33. Search in Rotated Sorted Array LeetCode Solution

In this guide we will provide 33. Search in Rotated Sorted Array LeetCode Solution with best time and space complexity. The solution to Search in Rotated Sorted Array problem is provided in various programming languages like C++, Java and python. This will be helpful for you if you are preparing for placements, hackathon, interviews or practice purposes. The solutions provided here are very easy to follow and with detailed explanations.

Problem Statement

Complexity Analysis

Search in Rotated Sorted Array solution in C++
Search in Rotated Sorted Array soution in Java
Search in Rotated Sorted Array solution Python
Additional Resources

33. Search in Rotated Sorted Array LeetCode Solution image

Problem Statement of Search in Rotated Sorted Array

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

Complexity Analysis

Time Complexity: O(\log n)
Space Complexity: O(1)

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

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) / 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;
  }
};
/* code provided by PROGIEZ */

33. Search in Rotated Sorted Array LeetCode Solution in Java

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) / 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;
  }
}
// code provided by PROGIEZ

33. Search in Rotated Sorted Array LeetCode Solution in Python

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
 #code by PROGIEZ