33. Search in Rotated Sorted Array

Explanation

To solve this problem in O(log n) runtime complexity, we can use a modified binary search algorithm. We can divide the rotated array into two halves. One of the halves must be sorted. We can then check if the target value lies within the sorted half or the unsorted half based on some conditions. We adjust the binary search accordingly based on which half the target value might lie in.

We compare the target value with the values at the start, middle, and end indices of the array to determine which half is sorted.
Based on the sorted half, we check if the target value lies within the range of the sorted half. If it does, we perform binary search in that half. Otherwise, we search in the other half.
We repeat this process until we find the target value or determine that it does not exist in the array.

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

View Leetcode Problem

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;
            }
            
            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 -1;
    }
}

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