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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 help you crack a number of complex problems — searching an element in a sorted and rotated array.

We solve it using “linear search.” **This is approach 1 of 4.**

To check out the other approaches, follow the links:

- Approach 2) Two-pass binary search
- Approach 3) Single-pass binary search using casework
- 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.

It is the most straightforward possible approach to solve this problem. Basically, we use brute force here. In this approach, we will linearly scan the array and check whether the target exists or not.

- Run a loop from index = 0 to arr.length and do:

If arr[index] == target then return index.

- After the termination of the loop, return -1 as the target was not found.

Here’s the C++ code to implement the algorithm above:

/*

Time Complexity: O(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)

{

// Linearly check every element if its equal to target

for (int i = 0; i < n; i++)

{

if (arr[i] == target)

{

return i;

}

}

// We are not able to find target in array so returning -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(N), where "N" is the total number of elements in the given array.

**Best Case: O(1)**

When the target is the first element of the array.

**Worst Case: O(N)**

When the target is not present in the array, we need to check O(N) elements. ****

**Average Case: O(N)**

On average, we need to check O(N) elements to see if they match the target or not.

**Best Case = Worst Case: O(1) = Average Case = O(1)**We are using a few extra variables that take O(1) space. Hence, the space complexity will be O(1).

This method works, but the brute-force approach is a bit trivial. Let's figure out a better way. The given array is **sorted. **What would we do if the given input was sorted but not rotated? Can we apply binary search? Read this article to find out:

*Searching an Element in a Sorted and Rotated Array: Two-pass Binary Search*

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*Article contributed by Pankaj Sharma*