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LeetCode 81: Search in Rotated Sorted Array II Solution

Master LeetCode problem 81 (Search in Rotated Sorted Array II), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

81. Search in Rotated Sorted Array II

Medium
ArrayBinary Search

Problem Explanation

Explanation

To solve this problem efficiently, we can modify the binary search algorithm. The idea is to check which part of the rotated array is sorted and then perform binary search on that part. We need to handle the case where there are duplicates in the array.

  1. Initialize two pointers, left at index 0 and right at index nums.length - 1.
  2. While left <= right, calculate the middle index mid.
  3. If nums[mid] == target, return true.
  4. If nums[left] == nums[mid], increment left and continue the loop.
  5. Check if the left half (nums[left] <= nums[mid]) is sorted:
    • If nums[left] <= nums[mid] and nums[left] <= target < nums[mid], update right = mid - 1.
    • Otherwise, update left = mid + 1.
  6. If the left half is not sorted, the right half must be sorted:
    • If nums[mid] < target <= nums[right], update left = mid + 1.
    • Otherwise, update right = mid - 1.
  7. If target is not found after the loop, return false.

Time complexity: O(log n)
Space complexity: O(1)

Solution Code

class Solution {
    public boolean search(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                return true;
            }
            
            if (nums[left] == nums[mid]) {
                left++;
            } else if (nums[left] <= nums[mid]) {
                if (nums[left] <= target && target < nums[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            } else {
                if (nums[mid] < target && target <= nums[right]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            }
        }
        return false;
    }
}

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