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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 using namespace std; // Function to return the count// of rotationsint 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 Codeint 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 rotationspublic 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 rotationsdef 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 Codeif __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 rotationsusing System;class GFG{ // Function to return the count of// rotationspublic 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 Codepublic 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


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 using namespace std; // Function to return the// count of rotationsint 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 rotationsdef 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 codeif __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 approachusing System; class GFG{ // Function to return the// count of rotationspublic 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 codepublic 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


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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