[Pangram]: fix redundant double iteration in dig deeper solution

exercises/practice/pangram/.approaches/config.json

@@ -1,6 +1,7 @@
 {
   "introduction": {
-    "authors": ["bobahop"]
+    "authors": ["bobahop"],
+    "contributors": ["princemuel"]
   },
   "approaches": [
     {
@@ -22,7 +23,8 @@
       "slug": "set-len",
       "title": "set with len",
       "blurb": "Use set with len.",
-      "authors": ["bobahop"]
+      "authors": ["bobahop"],
+      "contributors": ["princemuel"]
     },    
     {
       "uuid": "0a6d1bbf-6d60-4489-b8d9-b8375894628b",

exercises/practice/pangram/.approaches/introduction.md

@@ -42,9 +42,7 @@ For more information, check the [`set` with `issubset()` approach][approach-set-
 
 ```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
 ```
 
 For more information, check the [`set` with `len()` approach][approach-set-len].

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

@@ -2,20 +2,18 @@
 
 ```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`.
+- The characters are filtered using a [set comprehension][set-comprehension] with an `if` [`isalpha()`][isalpha] statement, so that only alphabetic characters make it into the set.
+- The function returns whether the [`len()`][len] of the [`set`][set] is `26`.
+  If the number of unique [ASCII][ascii] (American Standard Code for Information Interchange) letters in the `set` is equal to the `26` letters in the [ASCII][ascii] alphabet, then the function will return `True`.
+- This approach is efficient because it uses a set to eliminate duplicates and directly checks the length, which is a constant time operation.
 
 [lower]: https://docs.python.org/3/library/stdtypes.html?#str.lower
 [set]: https://docs.python.org/3/library/stdtypes.html?#set
-[list-comprehension]: https://docs.python.org/3/tutorial/datastructures.html#list-comprehensions
+[set-comprehension]: https://realpython.com/python-set-comprehension/#introducing-set-comprehensions
 [isalpha]: https://docs.python.org/3/library/stdtypes.html?highlight=isalpha#str.isalpha
 [len]: https://docs.python.org/3/library/functions.html?#len
-[list]: https://docs.python.org/3/library/stdtypes.html?#list
+[ascii]: https://en.wikipedia.org/wiki/ASCII

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/config.json

@@ -5,7 +5,8 @@
       "slug": "performance",
       "title": "Performance deep dive",
       "blurb": "Deep dive to find out the most performant approach to determining a pangram.",
-      "authors": ["bobahop"]
+      "authors": ["bobahop"],
+      "contributors": ["princemuel"]
     }
   ]
 }

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
 

exercises/practice/pangram/.articles/performance/content.md

@@ -15,18 +15,18 @@ For our performance investigation, we'll also include a fourth approach that [us
 To benchmark the approaches, we wrote a [small benchmark application][benchmark-application] using the [`timeit`][timeit] library.
 
 ```
-all:   1.505466179997893e-05
-all:   1.6063886400021147e-05  // with sentence.casefold()
-set:   1.950172399985604e-06
-len:   3.7158977999933994e-06
-bit:   8.75982620002469e-06
+all:   1.8692991019000146e-05
+all:   1.686682232399926e-05  // with sentence.casefold()
+set:   2.5181135439997888e-06
+len:   5.848111433000668e-06
+bit:   1.2118699087000096e-05
 ```
 
 - The `set` `len()` approach is not as fast as the `set` `issubset()` approach.
-- The `all()` approach is slower than either `set` approach.
-Using `casefold` was slower than using `lower`.
+- The `all()` approach is significantly slower than either `set` approach (approximately 6-8x slower).
+  Using `casefold()` versus `lower()` showed variable performance, with each being faster in different runs.
 - Although the bit field approach may be faster in other languages, it is significantly slower in Python.
-It is faster than the `all()` approach, but much slower than either `set` approach.
+  It is faster than the `all()` approach, but much slower than either `set` approach.
 
 [benchmark-application]: https://github.com/exercism/python/blob/main/exercises/practice/pangram/.articles/performance/code/Benchmark.py
 [timeit]: https://docs.python.org/3/library/timeit.html

exercises/practice/pangram/.articles/performance/snippet.md

@@ -1,7 +1,7 @@
 ```
-all:   1.505466179997893e-05
-all:   1.6063886400021147e-05  // with sentence.casefold()
-set:   1.950172399985604e-06
-len:   3.7158977999933994e-06
-bit:   8.75982620002469e-06
+all:   1.8692991019000146e-05
+all:   1.686682232399926e-05  // with sentence.casefold()
+set:   2.5181135439997888e-06
+len:   5.848111433000668e-06
+bit:   1.2118699087000096e-05
 ```