Problem description:

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Your algorithm’s runtime complexity must be in the order of O(log n).

Example 1:

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

Example 2:

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

Solution:

We can use binary search to solve this question. The important part is to find the ascending part in the array.

For a Rotated Sorted Array, it must at some point follow the ascending order. We can check if nums[mid] >= nums[left], the first half would be a sorted array if it’s true. 

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class Solution {
public:
    int search(vector<int>& nums, int target) {
        int left= 0, right= nums.size()-1;
        
        while(left <= right){
            int mid = (left + right)/2;
            
            if(nums[mid] == target)
                return mid;
            
            if(nums[mid] >= nums[left])  
            {  
                if(nums[mid] > target && nums[left] <= target)  
                    right = mid - 1;  
                else  
                    left = mid + 1;  
            }  
            else{
                if(nums[mid] < target && target <= nums[right])
                    left = mid+1;
                else
                    right = mid-1;
            }
        }
        return -1;
    }
};

time complexity: $O(logn)$
space complexity: $O(1)$
reference: