Set intersection size

Compute the number of common elements between two lists.

sets intersection counting

Write a function `def intersection_size(list_a, list_b):` that returns the number of distinct elements that appear in both `list_a` and `list_b`.

- The lists can contain integers, strings, or any hashable values.
- Duplicates within a single list should be counted only once.
- The order of elements does not matter.

For example, given `[1, 2, 2, 3]` and `[2, 3, 4]`, the distinct common elements are `2` and `3`, so the answer is `2`.

Constraints

- `0 <= len(list_a), len(list_b) <= 10^5`
- Elements are hashable.
- Time: O(n + m) on average.
- Space: O(min(n, m)) or O(n) depending on implementation.

Example

>>> intersection_size([1, 2, 2, 3], [2, 3, 4])
2
>>> intersection_size([], [1, 2])
0
>>> intersection_size(['a', 'b'], ['b', 'c'])
1

5 points ~8 min

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Hints

Convert one list to a set to make membership checks fast.

Then count how many elements of the other list are in that set.

Remember to count each distinct common element only once — using a set for the result will help.

Test Results

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