he-005

# HumanEval/5

Implement a file **`solution.py`** that completes the function below. Keep the given name and signature; your file must define it at module level.

```python
from typing import List


def intersperse(numbers: List[int], delimeter: int) -> List[int]:
    """ Insert a number 'delimeter' between every two consecutive elements of input list `numbers'
    >>> intersperse([], 4)
    []
    >>> intersperse([1, 2, 3], 4)
    [1, 4, 2, 4, 3]
    """
```

<!-- imported from OpenAI HumanEval (HumanEval/5) -->

# Auto-generated from OpenAI HumanEval HumanEval/5. Do not edit by hand.
from solution import *  # noqa: F401,F403 (prompt helpers may be referenced by tests)
from solution import intersperse as candidate

METADATA = {
    'author': 'jt',
    'dataset': 'test'
}


def check(candidate):
    assert candidate([], 7) == []
    assert candidate([5, 6, 3, 2], 8) == [5, 8, 6, 8, 3, 8, 2]
    assert candidate([2, 2, 2], 2) == [2, 2, 2, 2, 2]


def test_he_005():
    check(candidate)

You are an expert programmer. Solve the task exactly as specified. Output your solution as fenced code blocks using the required file name(s) and the exact function/type signatures requested. Prefer correctness; do not include prose outside code unless asked.

```python
from typing import List


def intersperse(numbers: List[int], delimeter: int) -> List[int]:
    """ Insert a number 'delimeter' between every two consecutive elements of input list `numbers'
    >>> intersperse([], 4)
    []
    >>> intersperse([1, 2, 3], 4)
    [1, 4, 2, 4, 3]
    """
    result = []
    for i, num in enumerate(numbers):
        result.append(num)
        if i < len(numbers) - 1:
            result.append(delimeter)
    return result
```

The user wants a Python function `intersperse` that takes a list of integers `numbers` and an integer `delimeter`.
It should insert `delimeter` between every two consecutive elements of `numbers`.

Examples:
`intersperse([], 4)` -> `[]`
`intersperse([1, 2, 3], 4)` -> `[1, 4, 2, 4, 3]`

Let's think about the logic:
If `numbers` is empty, return `[]`.
If `numbers` has one element, return `numbers`.
Otherwise, for each element except the last, append the element and then the delimiter. Finally, append the last element.

Alternatively, we can use a list comprehension or `itertools`.
A simple loop:
```python
result = []
for i, num in enumerate(numbers):
    result.append(num)
    if i < len(numbers) - 1:
        result.append(delimeter)
return result
```

Or using `sum` with list of lists:
```python
return sum([[num, delimeter] for num in numbers[:-1]], []) + numbers[-1:] if numbers else []
```
But the loop is cleaner and more efficient.

Let's stick to the simple loop or a list comprehension.
Actually, a very pythonic way:
```python
if not numbers:
    return []
result = []
for i in range(len(numbers) - 1):
    result.append(numbers[i])
    result.append(delimeter)
result.append(numbers[-1])
return result
```

Wait, `numbers` could be empty. The problem says `intersperse([], 4) -> []`.
So handle empty case.

Let's write it out carefully.

```python
from typing import List

def intersperse(numbers: List[int], delimeter: int) -> List[int]:
    if not numbers:
        return []
    result = []
    for i in range(len(numbers) - 1):
        result.append(numbers[i])
        result.append(delimeter)
    result.append(numbers[-1])
    return result
```

Check constraints/edge cases:
`numbers = [1]` -> loop range(0) doesn't run, appends `numbers[-1]` which is `[1]`. Correct.
`numbers = [1, 2]` -> loop runs for i=0, appends 1, delimeter. Then appends 2. Result `[1, delimeter, 2]`. Correct.
`numbers = []` -> returns `[]`. Correct.

Could also be written as:
```python
def intersperse(numbers: List[int], delimeter: int) -> List[int]:
    res = []
    for i, x in enumerate(numbers):
        res.append(x)
        if i < len(numbers) - 1:
            res.append(delimeter)
    return res
```
This handles empty list correctly too (loop doesn't run, returns `[]`).

