perf: optimize is_pangram by fixing unnecessary list creation

Issue: Current implementation creates an intermediate list from a set, which is redundant and inefficient.

**Before:**
```python
return len([ltr for ltr in set(sentence.lower()) if ltr.isalpha()]) == 26
```
**After:**
```python
return len(set(ltr for ltr in sentence.lower() if ltr.isalpha())) == 26
```
Why this is better:
- Eliminates double iteration: Old version iterates once for set(), again for list comprehension
- Removes unnecessary list creation: No need to convert set → list just to count
- Better memory usage: Generator expression feeds directly into set constructor
- Same time complexity but more efficient constant factors

The old approach first deduplicates all the characters, then filters alphabetic ones. The new approach filters while building the set, which is the natural order of operations for this problem.

Author: princemuel

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].