Number of Fractions that Sum to 1 - Microsoft 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 the number of pairs of fractions there are that sums to 1.

Constraints

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

Example 1

Input

fractions = [

    [1, 4],

    [2, 5],

    [3, 4],

    [3, 5],

    [5, 10],

    [1, 2],

    [1, 2]

]

Output

5

Explanation

1/4 + 3/4, 2/5 + 3/5, 5/10 + 1/2, 5/10 + 1/2, 1/2 + 1/2 are the five pairs which sum to 1


Solution :



title-img 




                        Solution in C++ :

struct fraction {
    long long up, down;
    fraction(int u, int d) {
        long long gcd = __gcd(u, d);
        up = u / gcd, down = d / gcd;
    }

    const bool operator==(const fraction& that) const {
        return up == that.up and down == that.down;
    }
};

template <>
struct std::hash<fraction> {
    size_t operator()(const fraction& k) const {
        return std::hash<int>()(k.up) ^ std::hash<int>()(k.down);
    }
};

fraction partner(const fraction& a) {
    return fraction(a.down - a.up, a.down);
}

int solve(vector<vector<int>>& fractions) {
    int res = 0;
    unordered_map<fraction, int> fre;

    for (const vector<int>& f : fractions) {
        auto a = fraction(f[0], f[1]);
        auto b = partner(a);  // b = 1 - a;
        res += fre[b];
        fre[a]++;
    }
    return res;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    int gcd(int a, int b) {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
    public int solve(int[][] fractions) {
        int res = 0;
        HashMap<Pair<Integer, Integer>, Integer> hm = new HashMap();
        for (int[] f : fractions) {
            int a = f[0], b = f[1];
            int g = gcd(a, b);
            a /= g;
            b /= g;
            Pair cur = new Pair(a, b);
            Pair match = new Pair(b - a, b);
            res += hm.getOrDefault(match, 0);
            hm.put(cur, hm.getOrDefault(cur, 0) + 1);
        }
        return res;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, fractions):
        def find_gcd(a, b):
            if b == 0:
                return a
            return find_gcd(b, a % b)

        seen = {}
        res = 0
        for n, d in fractions:
            gcd = find_gcd(d, n)
            n //= gcd
            d //= gcd
            res += seen.get((d - n, d), 0)
            seen[(n, d)] = seen.get((n, d), 0) + 1
        return res


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