**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 :

` ````
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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