Search in Rotated Sorted Array LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Search in Rotated Sorted Array in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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About LeetCode

LeetCode is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews. 

LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either Easy, Medium, or Hard.

LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:

  • Mathematics/Basic Logical Based Questions
  • Arrays
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  • Dynamic Programming
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  • Trees & Graphs
  • Greedy Algorithms
  • Breadth-First Search
  • Depth-First Search
  • Sorting & Searching
  • BST (Binary Search Tree)
  • Database
  • Linked List
  • Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the Problem – Search in Rotated Sorted Array– LeetCode Problem

Search in Rotated Sorted Array– LeetCode Problem

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
Search in Rotated Sorted Array– LeetCode Solutions
class Solution {
 public:
  int search(vector<int>& nums, int target) {
    int l = 0;
    int r = nums.size() - 1;

    while (l <= r) {
      const int m = l + (r - l) / 2;
      if (nums[m] == target)
        return m;
      if (nums[l] <= nums[m]) {  // nums[l..m] are sorted
        if (nums[l] <= target && target < nums[m])
          r = m - 1;
        else
          l = m + 1;
      } else {  // nums[m..n - 1] are sorted
        if (nums[m] < target && target <= nums[r])
          l = m + 1;
        else
          r = m - 1;
      }
    }

    return -1;
  }
};
class Solution {
  public int search(int[] nums, int target) {
    int l = 0;
    int r = nums.length - 1;

    while (l <= r) {
      final int m = l + (r - l) / 2;
      if (nums[m] == target)
        return m;
      if (nums[l] <= nums[m]) { // nums[l..m] are sorted
        if (nums[l] <= target && target < nums[m])
          r = m - 1;
        else
          l = m + 1;
      } else { // nums[m..n - 1] are sorted
        if (nums[m] < target && target <= nums[r])
          l = m + 1;
        else
          r = m - 1;
      }
    }

    return -1;
  }
}
class Solution:
  def search(self, nums: List[int], target: int) -> int:
    l = 0
    r = len(nums) - 1

    while l <= r:
      m = (l + r) // 2
      if nums[m] == target:
        return m
      if nums[l] <= nums[m]:  # nums[l..m] are sorted
        if nums[l] <= target < nums[m]:
          r = m - 1
        else:
          l = m + 1
      else:  # nums[m..n - 1] are sorted
        if nums[m] < target <= nums[r]:
          l = m + 1
        else:
          r = m - 1

    return -1

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