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: 10/3Input: arr1[] = { 1, 1 } arr2[] = { 2, 2 }
Output: 1/1
Approach:
- Find the Least Common Multiple(LCM) of all the denominators stored in arr2[].
- 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 value of each numerator must be changed to:
- Find the sum of new Numerator(say sumN) formed after above step.
- 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 <bits/stdc++.h>using namespace std;// Function to find GCD of a & b// using Euclid Lemmaint 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 formvoid 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 Codeint 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 Lemmastatic 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 formstatic 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 codepublic 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 |
Python3
# Python3 program for the above approach # Function to find GCD of a & b# using Euclid Lemmadef gcd(a, b): # Base Case if (b == 0): return a return gcd(b, a % b)# Function to find the LCM of all# elements in arr[]def findlcm(arr, n): # Initialize result ans = arr[0] # Iterate arr[] to find LCM for i in range(1, n): ans = (((arr[i] * ans)) // (gcd(arr[i], ans))) # Return the final LCM return ans# Function to find the sum of N# fraction in reduced formdef addReduce(n, num, den): # To store the sum of all # final numerators final_numerator = 0 # Find the LCM of all denominator final_denominator = findlcm(den, n) # Find the sum of all N # numerators & denominators for i in range(n): # Add each fraction one by one final_numerator = (final_numerator + (num[i]) * (final_denominator // den[i])) # Find GCD of final numerator and # denominator GCD = gcd(final_numerator, final_denominator) # Convert into reduced form # by dividing from GCD final_numerator //= GCD final_denominator //= GCD # Print the final fraction print(final_numerator, "/", final_denominator)# Driver Code# Given NN = 3 # Given Numeratorarr1 = [ 1, 2, 5 ] # Given Denominatorarr2 = [ 2, 1, 6 ] # Function calladdReduce(N, arr1, arr2)# This code is contributed by code_hunt |
C#
// C# program for the above approach using System;class GFG{ // Function to find GCD of a & b// using Euclid Lemmastatic 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 formstatic 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 codepublic 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 |
Javascript
<script>// Javascript program for the above approach// Function to find GCD of a & b// using Euclid Lemmafunction gcd( a, b){ // Base Case if (b == 0) { return a; } return gcd(b, a % b);}// Function to find the LCM of all// elements in arr[]function findlcm( arr, n){ // Initialize result var ans = arr[0]; // Iterate arr[] to find LCM for (var 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 formfunction addReduce(n, num, den){ // To store the sum of all // final numerators var final_numerator = 0; // Find the LCM of all denominator var final_denominator = findlcm(den, n); // Find the sum of all N // numerators & denominators for (var i = 0; i < n; i++) { // Add each fraction one by one final_numerator = final_numerator + (num[i]) * parseInt(final_denominator / den[i]); } // Find GCD of final numerator and // denominator var GCD = gcd(final_numerator, final_denominator); // Convert into reduced form // by dividing from GCD final_numerator = parseInt(final_numerator / GCD); final_denominator = parseInt(final_denominator / GCD); // Print the final fraction document.write( final_numerator + "/" + final_denominator + "<br>");}// Driven Code// Given Nvar N = 3;// Given Numeratorvar arr1 = [ 1, 2, 5 ];// Given Denominatorvar arr2 = [ 2, 1, 6 ];// Function CalladdReduce(N, arr1, arr2);// This code is contributed by noob2000.</script> |
Output:
10/3
Time Complexity: O(N * log(max(a, b)), where N represents the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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