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 :



title-img 




                        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};
            set.add(result);
        }
        int[][] ans = new int[set.size()][2];
        int index = 0;
        for (int[] answer : set) {
            ans[index++] = answer;
        }
        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:
                s.add(n / d)
                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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