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# Search in Rotated Sorted Array LeetCode Solution

## Problem – Search in Rotated Sorted Array LeetCode Solution

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, nums, ..., 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 = 0
Output: 4
``````

Example 2:

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

Example 3:

``````Input: nums = , target = 0
Output: -1
``````

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`

### Search in Rotated Sorted Array LeetCode Solution in Java

``````public class Solution {
public int search(int[] A, int target) {
int lo = 0;
int hi = A.length - 1;
while (lo < hi) {
int mid = (lo + hi) / 2;
if (A[mid] == target) return mid;

if (A[lo] <= A[mid]) {
if (target >= A[lo] && target < A[mid]) {
hi = mid - 1;
} else {
lo = mid + 1;
}
} else {
if (target > A[mid] && target <= A[hi]) {
lo = mid + 1;
} else {
hi = mid - 1;
}
}
}
return A[lo] == target ? lo : -1;
}
``````

### Search in Rotated Sorted Array LeetCode Solution in Python

``````class Solution:
# @param {integer[]} numss
# @param {integer} target
# @return {integer}
def search(self, nums, target):
if not nums:
return -1

low, high = 0, len(nums) - 1

while low <= high:
mid = (low + high) / 2
if target == nums[mid]:
return mid

if nums[low] <= nums[mid]:
if nums[low] <= target <= nums[mid]:
high = mid - 1
else:
low = mid + 1
else:
if nums[mid] <= target <= nums[high]:
low = mid + 1
else:
high = mid - 1

return -1
``````

### Search in Rotated Sorted Array LeetCode Solution in C++

``````class Solution {
public:
int search(vector<int>& nums, int target) {
int l = 0, r = nums.size()-1;
while (l <= r) {
int mid = (l+r) / 2;
if (target == nums[mid])
return mid;
// there exists rotation; the middle element is in the left part of the array
if (nums[mid] > nums[r]) {
if (target < nums[mid] && target >= nums[l])
r = mid - 1;
else
l = mid + 1;
}
// there exists rotation; the middle element is in the right part of the array
else if (nums[mid] < nums[l]) {
if (target > nums[mid] && target <= nums[r])
l = mid + 1;
else
r = mid - 1;
}
// there is no rotation; just like normal binary search
else {
if (target < nums[mid])
r = mid - 1;
else
l = mid + 1;
}
}
return -1;
}
};
``````

### Search in Rotated Sorted Array LeetCode Solution in Ruby

``````def search(nums, target)
i = (0...nums.size).bsearch { |i|
(nums <= target) ^ (nums > nums[i]) ^ (target > nums[i])
}
nums[i || 0] == target ? i : -1
end
``````
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