# Sum of given N fractions in reduced form

Given two arrays arr1[] and arr2[] of length N which contains Numerator and Denominator of N fractions respectively, the task is to find the sum of the given N fractions in reduced form.
Examples:

Input: arr1[] = { 1, 2, 5 }, arr2[] = { 2, 1, 6 }
Output: 10/3
Input: arr1[] = { 1, 1 } arr2[] = { 2, 2 }
Output: 1/1

Approach:

1. Find the Least Common Multiple(LCM) of all the denominators stored in arr2[].
2. Change numerator of every fraction stored in arr1[] as:

Let L be the resultant LCM of all denominator(say L) and Numerator and Denominator of the fraction be N and D respectively.
Then then value of each numerator must be changed to:

1. Find the sum of new Numerator(say sumN) formed after above step.
2. Divide the sumL and L by the GCD of sumL and L to get the resultant fraction in reduced form.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach #include  using namespace std;   // Function to find  GCD of a & b // using Euclid Lemma int gcd(int a, int b) {     // Base Case     if (b == 0) {         return a;     }       return gcd(b, a % b); }   // Function to find the LCM of all // elements in arr[] int findlcm(int arr[], int n) {     // Initialize result     int ans = arr[0];       // Iterate arr[] to find LCM     for (int i = 1; i < n; i++) {         ans = (((arr[i] * ans)) / (gcd(arr[i], ans)));     }       // Return the final LCM     return ans; }   // Function to find the sum of N // fraction in reduced form void addReduce(int n, int num[],                int den[]) {       // To store the sum of all     // final numerators     int final_numerator = 0;       // Find the LCM of all denominator     int final_denominator = findlcm(den, n);       // Find the sum of all N     // numerators & denominators     for (int i = 0; i < n; i++) {           // Add each fraction one by one         final_numerator = final_numerator                           + (num[i]) * (final_denominator                                         / den[i]);     }       // Find GCD of final numerator and     // denominator     int GCD = gcd(final_numerator,                   final_denominator);       // Convert into reduced form     // by dividing from GCD     final_numerator /= GCD;     final_denominator /= GCD;       // Print the final fraction     cout << final_numerator          << "/"          << final_denominator          << endl; }   // Driven Code int main() {     // Given N     int N = 3;       // Given Numerator     int arr1[] = { 1, 2, 5 };       // Given Denominator     int arr2[] = { 2, 1, 6 };       // Function Call     addReduce(N, arr1, arr2);     return 0; }

## Java

 // Java program for the above approach  import java.util.*;   class GFG{   // Function to find GCD of a & b // using Euclid Lemma static int gcd(int a, int b) {           // Base case     if (b == 0)     {         return a;     }       return gcd(b, a % b); }   // Function to find the LCM of all // elements in arr[] static int findlcm(int arr[], int n) {           // Initialize result     int ans = arr[0];       // Iterate arr[] to find LCM     for(int i = 1; i < n; i++)      {         ans = (((arr[i] * ans)) /               (gcd(arr[i], ans)));     }       // Return the final LCM     return ans; }   // Function to find the sum of N // fraction in reduced form static void addReduce(int n, int num[],                              int den[]) {           // To store the sum of all     // final numerators     int final_numerator = 0;       // Find the LCM of all denominator     int final_denominator = findlcm(den, n);       // Find the sum of all N     // numerators & denominators     for(int i = 0; i < n; i++)     {           // Add each fraction one by one         final_numerator = final_numerator + (num[i]) *                          (final_denominator / den[i]);     }       // Find GCD of final numerator and     // denominator     int GCD = gcd(final_numerator,                   final_denominator);       // Convert into reduced form     // by dividing from GCD     final_numerator /= GCD;     final_denominator /= GCD;       // Print the final fraction     System.out.println(final_numerator + "/" +                         final_denominator); }   // Driver code public static void main(String[] args) {           // Given N     int N = 3;           // Given numerator     int arr1[] = { 1, 2, 5 };           // Given denominator     int arr2[] = { 2, 1, 6 };           // Function call     addReduce(N, arr1, arr2); } }   // This code is contributed by offbeat

## C#

 // C# program for the above approach  using System; class GFG{    // Function to find GCD of a & b // using Euclid Lemma static int gcd(int a, int b) {            // Base case     if (b == 0)     {         return a;     }        return gcd(b, a % b); }    // Function to find the LCM of all // elements in arr[] static int findlcm(int []arr, int n) {            // Initialize result     int ans = arr[0];        // Iterate arr[] to find LCM     for(int i = 1; i < n; i++)      {         ans = (((arr[i] * ans)) /               (gcd(arr[i], ans)));     }        // Return the final LCM     return ans; }    // Function to find the sum of N // fraction in reduced form static void addReduce(int n, int []num,                              int []den) {            // To store the sum of all     // final numerators     int final_numerator = 0;        // Find the LCM of all denominator     int final_denominator = findlcm(den, n);        // Find the sum of all N     // numerators & denominators     for(int i = 0; i < n; i++)     {            // Add each fraction one by one         final_numerator = final_numerator + (num[i]) *                          (final_denominator / den[i]);     }        // Find GCD of final numerator and     // denominator     int GCD = gcd(final_numerator,                   final_denominator);        // Convert into reduced form     // by dividing from GCD     final_numerator /= GCD;     final_denominator /= GCD;        // Print the final fraction     Console.Write(final_numerator + "/" +                    final_denominator); }    // Driver code public static void Main(string[] args) {            // Given N     int N = 3;            // Given numerator     int []arr1 = { 1, 2, 5 };            // Given denominator     int []arr2 = { 2, 1, 6 };            // Function call     addReduce(N, arr1, arr2); } }    // This code is contributed by Ritik Bansal

Output:

10/3



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