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

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

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


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)"


  • -231 <= numerator, denominator <= 231 - 1
  • denominator != 0
Fraction to Recurring Decimal– LeetCode Solutions
Fraction to Recurring Decimal Solution in C++:
class Solution {
  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 += ')';
      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)

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

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

    Map<Long, Integer> seen = new HashMap<>();

    for (long r = n % d; r > 0; r %= d) {
      if (seen.containsKey(r)) {
        sb.insert(seen.get(r), "(");
      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]:] + ')'
      dict[remainder] = len(ans)
      remainder *= 10
      ans += str(remainder // denominator)
      remainder %= denominator

    return ans

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