Hello Everyone,

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;`

`}`

**Output:**

2

**Time Complexity :** O(n)

**Auxiliary Space :** O(1)

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

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

- The minimum element is the only element whose previous is greater than it. If there is no previous 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 the minimum element is not at the 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:

- 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 the 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;`

`}`

**Output:**

2

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

**Auxiliary Space :** O(1) if we use iterative Binary Search is used.