Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Unique Fractions in C++
Suppose we have a list of fractions where each fraction contains [numerator, denominator] (numerator / denominator). We have ti find a new list of fractions such that the numbers in fractions are −
In their most reduced terms. (20 / 14 becomes 10 / 7).
Any duplicate fractions (after reducing) will be removed.
Sorted in ascending order by their actual value.
If the number is negative, the '-' sign will be with the numerator.
So, if the input is like {{16, 8},{4, 2},{7, 3},{14, 6},{20, 4},{-6, 12}}, then the output will be [[-1, 2],[2, 1],[7, 3],[5, 1]]
To solve this, we will follow these steps −
Define one set s
n := size of v
make an array r
-
for initialize i := 0, when i < n, update (increase i by 1), do −
c := gcd of |v[i, 0]| and |v[i, 1]|
v[i, 0] := v[i, 0] / c
v[i, 1] := v[i, 1] / c
insert {v[i, 0], v[i, 1]} at the end of r
sort the array r based on their values
make an array ret
-
for initialize i := 0, when i < size of r, update (increase i by 1), do −
-
if ret is not empty and last element of ret is same as r[i], then −
insert r[i] at the end of ret
-
return ret
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<vector<auto>> v) {
cout << "[";
for (int i = 0; i < v.size(); i++) {
cout << "[";
for (int j = 0; j < v[i].size(); j++) {
cout << v[i][j] << ", ";
}
cout << "],";
}
cout << "]" << endl;
}
class Solution {
public:
static bool cmp(vector <int>& a, vector <int>& b){
double aa = (double)a[0] / (double)a[1];
double bb = (double)b[0] / (double)b[1];
return aa < bb;
}
vector<vector<int>> solve(vector<vector<int>>& v) {
set < vector <int> > s;
int n = v.size();
vector < vector <int> > r;
for(int i = 0; i < n; i++){
int c = __gcd(abs(v[i][0]), abs(v[i][1]));
v[i][0] /= c;
v[i][1] /= c;
r.push_back({v[i][0], v[i][1]});
}
sort(r.begin(), r.end(), cmp);
vector < vector <int> > ret;
for(int i = 0; i < r.size(); i++){
if(!ret.empty() && ret.back() == r[i]) continue;
ret.push_back(r[i]);
}
return ret;
}
};
int main(){
vector<vector<int>> v = {{16, 8},{4, 2},{7, 3},{14, 6},{20, 4},{-
6, 12}};
Solution ob;
print_vector(ob.solve(v));
}
Input
{{16, 8},{4, 2},{7, 3},{14, 6},{20, 4},{-6, 12}}
Output
[[-1, 2, ],[2, 1, ],[7, 3, ],[5, 1, ],]