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
                    


View More Similar Problems

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

View Solution →

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 <

View Solution →

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

View Solution →

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

View Solution →

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

View Solution →

Kitty's Calculations on a Tree

Kitty has a tree, T , consisting of n nodes where each node is uniquely labeled from 1 to n . Her friend Alex gave her q sets, where each set contains k distinct nodes. Kitty needs to calculate the following expression on each set: where: { u ,v } denotes an unordered pair of nodes belonging to the set. dist(u , v) denotes the number of edges on the unique (shortest) path between nodes a

View Solution →