0033-search-in-rotated-sorted-array.rb

# frozen_string_literal: true

# 33. Search in Rotated Sorted Array
# https://leetcode.com/problems/search-in-rotated-sorted-array
# Medium

=begin
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
=end

# @param {Integer[]} nums
# @param {Integer} target
# @return {Integer}
def search(nums, target)
  left, right = 0, nums.size - 1

  while left <= right
    mid = (left + right) / 2

    return mid if nums[mid] == target

    if nums[mid] >= nums[left]
      if nums[mid] >= target && target >= nums[left]
        right = mid - 1
      else
        left = mid + 1
      end
    else
      if nums[mid] <= target && target <= nums[right]
        left = mid + 1
      else
        right = mid - 1
      end
    end
  end

  -1
end

# **************** #
#       TEST       #
# **************** #

require "test/unit"
class Test_search < Test::Unit::TestCase
  def test_
    assert_equal 4, search([4, 5, 6, 7, 0, 1, 2], 0)
    assert_equal(-1, search([4, 5, 6, 7, 0, 1, 2], 3))
    assert_equal(-1, search([1], 0))
  end
end