follows the same previous pattern. except that this covers an edge case as described below.

from heapq import *

class Solution:
  def findClosestElements(self, arr, K, X):
    maxHeap = []

    for i in range(len(arr)):
      difference = -abs(arr[i] - X)
      tuple = (difference, arr[i])

      # for edge cases where maxHeap is full already has K elements and there
      # already exists an element with same difference and is smaller number
      prevMaxDiff = maxHeap[0][0] if len(maxHeap) > 0 else None 
      if len(maxHeap) == K and difference == prevMaxDiff:
        continue

      heappush(maxHeap, tuple)

      if i >= K:
          heappop(maxHeap)
          
    result = []
    for i in range(len(maxHeap)):
      result.append(maxHeap[i][1])
    
    result.sort()
    return result