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#33MediumBinary search

Search in Rotated Sorted Array

The tell: Search a rotated sorted array in logarithmic time.

How you get there

  1. 1At any midpoint, at least one half is properly sorted.
  2. 2Determine which half is sorted by comparing nums[lo] to nums[mid].
  3. 3If the target lies inside that sorted half's range, search it; otherwise search the other.

Solution

Python
def search(nums: list[int], target: int) -> int:
    lo, hi = 0, len(nums) - 1
    while lo <= hi:
        mid = (lo + hi) // 2
        if nums[mid] == target:
            return mid
        if nums[lo] <= nums[mid]:                 # left half is sorted
            if nums[lo] <= target < nums[mid]:
                hi = mid - 1
            else:
                lo = mid + 1
        else:                                      # right half is sorted
            if nums[mid] < target <= nums[hi]:
                lo = mid + 1
            else:
                hi = mid - 1
    return -1
Time
O(log n)
Space
O(1)

What goes wrong

Using < instead of <= when testing nums[lo] <= nums[mid] mishandles the two-element case.

Test yourself on DSANext: Find Minimum in Rotated Sorted Array →