In this post, we are going to solve the 33. Search in Rotated Sorted Array problem of Leetcode. This problem 33. Search in Rotated Sorted Array is a Leetcode medium level problem. Let’s see the code, 33. Search in Rotated Sorted Array – Leetcode Solution.

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[0], nums[1], …, nums[k-1]. 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 = [1], 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

Now, let’s see the code of 33. Search in Rotated Sorted Array – Leetcode Solution.

Search in Rotated Sorted Array – Leetcode Solution

33. Search in Rotated Sorted Array – Solution in Java

class Solution {
    public int search(int[] nums, int target) {
        
        int s = 0;
        int e = nums.length-1;
        
        while(s <= e){
            int mid = s - (s - e)/2;
            if(nums[mid] == target) return mid;
            
            if(nums[s] <= nums[mid]){
                if(target >= nums[s] && target <= nums[mid])  e = mid - 1;
                else s = mid+1;
            }
            if(nums[mid] <= nums[e]){
                if(target <= nums[e] && target >= nums[mid])  s = mid + 1;
                else e = mid - 1;
            }
        }
        
        return -1;
    }
}

33. Search in Rotated Sorted Array – Solution in C++

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

33. Search in Rotated Sorted Array – Solution in Python

class Solution:

    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

Note: This problem 33. Search in Rotated Sorted Array is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.