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# Convert given Decimal number into an irreducible Fraction

Given a decimal number as N, the task is to convert N into an equivalent irreducible fraction.

An irreducible fraction is a fraction in which numerator and denominator are co-primes i.e., they have no other common divisor other than 1.

Examples:

Input: N = 4.50
Output: 9/2
Explanation:
9/2 in decimal is written as 4.5

Input: N = 0.75
Output: 3/4
Explanation:
3/4 in decimal is written as 0.75

Approach: Follow the steps given below to solve the problem.

• Fetch integral value and fractional part of the decimal number ‘n’.
• Consider the precision value to be 109 to convert the fractional part of the decimal to an integral equivalent.
• Calculate GCD of the integral equivalent of fractional part and precision value.
• Calculate numerator by dividing the integral equivalent of fractional part by GCD value. Calculate the denominator by dividing the precision value by GCD value.
• From the obtained mixed fraction, convert it into an improper fraction.

For example N = 4.50, integral value = 4  and  fractional part = 0.50

Consider precision value to be (109) that is precision value = 1000000000

Calculate GCD(0.50 *  109, 109) = 500000000

Calculate numerator = (0.50 * 10^9) / 500000000 = 1 and denominator = 10^9/ 500000000 = 2

Convert mixed fraction into improper fraction that is fraction = ((4 * 2) + 1) / 2 = 9/2

Below is the implementation of the above approach:

## C++

 `// C++ program for above approach` `#include ` `using` `namespace` `std;`   `// Recursive function to` `// return GCD of a and b` `long` `long` `gcd(``long` `long` `a, ``long` `long` `b)` `{` `    ``if` `(a == 0)` `        ``return` `b;` `    ``else` `if` `(b == 0)` `        ``return` `a;` `    ``if` `(a < b)` `        ``return` `gcd(a, b % a);` `    ``else` `        ``return` `gcd(b, a % b);` `}`   `// Function to convert decimal to fraction` `void` `decimalToFraction(``double` `number)` `{` `    ``// Fetch integral value of the decimal` `    ``double` `intVal = ``floor``(number);`   `    ``// Fetch fractional part of the decimal` `    ``double` `fVal = number - intVal;`   `    ``// Consider precision value to` `    ``// convert fractional part to` `    ``// integral equivalent` `    ``const` `long` `pVal = 1000000000;`   `    ``// Calculate GCD of integral` `    ``// equivalent of fractional` `    ``// part and precision value` `    ``long` `long` `gcdVal` `        ``= gcd(round(fVal * pVal), pVal);`   `    ``// Calculate num and deno` `    ``long` `long` `num` `        ``= round(fVal * pVal) / gcdVal;` `    ``long` `long` `deno = pVal / gcdVal;`   `    ``// Print the fraction` `    ``cout << (intVal * deno) + num` `         ``<< ``"/"` `<< deno << endl;` `}`   `// Driver Code` `int` `main()` `{` `    ``double` `N = 4.5;`   `    ``decimalToFraction(N);`   `    ``return` `0;` `}`

## C

 `// C implementation of the above approach` `#include ` `#include `   `int` `gcfFinder(``int` `a, ``int` `b)` `{ ``// gcf finder` `    ``int` `gcf = 1;` `    ``for` `(``int` `i = 1; i <= a && i <= b; i++) ` `    ``{` `        ``if` `(a % i == 0 && b % i == 0) ` `        ``{` `            ``gcf = i;` `        ``}` `    ``}` `    ``return` `gcf;` `}`   `int` `shortform(``int``* a, ``int``* b)` `{` `    ``for` `(``int` `i = 2; i <= *a && i <= *b; i++) ` `    ``{` `        ``if` `(*a % i == 0 && *b % i == 0) ` `        ``{` `            ``*a = *a / i;` `            ``*b = *b / i;` `        ``}` `    ``}` `    ``return` `0;` `}`   `// Driver Code` `int` `main(``void``)` `{` `    ``// converting decimal into fraction.` `    ``double` `a = 4.50;`   `    ``int` `c = 10000;` `    ``double` `b = (a - ``floor``(a)) * c;` `    ``int` `d = (``int``)``floor``(a) * c + (``int``)(b + .5f);`   `    ``while` `(1) ` `    ``{` `        ``if` `(d % 10 == 0) ` `        ``{` `            ``d = d / 10;` `            ``c = c / 10;` `        ``}` `        ``else` `            ``break``;` `    ``}` `    ``int``* i = &d;` `    ``int``* j = &c;` `    ``int` `t = 0;` `    ``while` `(t != 1)` `    ``{` `        ``int` `gcf = gcfFinder(d, c);` `        ``if` `(gcf == 1) ` `        ``{` `            ``printf``(``"%d/%d\n"``, d, c);` `            ``t = 1;` `        ``}` `        ``else` `        ``{` `            ``shortform(i, j);` `        ``}` `    ``}` `    ``return` `0;` `}` `// this code is contributed by harsh sinha username-` `// harshsinha03`

