- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Simplified Fractions in C++
Suppose we have an integer n, we have to find a list of all simplified fractions between 0 and 1 (exclusive) such that the denominator <= n. Here the fractions can be in any order.
So, if the input is like n = 4, then the output will be ["1/2","1/3","1/4","2/3","3/4"] as "2/4" is not a simplified fraction because it can be simplified to "1/2".
To solve this, we will follow these steps −
Define an array ret
for initialize i := 2, when i <= n, update (increase i by 1), do −
for initialize j := 1, when j < i, update (increase j by 1), do −
c := gcd of i and j
a := j / c
b := i / c
insert (a as string concatenate "/" concatenate b as string) at the end of ret
return an array of all unique elements present in ret
Example
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h> using namespace std; void print_vector(vector<string> v){ cout << "["; for(int i = 0; i<v.size(); i++){ cout << v[i] << ", "; } cout << "]"<<endl; } class Solution { public: vector<string> simplifiedFractions(int n) { vector<string> ret; for (int i = 2; i <= n; i++) { for (int j = 1; j < i; j++) { int c = __gcd(i, j); int a = j / c; int b = i / c; ret.push_back(to_string(a) + "/" + to_string(b)); } } set<string> s(ret.begin(), ret.end()); return vector<string>(s.begin(), s.end()); } }; main(){ Solution ob; print_vector(ob.simplifiedFractions(4)); }
Input
4
Output
[1/2, 1/3, 1/4, 2/3, 3/4, ]