Filter by Age

Return names of people above a given age from a list of dictionaries.

loops conditionals dictionaries lists

Implement a function `filter_by_age(people, min_age)` that accepts a list of dictionaries, each with keys `'name'` (str) and `'age'` (int), and an integer `min_age`. It should return a new list containing the `'name'` of every person whose `'age'` is **greater than or equal to** `min_age`. The order of names in the output must match the order of people in the input. If no one meets the age threshold, return an empty list.

**Function signature:**
```python
def filter_by_age(people: list[dict], min_age: int) -> list[str]:
```

**Assumptions:**
- `people` is a list of dictionaries; each dictionary contains exactly the keys `'name'` (non-empty string) and `'age'` (integer).
- `min_age` is an integer.
- The input list must NOT be modified.

Constraints

0 <= len(people) <= 10^4
-10^3 <= age <= 10^3
-10^3 <= min_age <= 10^3

Example

>>> filter_by_age([{'name': 'Alice', 'age': 25}, {'name': 'Bob', 'age': 30}], 26)
['Bob']
>>> filter_by_age([], 18)
[]

8 points ~12 min

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Hints

Iterate over the list and check each person's age.

Use a list comprehension or a for loop with an if condition.

Append only the 'name' field when age >= min_age.

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