Search in Rotated Sorted Array
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# LeetCode #33: Search in Rotated Sorted Array (C/C++).

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source: https://leetcode.com/problems/search-in-rotated-sorted-array
C/C++ Solution to LeetCode problem 33. Search in Rotated Sorted Array.

## 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, 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.

## Examples

### 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

## Solution

The logic is as follows:

• Binary search splits the array in two.
• If the array is rotated, one part will be sorted, and the other not.
• First we identify the sorted half (the sorted part has the value at the lower index <= that the value at the middle index).
• If the target is in the sorted part, then we update our indexes to perform the next binary search in that part.
• If not, then it means the target is in the unsorted part, so we update indexes to perform next search there.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 class Solution { public: int search(vector<int>& nums, int target) { int l = 0; int h = nums.size() - 1; int m; while (l <= h) { m = (l + h) / 2; if (nums[m] == target) return m; if (nums[l] <= nums[m]) { if (target <= nums[m] && target >= nums[l]) h = m - 1; else l = m + 1; } else { if (target >= nums[m] && target <= nums[h]) l = m + 1; else h = m - 1; } } return -1; } };