller than it's previous
    // element
    if (mid > low && arr[mid] < arr[mid - 1])
    {
         
        // Current element is
        // the smallest
        return mid;
    }
 
    // Check if current element is
    // greater than lower bound
    if (arr[mid] > arr[low])
    {
         
        // The sequence is increasing
        // so far
        // Search for smallest
        // element on the right
        // subarray
        return countRotation(arr, mid + 1,
                             high);
    }
 
    if (arr[mid] < arr[high])
    {
         
        // Smallest element lies on the
        // left subarray
        return countRotation(arr, low,
                             mid - 1);
    }
    else
    {
         
        // Search for the smallest
        // element on both subarrays
        int rightIndex = countRotation(arr,
                                       mid + 1,
                                       high);
        int leftIndex = countRotation(arr, low,
                                      mid - 1);
        if (rightIndex == 0)
        {
            return leftIndex;
        }
        return rightIndex;
    }
}
 
// Driver code   
int main()
{
    int arr1[] = { 4, 5, 1, 2, 3 };
    int N = sizeof(arr1) / sizeof(arr1[0]);
     
    cout << countRotation(arr1, 0, N - 1);
     
    return 0;
}
 
// This code is contributed by divyeshrabadiya07

Java




// Java Program to implement
// the above approach
 
public class GFG {
 
    // Function to return the
    // count of rotations
    public static int countRotation(int[] arr,
                                    int low,
                                    int high)
    {
        // If array is not rotated
        if (low > high) {
            return 0;
        }
 
        int mid = low + (high - low) / 2;
 
        // Check if current element is
        // greater than the next
        // element
 
        if (mid < high
            && arr[mid] > arr[mid + 1]) {
            // the next element is
            // the smallest
            return mid + 1;
        }
 
        // Check if current element is
        // smaller than it's previous
        // element
        if (mid > low
            && arr[mid] < arr[mid - 1]) {
            // Current element is
            // the smallest
            return mid;
        }
 
        // Check if current element is
        // greater than lower bound
 
        if (arr[mid] > arr[low]) {
            // The sequence is increasing
            // so far
            // Search for smallest
            // element on the right
            // subarray
            return countRotation(arr,
                                mid + 1,
                                high);
        }
 
        if (arr[mid] < arr[high]) {
            // Smallest element lies on the
            // left subarray
            return countRotation(arr,
                                low,
                                mid - 1);
        }
 
        else {
            // Search for the smallest
            // element on both subarrays
            int rightIndex = countRotation(arr,
                                        mid + 1,
                                        high);
            int leftIndex = countRotation(arr,
                                        low,
                                        mid - 1);
 
            if (rightIndex == 0) {
                return leftIndex;
            }
 
            return rightIndex;
        }
    }
 
    // Driver Program
    public static void main(String[] args)
    {
        int[] arr1 = { 4, 5, 1, 2, 3 };
 
        System.out.println(
            countRotation(
                arr1,
                0, arr1.length
                    - 1));
    }
}

Python3




# Python3 program to implement the
# above approach
 
# Function to return the
# count of rotations
def countRotation(arr, low, high):
 
    # If array is not rotated
    if (low > high):
        return 0
 
    mid = low + (high - low) // 2
 
    # Check if current element is
    # greater than the next
    # element
    if (mid < high and arr[mid] > arr[mid + 1]):
 
        # The next element is
        # the smallest
        return mid + 1
 
    # Check if current element is
    # smaller than it's previous
    # element
    if (mid > low and arr[mid] < arr[mid - 1]):
 
        # Current element is
        # the smallest
        return mid
 
    # Check if current element is
    # greater than lower bound
    if (arr[mid] > arr[low]):
 
        # The sequence is increasing
        # so far
        # Search for smallest
        # element on the right
        # subarray
        return countRotation(arr, mid + 1, high)
 
    if (arr[mid] < arr[high]):
 
        # Smallest element lies on the
        # left subarray
        return countRotation(arr, low, mid - 1)
 
    else:
 
