LeetCode-in-Dart

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 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

Solution

class Solution {
  int search(List<int> nums, int target) {
    int mid;
    int lo = 0;
    int hi = nums.length - 1;

    while (lo <= hi) {
      mid = ((hi - lo) >> 1) + lo;

      if (target == nums[mid]) {
        return mid;
      }

      // Check if the left half is sorted
      if (nums[lo] <= nums[mid]) {
        // Check if the target is within the sorted left half
        if (nums[lo] <= target && target <= nums[mid]) {
          hi = mid - 1;
        } else {
          lo = mid + 1;
        }
      }
      // Otherwise, the right half is sorted
      else {
        // Check if the target is within the sorted right half
        if (nums[mid] <= target && target <= nums[hi]) {
          lo = mid + 1;
        } else {
          hi = mid - 1;
        }
      }
    }

    return -1;
  }
}

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