Consider an array of distinct numbers sorted in increasing order. The array has been rotated (clockwise) k number of times. Given such an array, find the value of k.

**Examples:**

Input : arr[] = {15, 18, 2, 3, 6, 12} Output: 2 Explanation : Initial array must be {2, 3, 6, 12, 15, 18}. We get the given array after rotating the initial array twice. Input : arr[] = {7, 9, 11, 12, 5} Output: 4 Input: arr[] = {7, 9, 11, 12, 15}; Output: 0

**Method 1 (Using linear search)**

If we take closer look at examples, we can notice that the number of rotations is equal to index of minimum element. A simple linear solution is to find minimum element and returns its index. Below is C++ implementation of the idea.

## C++

// C++ program to find number of rotations // in a sorted and rotated array. #include<bits/stdc++.h> using namespace std; // Returns count of rotations for an array which // is first sorted in ascending order, then rotated int countRotations(int arr[], int n) { // We basically find index of minimum // element int min = arr[0], min_index; for (int i=0; i<n; i++) { if (min > arr[i]) { min = arr[i]; min_index = i; } } return min_index; } // Driver code int main() { int arr[] = {15, 18, 2, 3, 6, 12}; int n = sizeof(arr)/sizeof(arr[0]); cout << countRotations(arr, n); return 0; }

## Java

// Java program to find number of // rotations in a sorted and rotated // array. import java.util.*; import java.lang.*; import java.io.*; class LinearSearch { // Returns count of rotations for an // array which is first sorted in // ascending order, then rotated static int countRotations(int arr[], int n) { // We basically find index of minimum // element int min = arr[0], min_index = -1; for (int i = 0; i < n; i++) { if (min > arr[i]) { min = arr[i]; min_index = i; } } return min_index; } // Driver program to test above functions public static void main (String[] args) { int arr[] = {15, 18, 2, 3, 6, 12}; int n = arr.length; System.out.println(countRotations(arr, n)); } } // This code is contributed by Chhavi

## Python3

# Python3 program to find number # of rotations in a sorted and # rotated array. # Returns count of rotations for # an array which is first sorted # in ascending order, then rotated def countRotations(arr, n): # We basically find index # of minimum element min = arr[0] for i in range(0, n): if (min > arr[i]): min = arr[i] min_index = i return min_index; # Driver code arr = [15, 18, 2, 3, 6, 12] n = len(arr) print(countRotations(arr, n)) # This code is contributed by Smitha Dinesh Semwal

## C#

// c# program to find number of // rotations in a sorted and rotated // array. using System; class LinearSearch { // Returns count of rotations for an // array which is first sorted in // ascending order, then rotated static int countRotations(int []arr, int n) { // We basically find index of minimum // element int min = arr[0], min_index = -1; for (int i = 0; i < n; i++) { if (min > arr[i]) { min = arr[i]; min_index = i; } } return min_index; } // Driver program to test above functions public static void Main () { int []arr = {15, 18, 2, 3, 6, 12}; int n = arr.Length; Console.WriteLine(countRotations(arr, n)); } } // This code is contributed by vt_m.

## PHP

<?php // PHP program to find number // of rotations in a sorted // and rotated array. // Returns count of rotations // for an array which is first // sorted in ascending order, // then rotated function countRotations($arr, $n) { // We basically find index // of minimum element $min = $arr[0]; $min_index; for ($i = 0; $i < $n; $i++) { if ($min > $arr[$i]) { $min = $arr[$i]; $min_index = $i; } } return $min_index; } // Driver code $arr = array(15, 18, 2, 3, 6, 12); $n = sizeof($arr); echo countRotations($arr, $n); // This code is contributed // by ajit ?>

**Output:**

2

**Time Complexity :** O(n)

**Auxiliary Space :** O(1)

**Method 2 (Efficient Using Binary Search)**

Here are also we find index of minimum element, but using Binary Search. The idea is based on below facts :

- The minimum element is the only element whose previous is greater than it. If there is no previous element element, then there is no rotation (first element is minimum). We check this condition for middle element by comparing it with (mid-1)’th and (mid+1)’th elements.
- If minimum element is not at middle (neither mid nor mid + 1), then minimum element lies in either left half or right half.
- If middle element is smaller than last element, then the minimum element lies in left half
- Else minimum element lies in right half.

Below is the implementation taken from here.

