# 1046-last-stone-weight


Try it on <a href='https://leetcode.com/problems/1046-last-stone-weight'>leetcode</a>

## Description
<div class="description">
<div><p>You are given an array of integers <code>stones</code> where <code>stones[i]</code> is the weight of the <code>i<sup>th</sup></code> stone.</p>

<p>We are playing a game with the stones. On each turn, we choose the <strong>heaviest two stones</strong> and smash them together. Suppose the heaviest two stones have weights <code>x</code> and <code>y</code> with <code>x &lt;= y</code>. The result of this smash is:</p>

<ul>
	<li>If <code>x == y</code>, both stones are destroyed, and</li>
	<li>If <code>x != y</code>, the stone of weight <code>x</code> is destroyed, and the stone of weight <code>y</code> has new weight <code>y - x</code>.</li>
</ul>

<p>At the end of the game, there is <strong>at most one</strong> stone left.</p>

<p>Return <em>the smallest possible weight of the left stone</em>. If there are no stones left, return <code>0</code>.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre><strong>Input:</strong> stones = [2,7,4,1,8,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong> 
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.
</pre>

<p><strong>Example 2:</strong></p>

<pre><strong>Input:</strong> stones = [1]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= stones.length &lt;= 30</code></li>
	<li><code>1 &lt;= stones[i] &lt;= 1000</code></li>
</ul>
</div>
</div>

## Solution(Python)
```Python
class Solution:
    # Time Complexity: O(nlogn)
    # Space Complexity: O(n)
    def lastStoneWeight(self, stones: List[int]) -> int:
        heap = [-stone for stone in stones]
        heapq.heapify(heap)

        while len(heap) > 1:
            diff = heapq.heappop(heap) - heapq.heappop(heap)

            if diff != 0:
                heapq.heappush(heap, diff)

        return -heap[0] if len(heap) == 1 else 0

```