# LeetCode 1086: High Five

## Problem Restatement

We are given a list of `[id, score]` pairs.

For each student, compute the average of their top 5 scores (integer division).

Return the result sorted by student ID.

The official constraints state that each student has at least 5 scores.

## Input and Output

| Item | Meaning |
|---|---|
| Input | List of `[id, score]` pairs |
| Output | `[[id, top5avg], ...]` sorted by id |

Function shape:

```python
def highFive(items: list[list[int]]) -> list[list[int]]:
    ...
```

## Examples

Example 1:

```python
items = [[1,91],[1,92],[2,93],[2,97],[1,60],[2,77],[1,65],[1,87],[1,100],[2,100],[2,76]]
```

Student 1 scores: `[91,92,60,65,87,100]`. Top 5: `[100,92,91,87,65]`. Avg = `87`.

Student 2 scores: `[93,97,77,100,76]`. Top 5: `[100,97,93,77,76]`. Avg = `88`.

Answer:

```python
[[1,87],[2,88]]
```

## Edge Cases

- Duplicates usually matter; store counts when a set would lose necessary multiplicity.
- Update the frequency structure in the same order the invariant assumes.
- Check empty or one-element inputs if the problem allows them.

## Common Pitfalls

- Do not optimize away the invariant; the code should still make it clear what condition is being maintained.
- Prefer problem-specific names over one-letter variables once the logic becomes stateful.

## Implementation

```python
from collections import defaultdict

class Solution:
    def highFive(self, items: list[list[int]]) -> list[list[int]]:
        scores = defaultdict(list)
        for id_, score in items:
            scores[id_].append(score)

        return [
            [id_, sum(sorted(s, reverse=True)[:5]) // 5]
            for id_, s in sorted(scores.items())
        ]
```

## Testing

```python
def run_tests():
    s = Solution()

    result = s.highFive([[1,91],[1,92],[2,93],[2,97],[1,60],[2,77],[1,65],[1,87],[1,100],[2,100],[2,76]])
    assert result == [[1,87],[2,88]]

    print("all tests passed")

run_tests()
```

| Test | Expected | Why |
|---|---|---|
| Two students | `[[1,87],[2,88]]` | Top 5 averages per student |