# Unique Fractions - Amazon Top Interview Questions

### Problem Statement :

```You are given a list of lists fractions where each list contains [numerator, denominator] which represents the number numerator / denominator.

Return a new list of lists such that the numbers in fractions are:

In their most reduced terms. E.g. 8 / 6 becomes 4 / 3.
Any duplicate fractions that represent the same value are removed.
Sorted in ascending order by their value.
If the number is negative, the - sign should go to the numerator (the input also follows this).

Constraints

n ≤ 100,000 where n is the length of fractions

Example 1

Input

fractions = [
[8, 4],
[2, 1],
[7, 3],
[14, 6],
[10, 2],
[-3, 6]
]

Output
[
[-1, 2],
[2, 1],
[7, 3],
[5, 1]
]

Explanation

Once we reduce the numbers they become [[2, 1], [2, 1], [7, 3], [7, 3], [5, 1], [-1, 2]]. The result then comes from deduping and sorting by value.```

### Solution :

```                        ```Solution in C++ :

bool comparator(vector<int> a, vector<int> b) {
double v1 = double(a[0]) / double(a[1]), v2 = double(b[0] / double(b[1]));
return v1 <= v2;
}
vector<vector<int>> solve(vector<vector<int>>& fractions) {
int n = fractions.size();
if (n == 0) {
return {{}};
}
for (int i = 0; i < n; i++) {
int m = __gcd(fractions[i][0], fractions[i][1]);
fractions[i][0] /= m;
fractions[i][1] /= m;
if (fractions[i][1] < 0) {
fractions[i][1] *= -1;
fractions[i][0] *= -1;
}
// cout<<fractions[i][0]<<" "<<fractions[i][1]<<endl;
}
set<pair<int, int>> st;
for (int i = 0; i < n; i++) {
st.insert({fractions[i][0], fractions[i][1]});
}
vector<vector<int>> ans;
for (auto x : st) {
vector<int> temp;
temp.push_back(x.first);
temp.push_back(x.second);
ans.push_back(temp);
}
sort(ans.begin(), ans.end(), comparator);
return ans;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int[][] solve(int[][] fractions) {
Set<int[]> set = new TreeSet<>(
(fraction1,
fraction2) -> (fraction1[0] * fraction2[1]) - (fraction2[0] * fraction1[1]));
for (int[] fraction : fractions) {
int a = fraction[0], b = fraction[1];
int gcd = gcd(Math.max(Math.abs(a), Math.abs(b)), Math.min(Math.abs(a), Math.abs(b)));
int[] result = new int[] {a / gcd, b / gcd};
}
int[][] ans = new int[set.size()][2];
int index = 0;
for (int[] answer : set) {
}
return ans;
}

private int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
}```
```

```                        ```Solution in Python :

class Solution:
def GCD(self, x, y):
if y == 0:
return x
return self.GCD(y, x % y)

def solve(self, fractions):
s = set()
vals = []
for num in fractions:
n = num[0]
d = num[1]
if n / d not in s:
vals.append(num)
vals = sorted(vals, key=lambda frac: frac[0] / frac[1])
for ind in range(len(vals)):
num = vals[ind]
gcd = self.GCD(num[0], num[1])
vals[ind][0] = num[0] // gcd
vals[ind][1] = num[1] // gcd
return vals```
```

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## Reverse a doubly linked list

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