Get Watched Videos by Your Friends

Problem Statement

Problem Description

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 exactly 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

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

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

Solution

Intuition

To solve this problem, we use a Breadth-First Search (BFS) approach to explore all friends up to the specified level. We start from the given id and use a queue to perform BFS. At each level, we gather the videos watched by the friends and keep track of the frequencies. We then sort the videos by frequency and alphabetically for those with the same frequency.

Time Complexity and Space Complexity Analysis

Time Complexity: The time complexity is $O(n \cdot m \cdot \log m)$ , where n is the number of people and m is the maximum number of videos watched by any person. This is due to the BFS traversal and the sorting of the video frequencies.
Space Complexity: The space complexity is $O(n \cdot m)$ due to the space needed for the BFS queue and the frequency map.

Code

Python

from collections import deque, Counter

class Solution:
    def watchedVideosByFriends(self, watchedVideos: List[List[str]], friends: List[List[int]], id: int, level: int) -> List[str]:
        video_frequency = Counter()
        visited = [False] * len(friends)
        q = deque([id])
        visited[id] = True
        current_level = 0

        while q:
            size = len(q)
            if current_level == level:
                break
            for _ in range(size):
                person = q.popleft()
                for friend in friends[person]:
                    if not visited[friend]:
                        visited[friend] = True
                        q.append(friend)
            current_level += 1

        while q:
            person = q.popleft()
            video_frequency.update(watchedVideos[person])

        return sorted(video_frequency.keys(), key=lambda x: (video_frequency[x], x))

C++

#include <vector>
#include <string>
#include <unordered_map>
#include <queue>
#include <algorithm>
using namespace std;

class Solution {
public:
    vector<string> watchedVideosByFriends(vector<vector<string>>& watchedVideos, vector<vector<int>>& friends, int id, int level) {
        vector<string> result;
        unordered_map<string, int> videoFrequency;
        vector<bool> visited(friends.size(), false);
        queue<int> q;
        q.push(id);
        visited[id] = true;
        int currentLevel = 0;

        while (!q.empty()) {
            int size = q.size();
            if (currentLevel == level) break;

            while (size--) {
                int person = q.front();
                q.pop();
                for (int friendId : friends[person]) {
                    if (!visited[friendId]) {
                        visited[friendId] = true;
                        q.push(friendId);
                    }
                }
            }
            currentLevel++;
        }

        while (!q.empty()) {
            int person = q.front();
            q.pop();
            for (const string& video : watchedVideos[person]) {
                videoFrequency[video]++;
            }
        }

        vector<pair<string, int>> videoList(videoFrequency.begin(), videoFrequency.end());
        sort(videoList.begin(), videoList.end(), [](const pair<string, int>& a, const pair<string, int>& b) {
            if (a.second == b.second) return a.first &lt; b.first;
            return a.second &lt; b.second;
        });

        for (const auto& [video, freq] : videoList) {
            result.push_back(video);
        }

        return result;
    }
};

Java

import java.util.*;

class Solution {
    public List<String> watchedVideosByFriends(List<List<String>> watchedVideos, List<List<Integer>> friends, int id, int level) {
        Map<String, Integer> videoFrequency = new HashMap<>();
        boolean[] visited = new boolean[friends.size()];
        Queue<Integer> queue = new LinkedList<>();
        queue.add(id);
        visited[id] = true;
        int currentLevel = 0;

        while (!queue.isEmpty()) {
            int size = queue.size();
            if (currentLevel == level) break;

            for (int i = 0; i &lt; size; i++) {
                int person = queue.poll();
                for (int friend : friends.get(person)) {
                    if (!visited[friend]) {
                        visited[friend] = true;
                        queue.add(friend);
                    }
                }
            }
            currentLevel++;
        }

        while (!queue.isEmpty()) {
            int person = queue.poll();
            for (String video : watchedVideos.get(person)) {
                videoFrequency.put(video, videoFrequency.getOrDefault(video, 0) + 1);
            }
        }

        List<Map.Entry<String, Integer>> videoList = new ArrayList<>(videoFrequency.entrySet());
        videoList.sort((a, b) -> {
            if (a.getValue().equals(b.getValue())) return a.getKey().compareTo(b.getKey());
            return Integer.compare(a.getValue(), b.getValue());
        });

        List<String> result = new ArrayList<>();
        for (Map.Entry<String, Integer> entry : videoList) {
            result.add(entry.getKey());
        }

        return result;
    }
}

Problem Statement​

Problem Description​

Example​

Constraints​

Solution​

Intuition​

Time Complexity and Space Complexity Analysis​

Code​

Python​

C++​

Java​