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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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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