## Java

 `// Java program for the above approach` `import` `java.util.*;`   `class` `GFG{` `    `  `// Recursive function to` `// return GCD of a and b` `static` `long` `gcd(``long` `a, ``long` `b)` `{` `    ``if` `(a == ``0``)` `        ``return` `b;` `    ``else` `if` `(b == ``0``)` `        ``return` `a;` `    ``if` `(a < b)` `        ``return` `gcd(a, b % a);` `    ``else` `        ``return` `gcd(b, a % b);` `}` `    `  `// Function to convert decimal to fraction` `static` `void` `decimalToFraction(``double` `number)` `{` `  `  `    ``// Fetch integral value of the decimal` `    ``double` `intVal = Math.floor(number);` ` `  `    ``// Fetch fractional part of the decimal` `    ``double` `fVal = number - intVal;` ` `  `    ``// Consider precision value to` `    ``// convert fractional part to` `    ``// integral equivalent` `    ``final` `long` `pVal = ``1000000000``;` ` `  `    ``// Calculate GCD of integral` `    ``// equivalent of fractional` `    ``// part and precision value` `    ``long` `gcdVal = gcd(Math.round(` `                      ``fVal * pVal), pVal);` `  `  `    ``// Calculate num and deno` `    ``long` `num = Math.round(fVal * pVal) / gcdVal;` `    ``long` `deno = pVal / gcdVal;` ` `  `    ``// Print the fraction` `    ``System.out.println((``long``)(intVal * deno) + ` `                           ``num + ``"/"` `+ deno);` `}` `  `  `// Driver Code` `public` `static` `void` `main(String s[])` `{` `    ``double` `N = ``4.5``;` `    ``decimalToFraction(N);` `}    ` `}`   `// This code is contributed by rutvik_56`

## Python3

 `# Python3 program for the above approach` `from` `math ``import` `floor`   `# Recursive function to` `# return GCD of a and b` `def` `gcd(a, b):` `    `  `    ``if` `(a ``=``=` `0``):` `        ``return` `b` `    ``elif` `(b ``=``=` `0``):` `        ``return` `a` `    ``if` `(a < b):` `        ``return` `gcd(a, b ``%` `a)` `    ``else``:` `        ``return` `gcd(b, a ``%` `b)`   `# Function to convert decimal to fraction` `def` `decimalToFraction(number):` `    `  `    ``# Fetch integral value of the decimal` `    ``intVal ``=` `floor(number)`   `    ``# Fetch fractional part of the decimal` `    ``fVal ``=` `number ``-` `intVal`   `    ``# Consider precision value to` `    ``# convert fractional part to` `    ``# integral equivalent` `    ``pVal ``=` `1000000000`   `    ``# Calculate GCD of integral` `    ``# equivalent of fractional` `    ``# part and precision value` `    ``gcdVal ``=` `gcd(``round``(fVal ``*` `pVal), pVal)`   `    ``# Calculate num and deno` `    ``num``=` `round``(fVal ``*` `pVal) ``/``/` `gcdVal` `    ``deno ``=` `pVal ``/``/` `gcdVal`   `    ``# Print the fraction` `    ``print``((intVal ``*` `deno) ``+` `num, ``"/"``, deno)`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``N ``=` `4.5`   `    ``decimalToFraction(N)`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach` `using` `System;`   `class` `GFG{` `    `  `// Recursive function to` `// return GCD of a and b` `static` `long` `gcd(``long` `a, ``long` `b)` `{` `    ``if` `(a == 0)` `        ``return` `b;` `    ``else` `if` `(b == 0)` `        ``return` `a;` `    ``if` `(a < b)` `        ``return` `gcd(a, b % a);` `    ``else` `        ``return` `gcd(b, a % b);` `}` `    `  `// Function to convert decimal to fraction` `static` `void` `decimalToFraction(``double` `number)` `{`   `    ``// Fetch integral value of the decimal` `    ``double` `intVal = Math.Floor(number);`   `    ``// Fetch fractional part of the decimal` `    ``double` `fVal = number - intVal;`   `    ``// Consider precision value to` `    ``// convert fractional part to` `    ``// integral equivalent` `    ``long` `pVal = 1000000000;`   `    ``// Calculate GCD of integral` `    ``// equivalent of fractional` `    ``// part and precision value` `    ``long` `gcdVal = gcd((``long``)Math.Round(` `                    ``fVal * pVal), pVal);`   `    ``// Calculate num and deno` `    ``long` `num = (``long``)Math.Round(fVal * pVal) / gcdVal;` `    ``long` `deno = pVal / gcdVal;`   `    ``// Print the fraction` `    ``Console.WriteLine((``long``)(intVal * deno) + ` `                          ``num + ``"/"` `+ deno);` `}`   `// Driver Code` `public` `static` `void` `Main(String []s)` `{` `    ``double` `N = 4.5;` `    `  `    ``decimalToFraction(N);` `} ` `}`   `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output:

