How do you solve Get Watched Videos by Your Friends on LeetCode?

Get Watched Videos by Your Friends (LeetCode #1311) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding BFS is crucial for traversing friend networks effectively.

What is the time complexity of Get Watched Videos by Your Friends?

The optimal solution for Get Watched Videos by Your Friends runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse each friend and their videos only once, making it efficient for larger datasets.

What is the brute force solution for Get Watched Videos by Your Friends?

The brute force approach for Get Watched Videos by Your Friends is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will manually traverse the friends' list to find the videos watched by friends at the specified level. This approach is straightforward but inefficient, especially for larger datasets. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Get Watched Videos by Your Friends?

Always clarify the problem requirements before jumping into coding. Think aloud during the interview to show your thought process. Practice BFS and graph traversal problems to build confidence.

What are common mistakes when solving Get Watched Videos by Your Friends?

Failing to track visited nodes can lead to infinite loops. Not sorting the results correctly can result in incorrect outputs.

#1311

Get Watched Videos by Your Friends

Medium

ArrayHash TableBreadth-First SearchGraph TheorySortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses BFS to efficiently traverse the friends' network and gather videos watched at the specified level. This method minimizes unnecessary checks and leverages data structures for quick lookups.

⚙️

Algorithm

3 steps

1Step 1: Initialize a queue for BFS and a set to track visited nodes.
2Step 2: Perform BFS until reaching the desired level, collecting friends at that level.
3Step 3: Count the frequency of videos watched by those friends and sort the results.

solution.py27 lines

1# Full working Python code
2from collections import defaultdict, deque
3
4def watchedVideosByFriends(watchedVideos, friends, id, level):
5    queue = deque([id])
6    visited = set([id])
7    current_level = 0
8    video_count = defaultdict(int)
9
10    while queue:
11        if current_level == level:
12            break
13        for _ in range(len(queue)):
14            person = queue.popleft()
15            for friend in friends[person]:
16                if friend not in visited:
17                    visited.add(friend)
18                    queue.append(friend)
19        current_level += 1
20
21    while queue:
22        person = queue.popleft()
23        for video in watchedVideos[person]:
24            video_count[video] += 1
25
26    result = sorted(video_count.items(), key=lambda x: (x[1], x[0]))
27    return [video for video, _ in result]

ℹ

Complexity note: The time complexity is O(n) because we traverse each friend and their videos only once, making it efficient for larger datasets.

1Understanding BFS is crucial for traversing friend networks effectively.
2Using data structures like HashMap can significantly simplify counting and sorting tasks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.