> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Last Stone Weight Python Solution with Tests

> Tested Python solution for LeetCode 1046 with 17 pytest cases. Generate a practice environment with lcpy.

LeetCode 1046, Easy. Topics: Array, Heap (Priority Queue). [View on LeetCode](https://leetcode.com/problems/last-stone-weight/description/).

Generate this problem as a practice environment: tested reference solution, 17 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 1046   # by problem number
lcpy gen -s last_stone_weight   # by problem name
```

## Problem

You are given an array of integers `stones` where `stones[i]` is the weight of the `i^th` stone.

We are playing a game with the stones. On each turn, we choose the **heaviest two stones** and smash them together. Suppose the heaviest two stones have weights `x` and `y` with `x <= y`. The result of this smash is:

* If `x == y`, both stones are destroyed, and
* If `x != y`, the stone of weight `x` is destroyed, and the stone of weight `y` has new weight `y - x`.

At the end of the game, there is **at most one** stone left.

Return *the weight of the last remaining stone*. If there are no stones left, return `0`.

### Examples

```
Input: stones = [2,7,4,1,8,1]
Output: 1
Explanation:
We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of the last stone.
```

```
Input: stones = [1]
Output: 1
```

### Constraints

* 1 \<= stones.length \<= 30
* 1 \<= stones\[i] \<= 1000

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/last_stone_weight/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/last_stone_weight/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
import heapq


class Solution:
    # Time: O(n log n)
    # Space: O(n)
    def last_stone_weight(self, stones: list[int]) -> int:
        max_heap = [-stone for stone in stones]
        heapq.heapify(max_heap)

        while len(max_heap) > 1:
            heaviest = -heapq.heappop(max_heap)
            second = -heapq.heappop(max_heap)
            if heaviest != second:
                heapq.heappush(max_heap, -(heaviest - second))

        return -max_heap[0] if max_heap else 0
```

## Complexity

| Time       | Space |
| ---------- | ----- |
| O(n log n) | O(n)  |

## Tags

[NeetCode 150](/catalog/neetcode-150), [NeetCode 250](/catalog/neetcode-250), [NeetCode All](/catalog/neetcode).