0153-find-minimum-in-rotated-sorted-array.rb

# frozen_string_literal: true

# https://leetcode.com/problems/find-minimum-in-rotated-sorted-array
# 153. Find Minimum in Rotated Sorted Array
# Medium

=begin
Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

[4,5,6,7,0,1,2] if it was rotated 4 times.
[0,1,2,4,5,6,7] if it was rotated 7 times.
Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

Example 1:
Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:
Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:
Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times.

Constraints:
* n == nums.length
* 1 <= n <= 5000
* -5000 <= nums[i] <= 5000
* All the integers of nums are unique.
* nums is sorted and rotated between 1 and n times.
=end

# @param {Integer[]} nums
# @return {Integer}
def find_min(nums)
  left = 0
  right = nums.length - 1

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

    if nums[mid] > nums[right]
      left = mid + 1
    else
      right = mid
    end
  end

  nums[left]
end

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

require "test/unit"
class Test_find_min < Test::Unit::TestCase
  def test_
    assert_equal 1, find_min([3, 4, 5, 1, 2])
    assert_equal 0, find_min([4, 5, 6, 7, 0, 1, 2])
    assert_equal 11, find_min([11, 13, 15, 17])
  end
end