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

Constraints:

-2³¹ <= numerator, denominator <= 2³¹ - 1
denominator != 0

23:T5d7,# Intuition\nUpvote if you find this helpful!!\n\n# Approach\nJust run this code and do dry run on any example you will easily get the approach behind these code.\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(N)\n\n# Code\n```\nclass Solution {\npublic:\n string fractionToDecimal(int numerator, int denominator) {\n ios_base::sync_with_stdio(false);\n cin.tie(0);\n cout.tie(0);\n\n if (numerator == 0)\n return "0";\n\n string ans;\n if ((numerator > 0) ^ (denominator > 0))\n ans += "-";\n cout << "ans : " << ans << endl;\n\n long n = labs(numerator);\n long d = labs(denominator);\n long r = n % d;\n ans += to_string(n / d);\n cout << "ans : " << ans << endl;\n\n if (r == 0)\n return ans;\n\n ans += ".";\n cout << "ans : " << ans << endl;\n unordered_map m;\n while (r)\n {\n cout << "rem : " << r << endl;\n if (m.find(r) != m.end())\n {\n ans.insert(m[r], "(");\n ans += ")";\n cout << "ans : " << ans << endl;\n break;\n }\n\n m[r] = ans.size();\n cout << "m[r] : " << m[r] << endl;\n r = r * 10;\n ans += to_string(r / d);\n cout << "ans : " << ans << endl;\n r = r % d;\n }\n return ans;\n }\n};\n```24:T457,# Intuition\n\n\n# Approach\n\n\n# Complexity\n- Time complexity:\n\n\n- Sp

Fraction to Recurring Decimal

Problem Description

Problem Description

Answer Panel