# 33. Search in Rotated Sorted Array

## Problem Statement

<br>

There is an integer array `nums` sorted in ascending order (with **distinct** values).

Prior to being passed to your function, `nums` is **possibly rotated** at an unknown pivot index `k` (`1 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]` (**0-indexed**). For example, `[0,1,2,4,5,6,7]` might be rotated at pivot index `3` and become `[4,5,6,7,0,1,2]`.

Given the array `nums` **after** the possible rotation and an integer `target`, return *the index of* `target` *if it is in* `nums`*, or* `-1` *if it is not in* `nums`.

You must write an algorithm with `O(log n)` runtime complexity.

&#x20;

**Example 1:**

<pre><code><strong>Input: nums = [4,5,6,7,0,1,2], target = 0
</strong><strong>Output: 4
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: nums = [4,5,6,7,0,1,2], target = 3
</strong><strong>Output: -1
</strong></code></pre>

**Example 3:**

<pre><code><strong>Input: nums = [1], target = 0
</strong><strong>Output: -1
</strong></code></pre>

&#x20;

**Constraints:**

* `1 <= nums.length <= 5000`
* `-104 <= nums[i] <= 104`
* All values of `nums` are **unique**.
* `nums` is an ascending array that is possibly rotated.
* `-104 <= target <= 104`

## Intuition

```
Approach:

Take the help of first and last element to judge the poistion of the target
element
```

### Links

<https://leetcode.com/problems/search-in-rotated-sorted-array/description/>

### Video Links

### Approach 1:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    int search(vector<int>& nums, int target) {
        int low=0, high=nums.size()-1;
        while(low<=high){
            int mid = low+(high-low)/2;
            if(nums[mid] == target)
                return mid;
            else if(nums[low] <= nums[mid]){
                if(nums[low] <= target and target <= nums[mid])
                    high = mid-1;
                else
                    low = mid+1;
            }
            else{
                if(nums[mid] <= target and target <= nums[high])
                    low = mid+1;
                else
                    high = mid-1;
            }
        }

        return -1;
    }
};
```

{% endcode %}

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Similar Problems

###