Given an integer array, return the k-th smallest distance among all the pairs. The distance of a pair (A, B) is defined as the absolute difference between A and B. 

Example 1:

Input:
nums = [1,3,1]
k = 1
Output: 0 
Explanation:
Here are all the pairs:
(1,3) -> 2
(1,1) -> 0
(3,1) -> 2
Then the 1st smallest distance pair is (1,1), and its distance is 0.



Note:

2 .
0 .
1 .

Difficulty: MEDIUM_HARD
Starter code: class Solution:
    def smallestDistancePair(self, nums: List[int], k: int) -> int:
        

Solutions:
class Solution:

	def smallestDistancePair(self, nums, k):
		nums.sort()
		(l, r) = (0, nums[-1] - nums[0])
		while l < r:
			m = l + (r - l) // 2
			count = 0
			left = 0
			for right in range(len(nums)):
				while nums[right] - nums[left] > m:
					left += 1
				count += right - left
			if count < k:
				l = m + 1
			else:
				r = m
		return l


class Solution(object):

	def smallestDistancePair(self, nums, k):

		def possible(guess):
			count = left = 0
			for (right, x) in enumerate(nums):
				while x - nums[left] > guess:
					left += 1
				count += right - left
			return count >= k
		nums.sort()
		lo = 0
		hi = nums[-1] - nums[0]
		while lo < hi:
			mi = (lo + hi) // 2
			if possible(mi):
				hi = mi
			else:
				lo = mi + 1
		return lo