33. Search In Rotated Sorted Array¶

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33. Search in Rotated Sorted Array

Medium

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is possibly left rotated at an unknown 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 left rotated by 3 indices 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

Solution¶

class Solution {
    public int search(int[] arr, int target) {
        int n = arr.length;
        int low = 0, high = n - 1, ans = -1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] == target) {
                return mid;
            }
            //need to figure it out which part is sorted, either left part or right part;
            if (arr[low] <= arr[mid]) {
                //left part is sorted;
                if (target >= arr[low] && target <= arr[mid]) {
                    high = mid - 1;
                }
                else low = mid + 1;
            }
            else {
                //right part is sorted;
                if (target >= arr[mid] && target <= arr[high]) {
                    low = mid + 1;
                }
                else high = mid - 1;
            }
        } 
        return -1;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here