33-search-in-rotated-sorted-array.java

/*
33. Search in Rotated Sorted Array
Medium - 43.6%

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
*/

class Solution {
    public int 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 mid;
            }
            
            // Check which half is sorted
            if (nums[left] <= nums[mid]) {
                // Left half is sorted
                if (nums[left] <= target && target < nums[mid]) {
                    right = mid - 1; // Target is in left half
                } else {
                    left = mid + 1; // Target is in right half
                }
            } else {
                // Right half is sorted
                if (nums[mid] < target && target <= nums[right]) {
                    left = mid + 1; // Target is in right half
                } else {
                    right = mid - 1; // Target is in left half
                }
            }
        }
        
        return -1; // Target not found
    }
}