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Do you even need to keep a character frequency count?

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#1

Is this not a simpler solution that just manipulates the k variable and tests whether a character can be found in a string using Python’s “in” functionality?

def longest_substring_with_k_distinct(str1, k):
  windowStart = 0
  best_result = ""
  result = ""
  k_orig = k
  for windowEnd in range(0, len(str1)):
    if str1[windowEnd] not in result and k > 0:
      result += str1[windowEnd]
      k -= 1
    elif str1[windowEnd] in result:
      result += str1[windowEnd]
    else:
      if len(result) > len(best_result):
        best_result = result
      result = ""
      k = k_orig

  return len(best_result)
#2

Hi @Matt_Timmons-Brown

Yes, your solution is simpler and working perfectly fine.