last-stone-weight

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Algorithms

You are given an array of integers <code>stones</code> where <code>stones[i]</code> is the weight of the <code>ith</code> 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 <code>x</code> and <code>y</code> with <code>x &lt;= y</code>. The result of this smash is:

<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>

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 <code>0</code>.

&nbsp;
Example 1:

<pre>
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&#39;s the value of the last stone.
</pre>

Example 2:

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

&nbsp;
Constraints:

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