# Simple Python solution

• The len < 6 and 6 <= len <= 20 cases are easy. When len > 20, we need to do len - 20 times deletion. Also we need to do a change for every three repeating characters.

For any repeating sequences with len % 3 == 0, we can reduce one replacement by deleting one character. For any repeating sequences with len % 3 == 1, we can reduce one replacement by deleting two character. For the remaining sequences, we can reduce every replacement by deleting three character.

``````class Solution(object):
"""
:type s: str
:rtype: int
"""
missing_type = 3
if any('a' <= c <= 'z' for c in s): missing_type -= 1
if any('A' <= c <= 'Z' for c in s): missing_type -= 1
if any(c.isdigit() for c in s): missing_type -= 1

change = 0
one = two = 0
p = 2
while p < len(s):
if s[p] == s[p-1] == s[p-2]:
length = 2
while p < len(s) and s[p] == s[p-1]:
length += 1
p += 1

change += length / 3
if length % 3 == 0: one += 1
elif length % 3 == 1: two += 1
else:
p += 1

if len(s) < 6:
return max(missing_type, 6 - len(s))
elif len(s) <= 20:
return max(missing_type, change)
else:
delete = len(s) - 20

change -= min(delete, one)
change -= min(max(delete - one, 0), two * 2) / 2
change -= max(delete - one - 2 * two, 0) / 3

return delete + max(missing_type, change)
``````

• It's the most concise solution I have seen so far. Thx

• this is really impressive

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