diff --git "a/leetcode3/\353\263\200\354\247\200\355\230\221/v3/1311. Get Watched Videos by Your Friends.py" "b/leetcode3/\353\263\200\354\247\200\355\230\221/v3/1311. Get Watched Videos by Your Friends.py"
new file mode 100644
index 00000000..8cb6046c
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+++ "b/leetcode3/\353\263\200\354\247\200\355\230\221/v3/1311. Get Watched Videos by Your Friends.py"	
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+'''
+1. 아이디어 :
+    id에서 출발해 BFS를 돌면서 거리(level)를 함께 큐에 저장한다.
+    거리가 level인 친구들을 모은 뒤, 그 친구들이 본 영상들을 세어서
+    (시청 횟수, 제목) 오름차순으로 정렬해 반환한다.
+2. 시간복잡도 :
+   o(V + E + N log N) — 그래프 탐색 V+E, 영상 정렬 N log N
+3. 자료구조/알고리즘 :
+   BFS, defaultdict, Counter
+
+'''
+
+from collections import defaultdict, deque, Counter
+
+class Solution:
+    def watchedVideosByFriends(self, watchedVideos: list[list[str]], friends: list[list[int]], id: int, level: int) -> list[str]:
+        n = len(friends)
+        friend_map = defaultdict(list)
+        for i, friend in enumerate(friends):
+            friend_map[i].extend(friend)
+
+        visited = [False] * n
+        queue = deque()
+        queue.append((0, id))
+        visited[id] = True
+
+        levels = []
+        while queue:
+            i, current = queue.pop()
+
+            if i == level:
+                levels.append(current)
+            else:
+                for nxt in friend_map[current]:
+                    if not visited[nxt]:
+                        visited[nxt] = True
+                        queue.append((i + 1, nxt))
+
+        ans = []
+        for i in levels:
+            for j in watchedVideos[i]:
+                ans.append(j)
+
+        return [key for key, value in sorted(Counter(ans).items(), key=lambda x: (x[1], x[0]))]