# Search in Rotated Sorted Array Problem

## Description

LeetCode Problem 33.

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:

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

Example 2:

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

Example 3:

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

## Sample C++ Code

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class Solution {
public:
int search(int A[], int n, int target) {
int lo=0, hi=n-1;
// find the index of the smallest value using binary search.
// Loop will terminate since mid < hi, and lo or hi will shrink by at least 1.
// Proof by contradiction that mid < hi: if mid==hi, then lo==hi and loop would have been terminated.
while(lo<hi){
int mid=(lo+hi)/2;
if(A[mid]>A[hi]) lo=mid+1;
else hi=mid;
}
// lo==hi is the index of the smallest value and also the number of places rotated.
int rot=lo;
lo=0; hi=n-1;
// The usual binary search and accounting for rotation.
while(lo<=hi){
int mid=(lo+hi)/2;
int realmid=(mid+rot)%n;
if (A[realmid]==target) return realmid;
if (A[realmid]<target) lo=mid+1;
else hi=mid-1;
}
return -1;
}
};
``````