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In this post, you will find the solution for the **Search in Rotated Sorted Array** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Search in Rotated Sorted Array– LeetCode Problem

Search in Rotated Sorted Array– LeetCode Problem

**Problem:**

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.

**Example 1:**

Input:nums = [4,5,6,7,0,1,2], target = 0Output:4

**Example 2:**

Input:nums = [4,5,6,7,0,1,2], target = 3Output:-1

**Example 3:**

Input:nums = [1], target = 0Output:-1

**Constraints:**

`1 <= nums.length <= 5000`

`-10`

^{4}<= nums[i] <= 10^{4}- All values of
`nums`

are**unique**. `nums`

is an ascending array that is possibly rotated.`-10`

^{4}<= target <= 10^{4}

Search in Rotated Sorted Array– LeetCode Solutions

class Solution { public: int search(vector& nums, int target) { int l = 0; int r = nums.size() - 1; while (l <= r) { const int m = l + (r - l) / 2; if (nums[m] == target) return m; if (nums[l] <= nums[m]) { // nums[l..m] are sorted if (nums[l] <= target && target < nums[m]) r = m - 1; else l = m + 1; } else { // nums[m..n - 1] are sorted if (nums[m] < target && target <= nums[r]) l = m + 1; else r = m - 1; } } return -1; } };

class Solution { public int search(int[] nums, int target) { int l = 0; int r = nums.length - 1; while (l <= r) { final int m = l + (r - l) / 2; if (nums[m] == target) return m; if (nums[l] <= nums[m]) { // nums[l..m] are sorted if (nums[l] <= target && target < nums[m]) r = m - 1; else l = m + 1; } else { // nums[m..n - 1] are sorted if (nums[m] < target && target <= nums[r]) l = m + 1; else r = m - 1; } } return -1; } }

class Solution: def search(self, nums: List[int], target: int) -> int: l = 0 r = len(nums) - 1 while l <= r: m = (l + r) // 2 if nums[m] == target: return m if nums[l] <= nums[m]: # nums[l..m] are sorted if nums[l] <= target < nums[m]: r = m - 1 else: l = m + 1 else: # nums[m..n - 1] are sorted if nums[m] < target <= nums[r]: l = m + 1 else: r = m - 1 return -1