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33. Search in Rotated Sorted Array — Step-by-Step Visualization

mediumLeetCode #33ArrayBinary Search

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`

Algorithm Pattern

Two-Sided Binary Search

Key Idea

A rotated sorted array always has at least one sorted half when split in the middle. We determine which half is sorted first, then check if our target is within that half to decide where to search next.

Step-by-Step Approach

  1. Initialize `L = 0` and `R = nums.length - 1`.
  2. Calculate `mid = (L + R) // 2`.
  3. If `nums[mid] == target`, we found it!
  4. Identify the sorted half: If `nums[L] <= nums[mid]`, the left half `[L...mid]` is sorted. Otherwise, the right half `[mid...R]` is sorted.
  5. Check if `target` is in the sorted half. If it is, move the search range to that half. Otherwise, move it to the other half.

Common Gotchas

  • Be careful with the boundaries of the conditions like `nums[L] <= target <= nums[mid]`. All comparisons should be distinct for clarity.
  • The rotation might be at index 0 (non-rotated), the algorithm still holds.

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