`9/2`

Time complexity: O(log min(a, b))
Auxiliary space: O(1)

Approach 2: Follow the steps given below to solve the problem.

For larger decimal values the float function automatically rounds off the input resulting in an incorrect response
Using inbuilt python libraries doesn’t round off the input for example for the below input the decimal value the code above rounds off the input

• Import library Decimal to convert a string input into decimal
• Import library Fraction to convert a Decimal input into a fraction
• Now convert the fraction into the string and give the output

For example N = “123456789.25252525”

Decimal(N) takes string input and converts it into decimal value = 123456789.25252525

Now we convert the fraction into string using type casting str(493827157010101/4000000)

result “493827157010101/4000000”

Below is the implementation of the above approach:

## Java

 `import` `java.math.BigDecimal;` `import` `java.math.BigInteger;`   `public` `class` `DecimalToFraction {` `    ``public` `static` `void` `main(String[] args) {` `        ``String N = ``"123456789.25252525"``;` `        ``decimalToFraction(N);` `    ``}`   `    ``public` `static` `void` `decimalToFraction(String number) {` `        ``BigDecimal decimal = ``new` `BigDecimal(number);` `        ``BigInteger numerator = decimal.unscaledValue();` `        ``int` `scale = decimal.scale();` `        ``BigInteger denominator = BigInteger.TEN.pow(scale);` `        ``BigInteger gcd = numerator.gcd(denominator);` `        ``numerator = numerator.divide(gcd);` `        ``denominator = denominator.divide(gcd);`   `        ``System.out.println(numerator + ``"/"` `+ denominator);` `    ``}` `}`

## Python3

 `from` `decimal ``import` `Decimal` `from` `fractions ``import` `Fraction`     `def` `decimalToFraction(number):` `    ``f ``=` `Fraction(Decimal(``str``(number)))` `    ``print``(``str``(f))`     `if` `__name__ ``=``=` `'__main__'``:`   `    ``N ``=` `"123456789.25252525"` `    ``decimalToFraction(N)`   `# This code is contributed by sonusahu050502`

## Javascript

 `// JavaScript program for the above approach`   `const Decimal = require(``'decimal.js'``);` `const { Fraction } = require(``'fractions'``);`   `function` `decimalToFraction(number) {` `    ``const f = ``new` `Fraction(``new` `Decimal(String(number)));` `    ``console.log(String(f));` `}`   `const N = ``"123456789.25252525"``;` `decimalToFraction(N);`   `// This code is contributed by codebraxnzt`

Output

`493827157010101/4000000`

Time complexity: O(k + log n)
Auxiliary space: O(1)

Explanation:
The time complexity of the decimalToFraction function depends on the length of the input number and the efficiency of the Fraction constructor. In this case, the input number is converted to a Decimal object using the Decimal constructor, which has a time complexity of O(k), where k is the number of digits in the input number. Then, the Fraction constructor is called with the Decimal object, which has a time complexity of O(log n), where n is the value of the input Decimal object. Therefore, the overall time complexity of decimalToFraction is O(k + log n).

The auxiliary space complexity of the decimalToFraction function is O(1), as it only uses a constant amount of additional memory to store the Fraction object and its string representation of it.

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