Python: can I shorten this code further for checking substrings

Question

So the golf question was like this:

You are given a number T and then a total of T number of 17-character strings. For each of the strings, output yes if it contains four consecutive identical characters or contains the substring DX, else output no.

The following constraints are promised for each of the 17-character strings:

1. The possible character set is DX2AKQJT9876543;
2. The input string is sorted in the above mentioned order, hence the number of times each character appears is exactly the size of the longest contiguous substring of the character;
3. No characters will appear more than 4 times;
4. X and D will appear no more than once each.

Test case:

3
X2AAKKKKQT9765433
DX22AKKQJTT884443
X2AAAKQQJT8554433

should have the output

yes
yes
no

---

Note:

• Output is not case sensitive, so YeS is also legal;
• You may assume an extra \n after the end of the input.

---

My attempt goes like this: Python, probably 84 85 Bytes

import re
input()
while n:=input():print("yes"if re.search(r"(.)\1{3}|DX",n)else"no")

Also, I have the following alternatives for line 3, while are approximately the same length:

while n:=input():print("yneos"[not re.search(r"(.)\1{3}|DX",n)::2])
while n:=input():print(["yes","no"][not re.search(r"(.)\1{3}|DX",n)])
while n:=input():print((not re.search(r"(.)\1{3}|DX",n))*"no"or"yes")

Is there anything I can do to make this shorter? Thanks in advance!

Source: Question made by a friend, problem ID on Luogu is P8466

Answer 1

Python, 72 bytes

for n in[*open(0)][1:]:print("yneos"[all(map(str.__ne__,' X '+n,n))::2])

Try it online!

Knocks 5 more bytes off of Jonathan Allan's answer.

Answer 2

82 bytes

Can probably be golfed more, but here is a first golf:

-2 bytes by changing

"yes"if re.search(r"(.)\1{3}|DX",n)else"no"

To:

re.search(r"(.)\1{3}|DX",n)and"yes"or"no"

Try it online.

And another -1 by changing (not sure about this one, but both your 85 bytes version and this one gives an EOFError on TIO):

input()
while n:=input():...n

To:

n=input()
while n:...input()

Try it online.

Answer 3

Python 3.8, 77 bytes

input()
while n:='X '+input():print("yneos"[all(map(str.__ne__,n,n[3:]))::2])

Try it online!

The check for four equal characters in a row can be achieved by checking for equality between a character and the character three positions to its right. This can be applied across a string with any(map(str.__eq__,n,n[3:]) which will yield True if so or False otherwise.

This is inverted in the code by checking that all such pairs of characters are not equal with all(map(str.__ne__,n,n[3:])) to allow the result to be used in a slice of "yneos" that steps by two characters. That is, "yneos"[False::2] is equivalent to "yneos"[0::2] which is "yes"; while "yneos"[True::2] is equivalent to "yneos"[1::2] which is "no".

To check for the presence of "DX" we only need to see if the string starts with "DX" due to the strict ordering and the rule that states each of "D" and "X" may only appear once. Moreover this means the string cannot have a second character of "X" unless it starts with "DX", so we actually only need to check if the second character is "X". Therefore prefixing the string with "X " allows a reuse of the check for equal characters being three apart for this too, hence n:='X '+input().