<p>Given two integers representing the <code>numerator</code> and <code>denominator</code> of a fraction, return <em>the fraction in string format</em>.</p>

<p>If the fractional part is repeating, enclose the repeating part in parentheses.</p>

<p>If multiple answers are possible, return <strong>any of them</strong>.</p>

<p>It is <strong>guaranteed</strong> that the length of the answer string is less than <code>10<sup>4</sup></code> for all the given inputs.</p>

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

<pre>
<strong>Input:</strong> numerator = 1, denominator = 2
<strong>Output:</strong> &quot;0.5&quot;
</pre>

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

<pre>
<strong>Input:</strong> numerator = 2, denominator = 1
<strong>Output:</strong> &quot;2&quot;
</pre>

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

<pre>
<strong>Input:</strong> numerator = 4, denominator = 333
<strong>Output:</strong> &quot;0.(012)&quot;
</pre>

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

<ul>
	<li><code>-2<sup>31</sup> &lt;=&nbsp;numerator, denominator &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>denominator != 0</code></li>
</ul>

class Solution:
    def fractionToDecimal(self, numerator: int, denominator: int) -> str:
        sign = '-' if numerator / denominator < 0 else ''
        n, d, s, ans = abs(numerator), abs(denominator), {}, str(abs(numerator)//abs(denominator))
        n = n % d
        if n == 0: return sign + ans
        ans, frac, fraccount = ans + '.', '', 0
        while n not in s and n != 0:
            s[n], n = fraccount, n * 10
            frac += str(n//d)
            n = n%d
            fraccount += 1
        if n == 0: return sign + ans + frac
        else:
            return sign + ans + frac[:s[n]] + '(' + frac[s[n]:] + ')'
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