Alternating Digit Sum
<p>You are given a positive integer <code>n</code>. Each digit of <code>n</code> has a sign according to the following rules:</p>

<ul>
	<li>The <strong>most significant digit</strong> is assigned a <strong>positive</strong> sign.</li>
	<li>Each other digit has an opposite sign to its adjacent digits.</li>
</ul>

<p>Return <em>the sum of all digits with their corresponding sign</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 521
<strong>Output:</strong> 4
<strong>Explanation:</strong> (+5) + (-2) + (+1) = 4.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 111
<strong>Output:</strong> 1
<strong>Explanation:</strong> (+1) + (-1) + (+1) = 1.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 886996
<strong>Output:</strong> 0
<strong>Explanation:</strong> (+8) + (-8) + (+6) + (-9) + (+9) + (-6) = 0.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
</ul>

<p>&nbsp;</p>
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Math
class Solution:
    def alternateDigitSum(self, n: int) -> int:
        return sum((-1) ** i * int(x) for i, x in enumerate(str(n)))
