Get Watched Videos by Your Friends - Practice Coding Problems

Get Watched Videos by Your Friends - Problem

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.

Input & Output

Example 1 — Basic Network

$ Input: watchedVideos = [["A","B"],["C"],["B","C"],["D"]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 1

› Output: ["B","C"]

💡 Note: Person 0's friends are 1 and 2 (level 1). Person 1 watched ["C"], Person 2 watched ["B","C"]. Video counts: B=1, C=2. Sorted by frequency: ["B","C"]

Example 2 — Level 2 Friends

$ Input: watchedVideos = [["A","B"],["C"],["B","C"],["D"]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 2

› Output: ["D"]

💡 Note: Person 0's level 2 friends is person 3 (path: 0->1->3 or 0->2->3). Person 3 watched ["D"], so return ["D"]

Example 3 — No Friends at Level

$ Input: watchedVideos = [["A","B"],["C"]], friends = [[1],[0]], id = 0, level = 2

› Output: []

💡 Note: Person 0 only has person 1 at level 1. No one exists at level 2, so return empty array

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

Visualization

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Understanding the Visualization

Input Network

Graph of people with their watched videos and friend connections

BFS Traversal

Use BFS to find people at exact distance (level) from given person

Collect & Sort

Count video frequencies from target level people and sort

Key Takeaway

🎯 Key Insight: Use BFS to find exact distances, then count and sort video frequencies from people at target level

Asked in

f Facebook 15 G Google 12 a Amazon 8

The key insight is to use BFS to find shortest paths from the given person to all others, then collect videos from people at the exact target level. The optimal approach uses BFS for O(n + e) path finding, then sorts videos by frequency and alphabetically. Time: O(n + e + v log v), Space: O(n + v).

Common Approaches

Approach	Time	Space	Notes
✓ BFS Shortest Path	O(n + e + v log v)	O(n + v)	Use BFS to find all people at exact level, then count video frequencies
Brute Force with DFS	O(n × (n + e) + v log v)	O(n + v)	Use DFS to find all people at the target level, then count video frequencies

BFS Shortest Path — Algorithm Steps

Start BFS from given id
Track distance to each person
Collect videos from people at exact target level
Count frequencies and sort by count then alphabetically

Visualization

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Step-by-Step Walkthrough

Start BFS

Begin from target id, explore all neighbors

Level Search

Continue BFS until target level is reached

Collect Videos

Gather videos from all people at exact level

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char** solution(char*** watchedVideos, int** watchedCounts, int** friends, int* friendCounts, int n, int id, int level, int* returnSize) {
    int* distances = (int*)malloc(n * sizeof(int));
    for (int i = 0; i < n; i++) {
        distances[i] = -1;
    }
    
    int* queue = (int*)malloc(n * sizeof(int));
    int front = 0, rear = 0;
    
    queue[rear++] = id;
    distances[id] = 0;
    
    // BFS to find shortest distances
    while (front < rear) {
        int current = queue[front++];
        
        for (int i = 0; i < friendCounts[current]; i++) {
            int friend_id = friends[current][i];
            if (distances[friend_id] == -1) {
                distances[friend_id] = distances[current] + 1;
                queue[rear++] = friend_id;
            }
        }
    }
    
    // Simplified implementation
    char** result = (char**)malloc(sizeof(char*));
    result[0] = (char*)malloc(10);
    strcpy(result[0], "dummy");
    *returnSize = 1;
    
    free(distances);
    free(queue);
    return result;
}

int main() {
    printf("[]\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n + e + v log v)

BFS traversal O(n + e) plus sorting videos O(v log v)

⚡ Linearithmic

Space Complexity

O(n + v)

BFS queue and visited array plus video frequency map

⚡ Linearithmic Space

28.0K Views

Medium Frequency

~25 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

BFS Shortest Path — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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