1311. Get Watched Videos by Your Friends

https://leetcode.com/problems/get-watched-videos-by-your-friends/

There are n people, each person has a unique id between 0 and n-1. Given the arrays watchedVideos and friends, where watchedVideos[i] and friends[i] contain the list of watched videos and the list of friends respectively for the person with id = i.

Level 1 of videos are all watched videos by your friends, level 2 of videos are all watched videos by the friends of your friends and so on. In general, the level k of videos are all watched videos by people with the shortest path equal to k with you. Given your id and the level of videos, return the list of videos ordered by their frequencies (increasing). For videos with the same frequency order them alphabetically from least to greatest.

Example 1:

Input: watchedVideos = [["A","B"],["C"],["B","C"],["D"]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 1
Output: ["B","C"] 
Explanation: 
You have id = 0 (green color in the figure) and your friends are (yellow color in the figure):
Person with id = 1 -> watchedVideos = ["C"] 
Person with id = 2 -> watchedVideos = ["B","C"] 
The frequencies of watchedVideos by your friends are: 
B -> 1 
C -> 2

Example 2:

Input: watchedVideos = [["A","B"],["C"],["B","C"],["D"]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 2
Output: ["D"]
Explanation: 
You have id = 0 (green color in the figure) and the only friend of your friends is the person with id = 3 (yellow color in the figure).

Constraints:

• n == watchedVideos.length == friends.length

• 2 <= n <= 100

• 1 <= watchedVideos[i].length <= 100

• 1 <= watchedVideos[i][j].length <= 8

• 0 <= friends[i].length < n

• 0 <= friends[i][j] < n

• 0 <= id < n

• 1 <= level < n

• if friends[i] contains j, then friends[j] contains i

每个人会收看一些电视节目，人和人之间的朋友关系表示成无向图，问从给定人开始，它的第K层朋友们都看哪些节目，并把节目按收看人数频次输出。第K层 => BFS，找到后把对应节目排序。

class Solution {
public:
    vector<string> watchedVideosByFriends(vector<vector<string>>& watchedVideos, vector<vector<int>>& friends, int id, int level) {
        queue<int> q;
        q.push(id);
        unordered_set<int> visited;
        visited.insert(id);
        for (int i = 0; i < level && !q.empty(); ++i) {
            for (int j = q.size(); j > 0; --j) {
                const auto t = q.front(); q.pop();
                for (auto &f : friends[t]) {
                    if (!visited.count(f)) {
                        q.push(f);
                        visited.insert(f);
                    }
                }
            }
        }
        unordered_map<string, int> freq;
        while (!q.empty()) {
            const auto t = q.front(); q.pop();
            for (const auto c : watchedVideos[t]) {
                ++freq[c];
            }
        }
        const auto cmp = [&](auto &a, auto &b) {
            if (a.second != b.second) return a.second > b.second;
            return a.first > b.first;
        };
        priority_queue<pair<string, int>, vector<pair<string, int>>, decltype(cmp)> pq(cmp);
        for (const auto &p : freq) {
            pq.push(p);
        }
        vector<string> res;
        while (!pq.empty()) {
            const auto t = pq.top(); pq.pop();
            res.push_back(t.first);
        }
        return res;
    }
};