# Product of given N fractions in reduced form

• Difficulty Level : Easy
• Last Updated : 23 Jun, 2022

Given the Numerator and Denominator of N fractions. The task is to find the product of N fraction and output the answer in reduced form.
Examples:

Input : N = 3
num[] = { 1, 2, 5 }
den[] = { 2, 1, 6 }
Output : 5/6
Product of 1/2 * 2/1 * 5/6 is 10/12.
Reduced form of 10/12 is 5/6.

Input : N = 4
num[] = { 1, 2, 5, 9 }
den[] = { 2, 1, 6, 27 }
Output : 5/18

The idea is to find the product of Numerator in a variable, say new_num. Now, find the product of Denominator in another variable, say new_den.
Now, to find the answer in Reduced form, find the Greatest Common Divisor of new_num and new_den and divide the new_num and new_den by the calculated GCD.
Below is the implementation of this approach:

## C++

 // CPP program to find product of N// fractions in reduced form.#include using namespace std; // Function to return gcd of a and bint gcd(int a, int b){    if (a == 0)        return b;    return gcd(b % a, a);} // Print the Product of N fraction in Reduced Form.void productReduce(int n, int num[], int den[]){    int new_num = 1, new_den = 1;     // finding the product of all N    // numerators and denominators.    for (int i = 0; i < n; i++) {        new_num *= num[i];        new_den *= den[i];    }     // Finding GCD of new numerator and    // denominator    int GCD = gcd(new_num, new_den);     // Converting into reduced form.    new_num /= GCD;    new_den /= GCD;     cout << new_num << "/" << new_den << endl;} // Driven Programint main(){    int n = 3;    int num[] = { 1, 2, 5 };    int den[] = { 2, 1, 6 };     productReduce(n, num, den);    return 0;}

## Java

 // Java program to find product of N// fractions in reduced form. import java.io.*; class GFG {// Function to return gcd of a and bstatic int gcd(int a, int b){    if (a == 0)        return b;    return gcd(b % a, a);} // Print the Product of N fraction in// Reduced Form.static void productReduce(int n, int num[],                                   int den[]){    int new_num = 1, new_den = 1;     // finding the product of all N    // numerators and denominators.    for (int i = 0; i < n; i++) {        new_num *= num[i];        new_den *= den[i];    }     // Finding GCD of new numerator and    // denominator    int GCD = gcd(new_num, new_den);     // Converting into reduced form.    new_num /= GCD;    new_den /= GCD;     System.out.println(new_num + "/" +new_den);}     // Driven Program    public static void main (String[] args) {                 int n = 3;        int num[] = { 1, 2, 5 };        int den[] = { 2, 1, 6 };             productReduce(n, num, den);             }} //This code is contributed by vt_m.

## Python3

 # Python3 program to find# product of N fractions# in reduced form. # Function to return# gcd of a and bdef gcd(a, b):    if (a == 0):        return b;    return gcd(b % a, a); # Print the Product of N# fraction in Reduced Form.def productReduce(n, num, den):    new_num = 1;    new_den = 1;         # finding the product    # of all N numerators    # and denominators.    for i in range(n):        new_num = new_num * num[i];        new_den = new_den * den[i];         # Finding GCD of    # new numerator    # and denominator    GCD = gcd(new_num, new_den);         # Converting into    # reduced form.    new_num = new_num / GCD;    new_den = new_den / GCD;         print(int(new_num), "/",              int(new_den)); # Driver Coden = 3;num = [1, 2, 5];den = [2, 1, 6];productReduce(n, num, den); # This code is contributed# by mits

## C#

 // C# program to find product of N// fractions in reduced form. using System; class GFG {             // Function to return gcd of a and b    static int gcd(int a, int b)    {        if (a == 0)            return b;                     return gcd(b % a, a);    }         // Print the Product of N fraction in    // Reduced Form.    static void productReduce(int n, int []num,                                    int []den)    {        int new_num = 1, new_den = 1;             // finding the product of all N        // numerators and denominators.        for (int i = 0; i < n; i++) {            new_num *= num[i];            new_den *= den[i];        }             // Finding GCD of new numerator and        // denominator        int GCD = gcd(new_num, new_den);             // Converting into reduced form.        new_num /= GCD;        new_den /= GCD;             Console.WriteLine(new_num + "/" +new_den);    }     // Driven Program    public static void Main () {                 int n = 3;        int []num = { 1, 2, 5 };        int []den = { 2, 1, 6 };             productReduce(n, num, den);             }} //This code is contributed by vt_m.

## PHP

 

## Javascript

 

Output :

5/6

Time Complexity: O(n + log(min(a,b)))
Auxiliary Space: O(log(min(a,b)))

How to avoid overflow?
The above solution causes overflow for large numbers. We can avoid overflow by first finding prime factors of all numerators and denominators. Once we have found prime factors, we can cancel common prime factors.
Note : When you are asked to represent the answer in form of {P \times {Q} ^ {-1}} .
For these questions, first convert the numerator and denominator in reducible form P / Q as explained above. Then, find modular multiplicative inverse of Q with respect to a prime number m (Generally, 10^9 + 7) which is given in question. After finding modular multiplicative inverse of Q, multiply it with P and take modulo with given prime number m which gives us our required output.
// Thanks VaiBzZk for suggesting this condition.

Please suggest if someone has a better solution which is more efficient in terms of space and time.