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Equal Rational Numbers - Solution & Explanation

HardMathString18 min readAsked at: Microsoft, Google
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Problem Statement

Given two strings s and t, each of which represents a non-negative rational number, return true if and only if they represent the same number. The strings may use parentheses to denote the repeating part of the rational number.

A rational number can be represented using up to three parts: <IntegerPart>, <NonRepeatingPart>, and a <RepeatingPart>. The number will be represented in one of the following three ways:

  • <IntegerPart>
    • For example, 12, 0, and 123.
  • <IntegerPart><.><NonRepeatingPart>
    • For example, 0.5, 1., 2.12, and 123.0001.
  • <IntegerPart><.><NonRepeatingPart><(><RepeatingPart><)>
    • For example, 0.1(6), 1.(9), 123.00(1212).

The repeating portion of a decimal expansion is conventionally denoted within a pair of round brackets. For example:

  • 1/6 = 0.16666666... = 0.1(6) = 0.1666(6) = 0.166(66).

 

Example 1:

Input: s = "0.(52)", t = "0.5(25)"
Output: true
Explanation: Because "0.(52)" represents 0.52525252..., and "0.5(25)" represents 0.52525252525..... , the strings represent the same number.

Example 2:

Input: s = "0.1666(6)", t = "0.166(66)"
Output: true

Example 3:

Input: s = "0.9(9)", t = "1."
Output: true
Explanation: "0.9(9)" represents 0.999999999... repeated forever, which equals 1.  [See this link for an explanation.]
"1." represents the number 1, which is formed correctly: (IntegerPart) = "1" and (NonRepeatingPart) = "".

 

Constraints:

  • Each part consists only of digits.
  • The <IntegerPart> does not have leading zeros (except for the zero itself).
  • 1 <= <IntegerPart>.length <= 4
  • 0 <= <NonRepeatingPart>.length <= 4
  • 1 <= <RepeatingPart>.length <= 4

Approach Overview

Problem Overview: Two strings represent rational numbers that may contain repeating decimals using parentheses, such as "0.(52)" or "0.5(25)". The task is to determine whether both strings represent the same numeric value even if their decimal representations look different.

Approach 1: Convert Strings to Fractions (O(n) time, O(1) space)

The reliable way to compare repeating decimals is to convert each string into an exact fraction. Parse the integer part, the non‑repeating decimal part, and the repeating section inside parentheses. A repeating decimal like a.b(c) can be converted using a formula that builds the numerator and denominator from powers of 10. For example, if the non‑repeating section has length k and the repeating section has length r, the denominator becomes 10^k * (10^r - 1). Reduce the resulting fraction using gcd and compare the normalized numerator and denominator for both inputs. This approach avoids floating‑point precision issues and works for all valid representations.

This method relies on careful string parsing and integer arithmetic. You iterate through the string once to extract segments, compute powers of ten, and normalize with gcd. Because only a few integers are stored, the space complexity stays constant. This approach fits well with problems involving math reasoning and precise decimal handling.

Approach 2: Simulate Repetition and Direct Comparison (O(n * k) time, O(k) space)

Another strategy expands the decimal strings by simulating the repeating portion. Parse the repeating block and append it repeatedly until the generated decimal reaches a fixed safe length (commonly around 20–30 digits after the decimal). Do this for both inputs, then compare the resulting normalized decimal strings directly. Because rational numbers eventually repeat, generating enough digits ensures that equivalent values align.

This approach treats the problem primarily as a string manipulation task. You build the decimal representation using concatenation or buffers, pad both results to the same length, and compare characters. It is easier to implement but relies on choosing a sufficiently large repetition length. Time complexity grows with the simulated digits, and extra memory is needed for the constructed strings.

Recommended for interviews: Converting the string representation into reduced fractions is the preferred solution. It runs in O(n) time, avoids precision errors, and demonstrates strong understanding of rational number representation. The simulation approach shows practical reasoning about repeating decimals but is typically considered less robust.

Approach 1: Convert Strings to Fractions

We can represent each decimal number, potentially with repeating decimals, as a fraction. By converting the number represented by the string into a fraction (numerator and denominator) in a consistent way, we can directly compare the fractions. This involves algebraic manipulation to handle the repeating parts.

For example, 0.1(6) can be represented as 1/6. By converting the repeating part into a fraction, you maintain precision without dealing with floating point inaccuracies. Once both numbers are represented as fractions, compare the fractions as ratios of their numerators and denominators to determine if they represent the same number.

