Search in Rotated Sorted Array Leetcode Solution

In this post, we are going to solve the Search in Rotated Sorted Array Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Table of Contents

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]] (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 = [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 leetcode solution of Search in Rotated Sorted Array Leetcode Solution.

Search in Rotated Sorted Array Leetcode 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

Search in Rotated Sorted Array Leetcode Solution in CPP

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;
    }
};

Search in Rotated Sorted Array Leetcode 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;
    }
}

Note: This problem Search in Rotated Sorted Array is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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