        # Search for the smallest
        # element on both subarrays
        rightIndex = countRotation(arr,
                                   mid + 1,
                                   high)
        leftIndex = countRotation(arr, low,
                                  mid - 1)
         
        if (rightIndex == 0):
            return leftIndex
 
        return rightIndex
 
# Driver code
if __name__ == '__main__':
     
    arr1 = [ 4, 5, 1, 2, 3 ]
    N = len(arr1)
 
    print(countRotation(arr1, 0, N - 1))
 
# This code is contributed by mohit kumar 29

C#




// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to return the
// count of rotations
public static int countRotation(int[] arr,
                                int low,
                                int high)
{
 
    // If array is not rotated
    if (low > high)
    {
        return 0;
    }
 
    int mid = low + (high - low) / 2;
 
    // Check if current element is
    // greater than the next
    // element
    if (mid < high &&
        arr[mid] > arr[mid + 1])
    {
         
        // The next element is
        // the smallest
        return mid + 1;
    }
 
    // Check if current element is
    // smaller than it's previous
    // element
    if (mid > low &&
        arr[mid] < arr[mid - 1])
    {
         
        // Current element is
        // the smallest
        return mid;
    }
 
    // Check if current element is
    // greater than lower bound
    if (arr[mid] > arr[low])
    {
         
        // The sequence is increasing
        // so far
        // Search for smallest
        // element on the right
        // subarray
        return countRotation(arr,
                             mid + 1,
                             high);
    }
 
    if (arr[mid] < arr[high])
    {
         
        // Smallest element lies on the
        // left subarray
        return countRotation(arr, low,
                             mid - 1);
    }
     
    else
    {
         
        // Search for the smallest
        // element on both subarrays
        int rightIndex = countRotation(arr,
                                       mid + 1,
                                       high);
        int leftIndex = countRotation(arr, low,
                                      mid - 1);
 
        if (rightIndex == 0)
        {
            return leftIndex;
        }
        return rightIndex;
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int[] arr1 = { 4, 5, 1, 2, 3 };
 
    Console.WriteLine(countRotation(arr1, 0,
                            arr1.Length - 1));
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
// Javascript Program to implement
// the above approach
  
 
// Function to return the
    // count of rotations
function countRotation(arr,low,high)
{
 
    // If array is not rotated
        if (low > high) {
            return 0;
        }
  
        let mid = low + Math.floor((high - low) / 2);
  
        // Check if current element is
        // greater than the next
        // element
  
        if (mid < high
            && arr[mid] > arr[mid + 1]) {
            // the next element is
            // the smallest
            return mid + 1;
        }
  
        // Check if current element is
        // smaller than it's previous
        // element
        if (mid > low
            && arr[mid] < arr[mid - 1]) {
            // Current element is
            // the smallest
            return mid;
        }
  
        // Check if current element is
        // greater than lower bound
  
        if (arr[mid] > arr[low]) {
            // The sequence is increasing
            // so far
            // Search for smallest
            // element on the right
            // subarray
            return countRotation(arr,
                                mid + 1,
                                high);
        }
  
        if (arr[mid] < arr[high])
        {
         
            // Smallest element lies on the
            // left subarray
            return countRotation(arr,
                                low,
                                mid - 1);
        }
  
        else
        {
         
            // Search for the smallest
            // element on both subarrays
            let rightIndex = countRotation(arr,
                                        mid + 1,
                                        high);
            let leftIndex = countRotation(arr,
                                        low,
                                        mid - 1);
  
            if (rightIndex == 0) {
                return leftIndex;
            }
  
            return rightIndex;
        }
}
 
 // Driver Program
let arr1=[4, 5, 1, 2, 3 ];
document.write(
            countRotation(
                arr1,
                0, arr1.length
                    - 1));
 
// This code is contributed by avanitrachhadiya2155
</script>
Output: 
2

 

Time Complexity: O(N) 
The complexity will be O(logN) for an array without duplicates. But if the array contains duplicates, then it will recursively call the search for both halves. So the worst-case complexity will be O(N).

Auxiliary Space:O(N) 
At worst case, the recursion call stack will have N/2 recursion calls at a time.
 

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