LeetCode #972 — HARD

Equal Rational Numbers

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

"0.(52)"
"0.5(25)"

Example 2

"0.1666(6)"
"0.166(66)"

Example 3

"0.9(9)"
"1."

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #972: Equal Rational Numbers
class Solution {
  public boolean isRationalEqual(String s, String t) {
    return Math.abs(valueOf(s) - valueOf(t)) < 1e-9;
  }

  private static double[] ratios = new double[] {1.0, 1.0 / 9, 1.0 / 99, 1.0 / 999, 1.0 / 9999};

  private double valueOf(final String s) {
    if (!s.contains("("))
      return Double.valueOf(s);

    // Get the indices..
    final int leftParenIndex = s.indexOf('(');
    final int rightParenIndex = s.indexOf(')');
    final int dotIndex = s.indexOf('.');

    // integerAndNonRepeating := <IntegerPart><.><NonRepeatingPart>
    final double nonRepeating = Double.valueOf(s.substring(0, leftParenIndex));
    final int nonRepeatingLength = leftParenIndex - dotIndex - 1;

    // repeating := <RepeatingPart>
    final int repeating = Integer.parseInt(s.substring(leftParenIndex + 1, rightParenIndex));
    final int repeatingLength = rightParenIndex - leftParenIndex - 1;
    return nonRepeating + repeating * Math.pow(0.1, nonRepeatingLength) * ratios[repeatingLength];
  }
}

// Accepted solution for LeetCode #972: Equal Rational Numbers
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #972: Equal Rational Numbers
// class Solution {
//   public boolean isRationalEqual(String s, String t) {
//     return Math.abs(valueOf(s) - valueOf(t)) < 1e-9;
//   }
// 
//   private static double[] ratios = new double[] {1.0, 1.0 / 9, 1.0 / 99, 1.0 / 999, 1.0 / 9999};
// 
//   private double valueOf(final String s) {
//     if (!s.contains("("))
//       return Double.valueOf(s);
// 
//     // Get the indices..
//     final int leftParenIndex = s.indexOf('(');
//     final int rightParenIndex = s.indexOf(')');
//     final int dotIndex = s.indexOf('.');
// 
//     // integerAndNonRepeating := <IntegerPart><.><NonRepeatingPart>
//     final double nonRepeating = Double.valueOf(s.substring(0, leftParenIndex));
//     final int nonRepeatingLength = leftParenIndex - dotIndex - 1;
// 
//     // repeating := <RepeatingPart>
//     final int repeating = Integer.parseInt(s.substring(leftParenIndex + 1, rightParenIndex));
//     final int repeatingLength = rightParenIndex - leftParenIndex - 1;
//     return nonRepeating + repeating * Math.pow(0.1, nonRepeatingLength) * ratios[repeatingLength];
//   }
// }

# Accepted solution for LeetCode #972: Equal Rational Numbers
class Solution:
  def isRationalEqual(self, s: str, t: str) -> bool:
    ratios = [1, 1 / 9, 1 / 99, 1 / 999, 1 / 9999]

    def valueOf(s: str) -> float:
      if s.find('(') == -1:
        return float(s)

      # Get the indices.
      leftParenIndex = s.find('(')
      rightParenIndex = s.find(')')
      dotIndex = s.find('.')

      # integerAndNonRepeating := <IntegerPart><.><NonRepeatingPart>
      integerAndNonRepeating = float(s[:leftParenIndex])
      nonRepeatingLength = leftParenIndex - dotIndex - 1

      # repeating := <RepeatingPart>
      repeating = int(s[leftParenIndex + 1:rightParenIndex])
      repeatingLength = rightParenIndex - leftParenIndex - 1
      return integerAndNonRepeating + repeating * 0.1**nonRepeatingLength * ratios[repeatingLength]

    return abs(valueOf(s) - valueOf(t)) < 1e-9

// Accepted solution for LeetCode #972: Equal Rational Numbers
struct Solution;

impl Solution {
    fn is_rational_equal(s: String, t: String) -> bool {
        let a = Self::parse(s);
        let b = Self::parse(t);
        (a - b).abs() < std::f64::EPSILON
    }

    fn parse(s: String) -> f64 {
        let n = s.len();
        if let Some(dot) = s.find('.') {
            let integer_part = &s[..dot];
            if let Some(lparen) = s.find('(') {
                let non_repeating_part = &s[dot + 1..lparen];
                let repeating_part = &s[lparen + 1..n - 1];
                Self::rational(integer_part, non_repeating_part, repeating_part)
            } else {
                let non_repeating_part = &s[dot + 1..];
                Self::rational(integer_part, non_repeating_part, "")
            }
        } else {
            s.parse::<f64>().unwrap()
        }
    }

    fn rational(integer_part: &str, non_repeating_part: &str, repeating_part: &str) -> f64 {
        let a = integer_part.parse::<f64>().unwrap_or(0.0);
        let n = non_repeating_part.len();
        let b = non_repeating_part.parse::<f64>().unwrap_or(0.0);
        let m = repeating_part.len();
        let c = repeating_part.parse::<f64>().unwrap_or(0.0);
        let mut d = 0.0;
        for _ in 0..m {
            d *= 10.0;
            d += 9.0;
        }
        let mut e = 1.0;
        for _ in 0..n {
            e *= 10.0;
        }
        if n == 0 && m == 0 {
            return a;
        }
        if n == 0 {
            return a + c / d;
        }
        if m == 0 {
            return a + b / e;
        }
        a + (b + c / d) / e
    }
}

#[test]
fn test() {
    let s = "0.(52)".to_string();
    let t = "0.5(25)".to_string();
    let res = true;
    assert_eq!(Solution::is_rational_equal(s, t), res);
    let s = "0.1666(6)".to_string();
    let t = "0.166(66)".to_string();
    let res = true;
    assert_eq!(Solution::is_rational_equal(s, t), res);
    let s = "0.9(9)".to_string();
    let t = "1.".to_string();
    let res = true;
    assert_eq!(Solution::is_rational_equal(s, t), res);
}

// Accepted solution for LeetCode #972: Equal Rational Numbers
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #972: Equal Rational Numbers
// class Solution {
//   public boolean isRationalEqual(String s, String t) {
//     return Math.abs(valueOf(s) - valueOf(t)) < 1e-9;
//   }
// 
//   private static double[] ratios = new double[] {1.0, 1.0 / 9, 1.0 / 99, 1.0 / 999, 1.0 / 9999};
// 
//   private double valueOf(final String s) {
//     if (!s.contains("("))
//       return Double.valueOf(s);
// 
//     // Get the indices..
//     final int leftParenIndex = s.indexOf('(');
//     final int rightParenIndex = s.indexOf(')');
//     final int dotIndex = s.indexOf('.');
// 
//     // integerAndNonRepeating := <IntegerPart><.><NonRepeatingPart>
//     final double nonRepeating = Double.valueOf(s.substring(0, leftParenIndex));
//     final int nonRepeatingLength = leftParenIndex - dotIndex - 1;
// 
//     // repeating := <RepeatingPart>
//     final int repeating = Integer.parseInt(s.substring(leftParenIndex + 1, rightParenIndex));
//     final int repeatingLength = rightParenIndex - leftParenIndex - 1;
//     return nonRepeating + repeating * Math.pow(0.1, nonRepeatingLength) * ratios[repeatingLength];
//   }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.