LeetCode 972 - Equal Rational Numbers

https://leetcode.com/problems/equal-rational-numbers/

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.

In general a rational number can be represented using up to three parts: an integer part, a non-repeating part, and a repeating part. The number will be represented in one of the following three ways:

• <IntegerPart> (e.g. 0, 12, 123)
• <IntegerPart><.><NonRepeatingPart>  (e.g. 0.5, 1., 2.12, 2.0001)
• <IntegerPart><.><NonRepeatingPart><(><RepeatingPart><)> (e.g. 0.1(6), 0.9(9), 0.00(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)

Both 0.1(6) or 0.1666(6) or 0.166(66) are correct representations of 1 / 6.

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) = "".

Note:

1. Each part consists only of digits.
2. The <IntegerPart> will not begin with 2 or more zeros.  (There is no other restriction on the digits of each part.)
3. 1 <= <IntegerPart>.length <= 4
4. 0 <= <NonRepeatingPart>.length <= 4
5. 1 <= <RepeatingPart>.length <= 4

https://leetcode.com/articles/equal-rational-numbers/

As both numbers represent a fraction, we need a fraction class to handle fractions. It should help us add two fractions together, keeping the answer in lowest terms.

Algorithm

We need to make sense of the fraction we are given. The hard part is the repeating part.

Say we have a string like S = "0.(12)". It represents (for $r = \frac{1}{100}$):

$S = \frac{12}{100} + \frac{12}{10000} + \frac{12}{10^6} + \frac{12}{10^8} + \frac{12}{10^{10}} + \cdots$

$S = 12 * (r + r^2 + r^3 + \cdots)$

$S = 12 * \frac{r}{1-r}$

as the sum $(r + r^2 + r^3 + \cdots)$ is a geometric sum.

In general, for a repeating part $x$ with length $k$, we have $r = 10^{-k}$ and the contribution is $\frac{xr}{1-r}$.

The other two parts are easier, as it is just a literal interpretation of the value.

  public boolean isRationalEqual(String S, String T) {

    Fraction f1 = convert(S);

    Fraction f2 = convert(T);

    return f1.n == f2.n && f1.d == f2.d;

  }

  public Fraction convert(String S) {

    int state = 0; // whole, decimal, repeating

    Fraction ans = new Fraction(0, 1);

    int decimal_size = 0;

    for (String part : S.split("[.()]")) {

      state++;

      if (part.isEmpty())

        continue;

      long x = Long.valueOf(part);

      int sz = part.length();

      if (state == 1) { // whole

        ans.iadd(new Fraction(x, 1));

      } else if (state == 2) { // decimal

        ans.iadd(new Fraction(x, (long) Math.pow(10, sz)));

        decimal_size = sz;

      } else { // repeating

        long denom = (long) Math.pow(10, decimal_size);

        denom *= (long) (Math.pow(10, sz) - 1);

        ans.iadd(new Fraction(x, denom));

      }

    }

    return ans;

  }

class Fraction {

  long n, d;

  Fraction(long n, long d) {

    long g = gcd(n, d);

    this.n = n / g;

    this.d = d / g;

  }

  public void iadd(Fraction other) {

    long numerator = this.n * other.d + this.d * other.n;

    long denominator = this.d * other.d;

    long g = Fraction.gcd(numerator, denominator);

    this.n = numerator / g;

    this.d = denominator / g;

  }

  static long gcd(long x, long y) {

    return x != 0 ? gcd(y % x, x) : y;

  }

}

https://medium.com/@poitevinpm/solution-to-leetcode-problem-972-equal-rational-numbers-efcccf0adce8

https://leetcode.com/problems/equal-rational-numbers/discuss/214205/Java-Math-explained

For 0 < q < 1, 1 + q + q^2 + ... = 1 / (1 - q)
Link: https://en.wikipedia.org/wiki/Geometric_progression#Infinite_geometric_series

0.(52) = 0 + 0.52 * (1 + 0.01 + 0.0001 + ...) = 0 + 0.52 / (1 - 0.01)
0.5(25) = 0.5 + 0.025 * (1 + 0.01 + 0.0001 + ...) = 0.5 + 0.025 / (1 - 0.01)

    private List<Double> ratios = Arrays.asList(1.0, 1.0 / 9, 1.0 / 99, 1.0 / 999, 1.0 / 9999);

    public boolean isRationalEqual(String S, String T) {
        return Math.abs(computeValue(S) - computeValue(T)) < 1e-9;
    }

    private double computeValue(String s) {
        if (!s.contains("(")) {
            return Double.valueOf(s);
        } else {
            double intNonRepeatingValue = Double.valueOf(s.substring(0, s.indexOf('(')));
            int nonRepeatingLength = s.indexOf('(') - s.indexOf('.') - 1;
            int repeatingLength = s.indexOf(')') - s.indexOf('(') - 1;
            int repeatingValue = Integer.parseInt(s.substring(s.indexOf('(') + 1, s.indexOf(')')));
            return intNonRepeatingValue + repeatingValue * Math.pow(0.1, nonRepeatingLength) * ratios.get(repeatingLength);
        }
    }

X.  https://zxi.mytechroad.com/blog/math/leetcode-972-equal-rational-numbers/
Expend the string
Extend the string to 16+ more digits and covert it to double.

0.9(9) => 0.99999999999999
0.(52) => 0.525252525252525
0.5(25) => 0.5252525252525

  bool isRationalEqual(string S, string T) {

    auto convert = [](string s) {

      // 0.1234(567)

      // 01234567890

      // i = 1, p = 6

      auto i = s.find('.');

      auto p = s.find('(');

      if (p != string::npos) {

        string r = s.substr(p + 1, s.length() - p - 2);

        s = s.substr(0, p);

        while (s.length() < 16)

          s += r;

      }

      return stod(s);

    };    

    return abs(convert(S) - convert(T)) < 1e-10;

X. https://leetcode.com/problems/equal-rational-numbers/discuss/214203/JavaC%2B%2BPython-Easy-Cheat
good idea but it won't pass interview

    public boolean isRationalEqual(String S, String T) {
        return f(S) == f(T);
    }

    public double f(String S) {
        int i = S.indexOf('(');
        if (i > 0) {
            String base = S.substring(0, i);
            String rep = S.substring(i + 1, S.length() - 1);
            for (int j = 0; j < 20; ++j) base += rep;
            return Double.valueOf(base);
        }
        return Double.valueOf(S);
    }