code: def bussiest_interval(arriving, leaving):
      # sort both arrays (don't care about pairs)
      arriving.sort()
      leaving.sort()
  
      n = len(arriving)
      i, j = 0, 0
      start, end = 0, 0
      overlapping = 0
      max_overlapping = 0
  
      # both arrays have same number of elements
      # but the biggest time is from the leaving array
      # becayse of that you're sure that 'i' will reach the end before 'j'
      while i < n:
          if arriving[i] <= leaving[j]:
              overlapping += 1
              if max_overlapping <= overlapping:
                  max_overlapping = overlapping
                  # save the start time if max_overlapping
                  start = arriving[i]
              i += 1
          else:
              if max_overlapping == overlapping:
                  # save the exit time if max_overlapping
                  end = leaving[j]
              overlapping -= 1
              j += 1
  
      # check again this to close the result interval because 'i' is completed and not 'j'
      if max_overlapping == overlapping:
          end = leaving[j]
  
      # return start&end or max_overlapping
      return (start, end)
  
label_desc: Use image node_id 1 for calling a global function with example usage: bussiest_interval(arriving, leaving) and returns: start,  end