## C++

// Binary Search based C++ program to find number // of rotations in a sorted and rotated array. #include<bits/stdc++.h> using namespace std; // Returns count of rotations for an array which // is first sorted in ascending order, then rotated int countRotations(int arr[], int low, int high) { // This condition is needed to handle the case // when array is not rotated at all if (high < low) return 0; // If there is only one element left if (high == low) return low; // Find mid int mid = low + (high - low)/2; /*(low + high)/2;*/ // Check if element (mid+1) is minimum element. // Consider the cases like {3, 4, 5, 1, 2} if (mid < high && arr[mid+1] < arr[mid]) return (mid+1); // Check if mid itself is minimum element if (mid > low && arr[mid] < arr[mid - 1]) return mid; // Decide whether we need to go to left half or // right half if (arr[high] > arr[mid]) return countRotations(arr, low, mid-1); return countRotations(arr, mid+1, high); } // Driver code int main() { int arr[] = {15, 18, 2, 3, 6, 12}; int n = sizeof(arr)/sizeof(arr[0]); cout << countRotations(arr, 0, n-1); return 0; }

## Java

// Java program to find number of // rotations in a sorted and rotated // array. import java.util.*; import java.lang.*; import java.io.*; class BinarySearch { // Returns count of rotations for an array // which is first sorted in ascending order, // then rotated static int countRotations(int arr[], int low, int high) { // This condition is needed to handle // the case when array is not rotated // at all if (high < low) return 0; // If there is only one element left if (high == low) return low; // Find mid // /*(low + high)/2;*/ int mid = low + (high - low)/2; // Check if element (mid+1) is minimum // element. Consider the cases like // {3, 4, 5, 1, 2} if (mid < high && arr[mid+1] < arr[mid]) return (mid + 1); // Check if mid itself is minimum element if (mid > low && arr[mid] < arr[mid - 1]) return mid; // Decide whether we need to go to left // half or right half if (arr[high] > arr[mid]) return countRotations(arr, low, mid - 1); return countRotations(arr, mid + 1, high); } // Driver program to test above functions public static void main (String[] args) { int arr[] = {15, 18, 2, 3, 6, 12}; int n = arr.length; System.out.println(countRotations(arr, 0, n-1)); } } // This code is contributed by Chhavi

## Python3

# Binary Search based Python3 # program to find number of # rotations in a sorted and # rotated array. # Returns count of rotations for # an array which is first sorted # in ascending order, then rotated def countRotations(arr,low, high): # This condition is needed to # handle the case when array # is not rotated at all if (high < low): return 0 # If there is only one # element left if (high == low): return low # Find mid # (low + high)/2 mid = low + (high - low)/2; mid = int(mid) # Check if element (mid+1) is # minimum element. Consider # the cases like {3, 4, 5, 1, 2} if (mid < high and arr[mid+1] < arr[mid]): return (mid+1) # Check if mid itself is # minimum element if (mid > low and arr[mid] < arr[mid - 1]): return mid # Decide whether we need to go # to left half or right half if (arr[high] > arr[mid]): return countRotations(arr, low, mid-1); return countRotations(arr, mid+1, high) # Driver code arr = [15, 18, 2, 3, 6, 12] n = len(arr) print(countRotations(arr, 0, n-1)) # This code is contributed by Smitha Dinesh Semwal

## C#

// C# program to find number of // rotations in a sorted and rotated // array. using System; class BinarySearch { // Returns count of rotations for an array // which is first sorted in ascending order, // then rotated static int countRotations(int []arr, int low, int high) { // This condition is needed to handle // the case when array is not rotated // at all if (high < low) return 0; // If there is only one element left if (high == low) return low; // Find mid // /*(low + high)/2;*/ int mid = low + (high - low)/2; // Check if element (mid+1) is minimum // element. Consider the cases like // {3, 4, 5, 1, 2} if (mid < high && arr[mid+1] < arr[mid]) return (mid + 1); // Check if mid itself is minimum element if (mid > low && arr[mid] < arr[mid - 1]) return mid; // Decide whether we need to go to left // half or right half if (arr[high] > arr[mid]) return countRotations(arr, low, mid - 1); return countRotations(arr, mid + 1, high); } // Driver program to test above functions public static void Main () { int []arr = {15, 18, 2, 3, 6, 12}; int n = arr.Length; Console.WriteLine(countRotations(arr, 0, n-1)); } } // This code is contributed by vt_m.

## PHP

<?php // Binary Search based PHP // program to find number // of rotations in a sorted // and rotated array. // Returns count of rotations // for an array which is // first sorted in ascending // order, then rotated function countRotations($arr, $low, $high) { // This condition is needed // to handle the case when // array is not rotated at all if ($high < $low) return 0; // If there is only // one element left if ($high == $low) return $low; // Find mid $mid = $low + ($high - $low) / 2; // Check if element (mid+1) // is minimum element. Consider // the cases like {3, 4, 5, 1, 2} if ($mid < $high && $arr[$mid + 1] < $arr[$mid]) return (int)($mid + 1); // Check if mid itself // is minimum element if ($mid > $low && $arr[$mid] < $arr[$mid - 1]) return (int)($mid); // Decide whether we need // to go to left half or // right half if ($arr[$high] > $arr[$mid]) return countRotations($arr, $low, $mid - 1); return countRotations($arr, $mid + 1, $high); } // Driver code $arr = array(15, 18, 2, 3, 6, 12); $n = sizeof($arr); echo countRotations($arr, 0, $n - 1); // This code is contributed bu ajit ?>

**Output:**

2

**Time Complexity : ** O(Log n)

**Auxiliary Space : ** O(1) if we use iterative Binary Search is used (Readers can refer Binary Search article for iterative Binary Search)

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