# Search In Rotated Array II Problem Solution

Search In Rotated Array II : There is an integer array `nums` sorted in non-decreasing order (not necessarily with distinct values).

Before being passed to your function, `nums` is rotated at an unknown pivot index `k` (`0 <= 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, `[0,1,2,4,4,4,5,6,6,7]` might be rotated at pivot index `5` and become `[4,5,6,6,7,0,1,2,4,4]`.

Given the array `nums` after the rotation and an integer `target`, return `true` if `target` is in `nums`, or `false` if it is not in `nums`.

Example :

```Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true```

## Problem Solution In Python

``````class Solution:
def search(self, nums: List[int], target: int) -> bool:
pivot=0
for ar in range(len(nums)-1):
if nums[ar]>nums[ar+1]:
pivot=ar+1
break
nums = nums[pivot:]+nums[0:pivot]
left=0
right=len(nums)-1
while left<=right:
mid = int((left+right)/2)
if nums[mid]==target:
return True
elif nums[mid]>target:
right=mid-1
else:
left=mid+1
return False
``````

## Problem Solution In Java

``````class Solution {
public boolean search(int[] nums, int target)
{

for(int i=0;i<nums.length;i++)
{
if(nums[i]==target)
return true;
}
return false;
}
}
``````

## Problem Solution In C++

``````class Solution {
public:
bool search(const vector<int>& nums, int target) {
for (const auto& i : nums) {
if (i == target) {
return true;
}
}
return false;
}
};

``````

## Problem Solution In C

``````bool search(int* nums, int numsSize, int target){
int n = 0;
while (n < numsSize) {
if (nums[n] < target) {
n++;
} else if (nums[n] == target) {
return true;
} else {
break;
}
}
if (n != 0) {
return false;
} else {
int end = numsSize - 1;
while (end > n) {
if (nums[end] > target) {
end--;
}else if (nums[end] == target) {
return true;
} else {
return false;
}
}
}
return false;
}``````