This solution defines a function toFraction that converts a string representation of a rational number into its (numerator, denominator) form by interpreting its integer, non-repeating, and repeating sections separately.

The method uses algebraic equivalents for fractions of repeating decimals to convert them. gcd is used to reduce the resulting fraction. Once both numbers are converted, they are compared directly as fractions.

Code

C

C++

Java

Python

C#

JavaScript

Complexity

Time Complexity: O(n), where n is the length of the maximum string length, since we handle each character a constant number of times.

Space Complexity: O(1), as we're using a constant amount of additional space for our number components and calculations.

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Approach 2: Simulate Repetition and Direct Comparison

Instead of converting into a fraction, simulate the actual number by expanding it up to a certain number of places where we are confident they will differ if they are indeed different numbers. Focus on accurately reproducing the decimal expansion including the repeating part, and use this expanded version for comparison.

You could use string manipulation to handle the repeat section and use a loop to add the repeating digits until you achieve a sufficient significant length (e.g., 20-30 digits) for a robust comparison.

This solution focuses on simulating the decimal expansion by repeating the portion inside parentheses enough times to ensure that, if the numbers are the same, they will coincide over a sufficient length to determine equality. The expand_decimal function repeats the periodic section until reaching a designated precision and then compares the results much like directly comparing two string values.

Code

Python

JavaScript

Complexity

Time Complexity: O(p), where p is the precision we choose (20 in this case), because we repeatedly generate strings to be compared until this precision is achieved.

Space Complexity: O(p), primarily used to store sequences of the generated decimal until precision is satisfied.

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Complexity Comparison

ApproachComplexity
Convert Strings to Fractions

Time Complexity: O(n), where n is the length of the maximum string length, since we handle each character a constant number of times.

Space Complexity: O(1), as we're using a constant amount of additional space for our number components and calculations.

Simulate Repetition and Direct Comparison

Time Complexity: O(p), where p is the precision we choose (20 in this case), because we repeatedly generate strings to be compared until this precision is achieved.

Space Complexity: O(p), primarily used to store sequences of the generated decimal until precision is satisfied.

Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Convert Strings to FractionsO(n)O(1)Best general solution; exact comparison without floating‑point errors
Simulate Repetition and CompareO(n * k)O(k)Useful for quick implementation using string expansion of repeating parts

Video Solution

花花酱 LeetCode 972. Equal Rational Numbers - 刷题找工作 EP239Hua Hua1,032 views views

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Frequently Asked Questions

Is Equal Rational Numbers easy or hard?
Equal Rational Numbers is classified as a Hard problem. The difficulty comes from correctly handling repeating decimal patterns and converting them into exact fractions without floating‑point rounding issues. Strong math reasoning and careful string parsing are required.
Equal Rational Numbers Python/Java solution
In Python or Java, parse the string sections and compute the fraction using integer math. Python benefits from arbitrary‑precision integers, while Java implementations typically use long values with gcd normalization. Both languages follow the same O(n) fraction conversion strategy.
How to solve Equal Rational Numbers in O(n)?
Parse the rational number into three parts: integer portion, non‑repeating decimal, and repeating cycle. Use a mathematical formula to convert the repeating decimal into a fraction with denominator 10^k * (10^r − 1). Reduce the fraction using gcd and compare both normalized fractions. The parsing step is linear in the string length, giving O(n) time.
What is the best approach for Equal Rational Numbers?
The best approach converts each rational number string into a reduced fraction. Parse the integer part, non‑repeating decimals, and repeating section, then compute the exact numerator and denominator using powers of 10. After reducing with gcd, simply compare the fractions. This runs in O(n) time and avoids floating‑point precision errors.
Is Equal Rational Numbers asked at Google/Amazon/Meta?
This problem is categorized as a hard math and string parsing problem and appears in advanced interview practice sets similar to those used by companies like Google and Meta. It tests understanding of rational number representation, precise arithmetic, and careful string parsing.
What data structure is used in Equal Rational Numbers?
The solution primarily uses string parsing and integer arithmetic rather than complex data structures. Variables store the integer, non‑repeating, and repeating segments, and mathematical operations convert them into a fraction. Some implementations also use helper functions like gcd for fraction reduction.
What is the time complexity of Equal Rational Numbers?
The optimal fraction conversion solution runs in O(n) time where n is the length of the input string. Each string is parsed once and constant‑time arithmetic operations are applied to compute the numerator and denominator. Space complexity is O(1) because only a few integers are stored.

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