Group names by first letter

Organize a list of names into a dictionary keyed by each name's first letter.

dictionaries strings grouping

Write a function `group_by_first_letter(names)` that takes a list of strings (names) and returns a dictionary where each key is a lowercase first letter (character) of a name, and the value is a list of all names (original case) that start with that letter, preserving the order they appear in the input.

- If the input list is empty, return an empty dictionary.
- Assume all names are non-empty strings.
- The keys in the result should be sorted alphabetically (standard Python dictionary order is fine).

Example: `group_by_first_letter(["Alice", "bob", "andy", "Bob", "carol"])` should return `{'a': ['Alice', 'andy'], 'b': ['bob', 'Bob'], 'c': ['carol']}`.

Constraints

0 <= len(names) <= 1000. Each name is a non-empty string of letters (at least one character).

Example

>>> group_by_first_letter(["Alice", "bob", "andy", "Bob", "carol"])
{'a': ['Alice', 'andy'], 'b': ['bob', 'Bob'], 'c': ['carol']}
>>> group_by_first_letter(["Zoe", "amy", "zack"])
{'a': ['amy'], 'z': ['Zoe', 'zack']}
>>> group_by_first_letter([])
{}

10 points ~10 min

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Hints

Use a dictionary to accumulate lists for each first letter.

Access the first character with `name[0]` and convert to lowercase using `.lower()`.

To preserve original order, append each name to the list for its key as you iterate.

No need to sort keys explicitly; Python keeps insertion order and keys will appear in first occurrence order.

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