Formatted question description: https://leetcode.ca/all/1311.html

# 1311. Get Watched Videos by Your Friends (Medium)

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 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

Related Topics:

## Solution 1. BFS

// OJ: https://leetcode.com/problems/get-watched-videos-by-your-friends/
// Time: O(N + VlogV) where N is the number of people, and V is the count of the result videos
// Space: O(N)
class Solution {
public:
vector<string> watchedVideosByFriends(vector<vector<string>>& watchedVideos, vector<vector<int>>& friends, int id, int level) {
unordered_map<string, int> m;
queue<pair<int, int>> q;
vector<bool> seen(watchedVideos.size(), false);
q.emplace(id, 0);
seen[id] = true;
while (q.size()) {
auto [i, lv] = q.front();
q.pop();
if (lv == level) {
for (auto &v : watchedVideos[i]) m[v]++;
} else {
for (int nei : friends[i]) {
if (seen[nei]) continue;
seen[nei] = true;
if (seen[nei]) continue;
q.emplace(nei, lv + 1);
}
}
}
vector<string> ans;
for (auto &[v, cnt] : m) ans.push_back(v);
sort(begin(ans), end(ans), [&](string &a, string &b) { return m[a] != m[b] ? m[a] < m[b] : a < b; });
return ans;
}
};


Java

• class Solution {
public List<String> watchedVideosByFriends(List<List<String>> watchedVideos, int[][] friends, int id, int level) {
int friendsCount = friends.length;
int[] colors = new int[friendsCount];
final int WHITE = 0;
final int GRAY = 1;
final int BLACK = 2;
queue.offer(id);
colors[id] = GRAY;
int friendLevel = 0;
while (!queue.isEmpty() && friendLevel < level) {
friendLevel++;
int size = queue.size();
for (int i = 0; i < size; i++) {
int curId = queue.poll();
int[] curFriends = friends[curId];
for (int friend : curFriends) {
if (colors[friend] == WHITE) {
colors[friend] = GRAY;
queue.offer(friend);
}
}
colors[curId] = BLACK;
}
}
Map<String, Integer> videosCountsMap = new HashMap<String, Integer>();
boolean flag = false;
while (!queue.isEmpty()) {
int curId = queue.poll();
List<String> curVideos = watchedVideos.get(curId);
for (String video : curVideos) {
int count = videosCountsMap.getOrDefault(video, 0);
count++;
videosCountsMap.put(video, count);
}
}
List<VideoFrequency> list = new ArrayList<VideoFrequency>();
Set<String> videosSet = videosCountsMap.keySet();
for (String video : videosSet) {
int count = videosCountsMap.get(video);
VideoFrequency videoFrequency = new VideoFrequency(video, count);
}
Collections.sort(list);
List<String> videos = new ArrayList<String>();
for (VideoFrequency videoFrequency : list)
return videos;
}
}

class VideoFrequency implements Comparable<VideoFrequency> {
String video;
int frequency;

public VideoFrequency() {

}

public VideoFrequency(String video, int frequency) {
this.video = video;
this.frequency = frequency;
}

public int compareTo(VideoFrequency videoFrequency2) {
if (this.frequency != videoFrequency2.frequency)
return this.frequency - videoFrequency2.frequency;
else
return this.video.compareTo(videoFrequency2.video);
}
}

• // OJ: https://leetcode.com/problems/get-watched-videos-by-your-friends/
// Time: O(N + VlogV) where N is the number of people, and V is the count of the result videos
// Space: O(N)
class Solution {
public:
vector<string> watchedVideosByFriends(vector<vector<string>>& watchedVideos, vector<vector<int>>& friends, int id, int level) {
unordered_map<string, int> m;
queue<pair<int, int>> q;
vector<bool> seen(watchedVideos.size(), false);
q.emplace(id, 0);
seen[id] = true;
while (q.size()) {
auto [i, lv] = q.front();
q.pop();
if (lv == level) {
for (auto &v : watchedVideos[i]) m[v]++;
} else {
for (int nei : friends[i]) {
if (seen[nei]) continue;
seen[nei] = true;
if (seen[nei]) continue;
q.emplace(nei, lv + 1);
}
}
}
vector<string> ans;
for (auto &[v, cnt] : m) ans.push_back(v);
sort(begin(ans), end(ans), [&](string &a, string &b) { return m[a] != m[b] ? m[a] < m[b] : a < b; });
return ans;
}
};

• # 1311. Get Watched Videos by Your Friends
# https://leetcode.com/problems/get-watched-videos-by-your-friends/

class Solution:
def watchedVideosByFriends(self, watchedVideos: List[List[str]], friends: List[List[int]], uid: int, level: int) -> List[str]:

visited = {uid}
deq = collections.deque([(uid, 0)])
res = set()

while deq:
node, l = deq.popleft()

if l > level: break

for nei in friends[node]:
if nei not in visited:
deq.append((nei, l + 1))