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For software engineers, preparing for a tech interview means a lot of practice! Not only must you practice several problems, but you also need to ensure that you cover a wide variety of topics.

In this article, we’re going to tackle a problem on search, which will, in turn, help you crack a number of complex problems — searching an element in a sorted and rotated array.

We solve it using “single-pass binary search using casework.” **This is approach 3 of 4.**

To check out the other approaches, follow the links:

- Approach 1) Linear search
- Approach 2) Two-pass binary search
- Approach 4) Single-pass binary search by comparing mid and target elements

**Here, we cover:**

- Problem statement
- Approach explanation
- Algorithm for the approach
- Code for the approach
- Time complexity of the approach
- Space complexity of the approach

You have been given a sorted and rotated array "arr" of "N" distinct integers. You are also given an integer "target." Your task is to print the index of the integer "target" in array "arr." If "target" does not exist in "arr," then you need to print -1.

**Note:**

- "arr" contains distinct elements. This means that there is NO pair of indices (i, j) such that 0 <= i < j < N and arr[i] equals arr[j].
- "arr" is a rotated sorted array.

The following array adheres to all the above conditions:

Now, if we rotate it three times towards the right, we get a sorted and rotated array:

**Let’s take a couple of input examples:**

**Input 1: **arr = [9, 11, 1, 3, 5 ,7] and target = 5

**Output 1: **4**Explanation 1: **5 is present at index 4, so we need to print 4.

**Input 2: **arr = [9, 11, 1, 3, 5 ,7] and target = 8**Output 2: **-1**Explanation 2: **8 is not present in "arr," so we need to print -1.

Now that we have a clear understanding of our problem, we will move to the solution.

Please make sure you’ve readApproach 1andApproach 2to understand the coverage better.

**Key observation: **If you divide the given array into two parts, there will be at least one part sorted in increasing order. So, we just need to do some casework to move our left and right pointers which are the two endpoints of our search space. How will we do this? Let’s take an example:

This is the array:

And this the target: **8**

**Step 1: **Initialize left = 0 and right = 7

Calculate mid as mid = (left + right)/2 = 3

Here, we can see that arr[left] <= arr[mid]. So, arr[left...mid] must be sorted. We can now check if the target lies between [left, mid]. If not, we need to move *left *to *mid + 1* as the target does not exist in [left, mid].

target >= arr[left] and target < arr[mid], i.e., target = 8 lies in [left, mid).

So, we will move *left *and set *right *to *right = mid - 1 = 2*.

**Step 2: Left = 0 and right = 2**mid = (left + right)/2 = 1

Here, arr[left] <= arr[mid]. So, arr[left ... mid] is sorted. Now, we check if the target lies in arr[left, mid) = [5]. But the target does not lie in this range. So, set *left = mid + 1 = 2*

**Step 3: **Left = 2 and right = 2

mid = (left + right)/2 = 2

Here. arr[mid] is equal to target. We’ve found our target, and we return it.

**1. **Declare three variables "left", "right," and "mid" for binary search, and initialize left = 0, right = size of array - 1.

**2. **Run a loop until left <= right (standard binary search).

** a.** Calculate mid as mid = (left + right)/2.

** b.** If arr[mid] == target, then return *mid *as we have found our required element.

** c. **If arr[left] <=arr[mid], then do:

** i.** If target >= arr[left] and target < arr[mid], then search in [left, mid) and set right := mid - 1.

** ii. **Else, search in (mid, right] and set left = mid + 1.

**d.** If arr[mid] <= arr[right], then do:

** i. **If target > arr[mid] and target <= arr[right], then search in (mid, right] and set left := mid + 1.

**ii. **Else, search in [left, mid) and set right := mid - 1.

**3. **Return -1 as we are not able to find the target in our array "arr."

We’ve implemented the algorithm in C++. Use this as a reference to code in a programming language of your choice.

/*

Time Complexity: O(log(N))

Space Complexity: O(1)

Where N is the total number of elements in the given array.

*/

#include<iostream>

using namespace std;

// Function to search target in array 'arr'

int searchInRotatedSortedArray(int arr[], int n, int target)

{

int left = 0, right = n - 1, mid;

while (left <= right)

{

int mid = (left + right) / 2;

// We have found our target

if (arr[mid] == target)

{

return mid;

}

// Check if arr[left . . . mid] is sorted

if (arr[left] <= arr[mid])

{

// If target lies in arr[left, mid)

if (target >= arr[left] and target < arr[mid])

{

right = mid - 1;

}

else

{

left = mid + 1;

}

}

// Check if arr[mid . . . right] is sorted

if (arr[mid] <= arr[right])

{

// If target lies in arr(mid, right]

if (target > arr[mid] and target <= arr[right])

{

left = mid + 1;

}

else

{

right = mid - 1;

}

}

}

// We are not able to find target in array so return -1

return -1;

}

int main()

{

int n = 10;

int arr[n] = {10, 13, 16, 20, 25, 1, 3, 4, 5, 8};

int target = 4;

cout << searchInRotatedSortedArray(arr, n, target);

return 0;

}

**O(log(N))**, where "N" is the total number of elements in the given array.

We are reducing our search space by half at each step. So, we will get a recurrence as T(N) = T(N / 2) + C, where C is constant. Look at the previous approach for its proof.

**Best Case: O(1)**Happens when arr[mid] equals to target for the first value of mid

**Worst Case: O(log(N))**Happens when the target is not present in the array

**Average case: O(log(N))**Happens when the target is present in the array. We need to do log(N) steps on average to search for a target.

**Best Case = Worst Case = Average Case = O(1)**

We are using a few extra variables that take O(1) space. Hence, the space complexity will be O(1).

While this approach works better than using linear search and two-pass binary search, there’s yet another approach that employs single-pass binary search by comparing mid and target elements. Read this article to learn more:

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*Article contributed by Pankaj Sharma*