Let's test mentally:
`numbers = []` -> `res = []`, loop doesn't run, returns `[]`. Correct.
`numbers = [1, 2, 3]`, `delimeter = 4`
i=0, x=1 -> res=[1], i<2 -> res=[1, 4]
i=1, x=2 -> res=[1, 4, 2], i<2 -> res=[1, 4, 2, 4]
i=2, x=3 -> res=[1, 4, 2, 4, 3], i<2 False
Returns `[1, 4, 2, 4, 3]`. Correct.

This is clean and efficient.

One minor point: the problem statement uses `delimeter` (typo in original, but we must keep it).
The signature is given.

I will produce the complete `solution.py` content.

```python
from typing import List


def intersperse(numbers: List[int], delimeter: int) -> List[int]:
    """ Insert a number 'delimeter' between every two consecutive elements of input list `numbers'
    >>> intersperse([], 4)
    []
    >>> intersperse([1, 2, 3], 4)
    [1, 4, 2, 4, 3]
    """
    result = []
    for i, num in enumerate(numbers):
        result.append(num)
        if i < len(numbers) - 1:
            result.append(delimeter)
    return result
```

Wait, should I include the docstring? The prompt says "Keep the given name and signature; your file must define it at module level." It's fine to include the docstring as given.

Let's double check if there's any simpler way using `itertools` or list comprehension, but the loop is perfectly fine and standard for HumanEval.

Another option:
```python
def intersperse(numbers: List[int], delimeter: int) -> List[int]:
    if not numbers:
        return []
    return [val for pair in zip(numbers, [delimeter] * (len(numbers) - 1) + [None]) for val in pair if val is not None] # too complex
```
Stick to the loop.

One more check: `numbers` could be large, but `append` is O(1) amortized, so O(N) total. Perfectly fine.

I will format the output as requested.
The prompt asks for a file `solution.py`.
I'll output exactly that.

Code:
```python
from typing import List


def intersperse(numbers: List[int], delimeter: int) -> List[int]:
    """ Insert a number 'delimeter' between every two consecutive elements of input list `numbers'
    >>> intersperse([], 4)
    []
    >>> intersperse([1, 2, 3], 4)
    [1, 4, 2, 4, 3]
    """
    result = []
    for i, num in enumerate(numbers):
        result.append(num)
        if i < len(numbers) - 1:
            result.append(delimeter)
    return result
```

Wait, `len(numbers)` is called in each iteration. It's better to compute it once or use a different approach, but for Python it's fine. To be strictly optimal:
```python
def intersperse(numbers: List[int], delimeter: int) -> List[int]:
    result = []
    n = len(numbers)
    for i, num in enumerate(numbers):
        result.append(num)
        if i < n - 1:
            result.append(delimeter)
    return result
```
Or even simpler:
```python
def intersperse(numbers: List[int], delimeter: int) -> List[int]:
    res = []
    for i in range(len(numbers)):
        res.append(numbers[i])
        if i < len(numbers) - 1:
            res.append(delimeter)
    return res
```
Actually, `itertools.chain` could be used but loop is fine.

I'll stick to the simple loop. It's robust.

One minor thing: The prompt has a typo in the docstring `delimeter` instead of `delimiter`. I will keep it as is.

Final check of the prompt: "Implement a file **`solution.py`** that completes the function below. Keep the given name and signature; your file must define it at module level."

I will provide the code block.
All good.
Proceeding. 
Let's generate it. 
I'll make sure the indentation and imports match.
The code is straightforward.
Done. 
Output matches request.✅
```python
from typing import List


def intersperse(numbers: List[int], delimeter: int) -> List[int]:
    """ Insert a number 'delimeter' between every two consecutive elements of input list `numbers'
    >>> intersperse([], 4)
    []
    >>> intersperse([1, 2, 3], 4)
    [1, 4, 2, 4, 3]
    """
    result = []
    for i, num in enumerate(numbers):
        result.append(num)
        if i < len(numbers) - 1:
            result.append(delimeter)
    return result
```
This looks solid. I will output it. 
One detail: `len(numbers)` inside the loop condition is evaluated each time. It's fine for typical HumanEval constraints, but I can optimize slightly if needed. Not necessary. I'll keep it simple.
Actually, `if i < len(numbers) - 1:` is fine.
I'll produce the final answer. 
All steps verified.✅
Output matches. 
Proceeds. 
[Done]