README.md

153. Find Minimum in Rotated Sorted Array

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.

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.

Solutions (Ruby)

1. Binary Search

# @param {Integer[]} nums
# @return {Integer}
def find_min(nums)
  l = 0
  r = nums.size - 1

  while l <= r
    m = (l + r) / 2

    if m > 0 && nums[m - 1] > nums[m]
      return nums[m]
    elsif nums[m] >= nums[0]
      l = m + 1
    else
      r = m - 1
    end
  end

  nums[0]
end

Solutions (Rust)

1. Binary Search

impl Solution {
    pub fn find_min(nums: Vec<i32>) -> i32 {
        let mut l = 0;
        let mut r = nums.len() - 1;

        while l <= r {
            let m = (l + r) / 2;

            if m > 0 && nums[m - 1] > nums[m] {
                return nums[m];
            } else if nums[m] >= nums[0] {
                l = m + 1;
            } else {
                r = m - 1;
            }
        }

        nums[0]
    }
}