<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 {
public:
	static std::string fractionToDecimal(int numerator, int denominator) noexcept {
		// 0 divided by anything is still 0
		if (!numerator) {
			return "0";
		}

		// check for result's sign
		std::string ans;
		bool isNegative = (numerator > 0) ^ (denominator > 0);
		if (isNegative) {
			ans.push_back('-');
		}

		// get rid off the negative numbers
		long long num = abs(1LL * numerator), den = abs(1LL * denominator);

		// if numerator is divisible by denominator, return the result
		ans += std::to_string(num / den);
		num %= den;
		if (!num) {
			return ans;
		}
		ans.push_back('.');

		// if numerator and denominator are not co-prime, divide them by their gcd
		long long gcd = std::gcd(num, den);
		num /= gcd;
		den /= gcd;

		// if it is possible to express the denominator as 2^x * 5^y (x, y are whole numbers), then there is no repeating part
		long long temp = den;
		while (temp & 1 ^ 1) {
			temp >>= 1;
		}
		while (!(temp % 5)) {
			temp /= 5;
		}
		bool isRepeating = temp != 1;

		// finding the repeating part
		std::map<long long, int> seen;
		std::string frac;
		int i = 1; // for emplacing '('
		while (true) {
			num *= 10;
			if (seen[num]) {
				i = seen[num];
				break;
			}

			seen[num] = i;
			long long div = num / den;
			frac += std::to_string(div);
			num %= den;
			++i;
		}
		--i;

		if (isRepeating) {
			frac.insert(frac.begin() + i, '(');
			frac.push_back(')');
		}
		else {
			frac.pop_back();
		}
		ans += frac;

		return ans;
	}
};
9300