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153. Find Minimum in Rotated Sorted Array

Medium

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.

Solution

# @param {Integer[]} nums
# @return {Integer}
def find_min_util(nums, l, r)
  return nums[l] if l == r

  mid = (l + r) / 2

  return nums[l] if mid == l && nums[mid] < nums[r]
  return nums[mid] if mid - 1 >= 0 && nums[mid - 1] > nums[mid]

  if nums[mid] < nums[l]
    find_min_util(nums, l, mid - 1)
  elsif nums[mid] > nums[r]
    find_min_util(nums, mid + 1, r)
  else
    find_min_util(nums, l, mid - 1)
  end
end

def find_min(nums)
  l = 0
  r = nums.length - 1
  find_min_util(nums, l, r)
end

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