feat(practice/pangram): update code snippets in approaches for set with len()

Fixes:
Cc:
Reviewed-by:

Author: princemuel

exercises/practice/pangram/.approaches/set-len/content.md

@@ -2,16 +2,14 @@
 
 ```python
 def is_pangram(sentence):
-    return len([ltr for ltr in set(sentence.lower()) if ltr.isalpha()]) \
-        == 26
-
+    return len(set(ltr for ltr in sentence.lower() if ltr.isalpha())) == 26
 ```
 
 - This approach first makes a [set][set] from the [`lower`][lower]cased characters of the `sentence`.
 - The characters in the `set`are then iterated in a [list comprehension][list-comprehension].
 - The characters are filtered by an `if` [`isalpha()`][isalpha] statement, so that only alphabetic characters make it into the list.
 - The function returns whether the [`len()`][len] of the [`list`][list] is `26`.
-If the number of unique letters in the `set` is equal to the `26` letters in the alphabet, then the function will return `True`.
+  If the number of unique letters in the `set` is equal to the `26` letters in the alphabet, then the function will return `True`.
 
 [lower]: https://docs.python.org/3/library/stdtypes.html?#str.lower
 [set]: https://docs.python.org/3/library/stdtypes.html?#set

exercises/practice/pangram/.approaches/set-len/snippet.txt

@@ -1,3 +1,2 @@
 def is_pangram(sentence):
-    return len([ltr for ltr in set(sentence.lower()) if ltr.isalpha()]) \
-        == 26
+    return len(set(ltr for ltr in sentence.lower() if ltr.isalpha())) == 26

exercises/practice/pangram/.articles/performance/code/Benchmark.py

@@ -28,7 +28,7 @@ def is_pangram(sentence):
 val = timeit.timeit("""is_pangram("Victor jagt zwölf_(12) Boxkämpfer quer über den großen Sylter Deich.")""",
                     """
 def is_pangram(sentence):
-    return len([ltr for ltr in set(sentence.lower()) if ltr.isalpha()]) == 26
+    return len(set(ltr for ltr in sentence.lower() if ltr.isalpha())) == 26
 
 """, number=loops) / loops