# 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) / double(a), v2 = double(b / double(b));
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], fractions[i]);
fractions[i] /= m;
fractions[i] /= m;
if (fractions[i] < 0) {
fractions[i] *= -1;
fractions[i] *= -1;
}
// cout<<fractions[i]<<" "<<fractions[i]<<endl;
}
set<pair<int, int>> st;
for (int i = 0; i < n; i++) {
st.insert({fractions[i], fractions[i]});
}
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 * fraction2) - (fraction2 * fraction1));
for (int[] fraction : fractions) {
int a = fraction, b = fraction;
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()];
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
d = num
if n / d not in s:
vals.append(num)
vals = sorted(vals, key=lambda frac: frac / frac)
for ind in range(len(vals)):
num = vals[ind]
gcd = self.GCD(num, num)
vals[ind] = num // gcd
vals[ind] = num // gcd
return vals```
```

## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

## Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

## Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

## Binary Search Tree : Lowest Common Ancestor

You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b

## Swap Nodes [Algo]

A binary tree is a tree which is characterized by one of the following properties: It can be empty (null). It contains a root node only. It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees. In-order traversal is performed as Traverse the left subtree. Visit root. Traverse the right subtree. For this in-order traversal, start from