Leetcode #1311: Get Watched Videos by Your Friends
In this guide, we solve Leetcode #1311 Get Watched Videos by Your Friends in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Breadth-First Search, Graph, Array, Hash Table, Sorting
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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
Python Solution
class Solution:
def watchedVideosByFriends(
self,
watchedVideos: List[List[str]],
friends: List[List[int]],
id: int,
level: int,
) -> List[str]:
q = deque([id])
vis = {id}
for _ in range(level):
for _ in range(len(q)):
i = q.popleft()
for j in friends[i]:
if j not in vis:
vis.add(j)
q.append(j)
cnt = Counter()
for i in q:
for v in watchedVideos[i]:
cnt[v] += 1
return sorted(cnt.keys(), key=lambda k: (cnt[k], k))
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.