# Fraction to Recurring Decimal LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

Fraction to Recurring Decimal LeetCode Problem | LeetCode Problems For Beginners | LeetCode Problems & Solutions | Improve Problem Solving Skills | LeetCode Problems Java | LeetCode Solutions in C++

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In this post, you will find the solution for the Fraction to Recurring Decimal in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemFraction to Recurring Decimal– LeetCode Problem

`Fraction to Recurring Decimal– LeetCode Problem`

### Problem:

Given two integers representing the `numerator` and `denominator` of a fraction, return the fraction in string format.

If the fractional part is repeating, enclose the repeating part in parentheses.

If multiple answers are possible, return any of them.

It is guaranteed that the length of the answer string is less than `104` for all the given inputs.

Example 1:

```Input: numerator = 1, denominator = 2
Output: "0.5"
```

Example 2:

```Input: numerator = 2, denominator = 1
Output: "2"
```

Example 3:

```Input: numerator = 4, denominator = 333
Output: "0.(012)"
```

Constraints:

• `-231 <= numerator, denominator <= 231 - 1`
• `denominator != 0`
`Fraction to Recurring Decimal– LeetCode Solutions`
`Fraction to Recurring Decimal Solution in C++:`
```class Solution {
public:
string fractionToDecimal(int numerator, int denominator) {
if (numerator == 0)
return "0";

string ans;

if (numerator < 0 ^ denominator < 0)
ans += "-";

long n = labs(numerator);
long d = labs(denominator);
ans += to_string(n / d);

if (n % d == 0)
return ans;

ans += '.';
unordered_map<int, int> seen;

for (long r = n % d; r; r %= d) {
if (seen.count(r)) {
ans.insert(seen[r], 1, '(');
ans += ')';
break;
}
seen[r] = ans.size();
r *= 10;
ans += to_string(r / d);
}

return ans;
}
};
```
`Fraction to Recurring Decimal Solution in Java:`
```class Solution {
public String fractionToDecimal(int numerator, int denominator) {
if (numerator == 0)
return "0";

StringBuilder sb = new StringBuilder();

if (numerator < 0 ^ denominator < 0)
sb.append("-");

long n = Math.abs((long) numerator);
long d = Math.abs((long) denominator);
sb.append(n / d);

if (n % d == 0)
return sb.toString();

sb.append(".");
Map<Long, Integer> seen = new HashMap<>();

for (long r = n % d; r > 0; r %= d) {
if (seen.containsKey(r)) {
sb.insert(seen.get(r), "(");
sb.append(")");
break;
}
seen.put(r, sb.length());
r *= 10;
sb.append(r / d);
}

return sb.toString();
}
}
```
`Fraction to Recurring Decimal Solution in Python:`
```class Solution:
def fractionToDecimal(self, numerator: int, denominator: int) -> str:
if numerator == 0:
return '0'

ans = ''

if (numerator < 0) ^ (denominator < 0):
ans += '-'

numerator = abs(numerator)
denominator = abs(denominator)
ans += str(numerator // denominator)

if numerator % denominator == 0:
return ans

ans += '.'
dict = {}

remainder = numerator % denominator
while remainder:
if remainder in dict:
ans = ans[:dict[remainder]] + '(' + ans[dict[remainder]:] + ')'
break
dict[remainder] = len(ans)
remainder *= 10
ans += str(remainder // denominator)
remainder %= denominator

return ans
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