2456. Most Popular Video Creator LeetCode Solution
In this guide, you will get 2456. Most Popular Video Creator LeetCode Solution with the best time and space complexity. The solution to Most Popular Video Creator problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
You are given two string arrays creators and ids, and an integer array views, all of length n. The ith video on a platform was created by creators[i], has an id of ids[i], and has views[i] views.
The popularity of a creator is the sum of the number of views on all of the creator’s videos. Find the creator with the highest popularity and the id of their most viewed video.
If multiple creators have the highest popularity, find all of them.
If multiple videos have the highest view count for a creator, find the lexicographically smallest id.
Note: It is possible for different videos to have the same id, meaning that ids do not uniquely identify a video. For example, two videos with the same ID are considered as distinct videos with their own viewcount.
Return a 2D array of strings answer where answer[i] = [creatorsi, idi] means that creatorsi has the highest popularity and idi is the id of their most popular video. The answer can be returned in any order.
Input: creators = [“alice”,”bob”,”alice”,”chris”], ids = [“one”,”two”,”three”,”four”], views = [5,10,5,4]
Output: [[“alice”,”one”],[“bob”,”two”]]
Explanation:
The popularity of alice is 5 + 5 = 10.
The popularity of bob is 10.
The popularity of chris is 4.
alice and bob are the most popular creators.
For bob, the video with the highest view count is “two”.
For alice, the videos with the highest view count are “one” and “three”. Since “one” is lexicographically smaller than “three”, it is included in the answer.
Example 2:
Input: creators = [“alice”,”alice”,”alice”], ids = [“a”,”b”,”c”], views = [1,2,2]
Output: [[“alice”,”b”]]
Explanation:
The videos with id “b” and “c” have the highest view count.
Since “b” is lexicographically smaller than “c”, it is included in the answer.
Constraints:
n == creators.length == ids.length == views.length
1 <= n <= 105
1 <= creators[i].length, ids[i].length <= 5
creators[i] and ids[i] consist only of lowercase English letters.
0 <= views[i] <= 105
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(n)
2456. Most Popular Video Creator LeetCode Solution in C++
struct Creator {
long popularity; // the popularity sum
string videoId; // the video id that has the maximum view
int maxView; // the maximum view of the creator
};
class Solution {
public:
vector<vector<string>> mostPopularCreator(vector<string>& creators,
vector<string>& ids,
vector<int>& views) {
vector<vector<string>> ans;
unordered_map<string, Creator> nameToCreator;
long maxPopularity = 0;
for (int i = 0; i < creators.size(); ++i) {
if (!nameToCreator.contains(creators[i])) {
nameToCreator[creators[i]] = Creator{
.popularity = views[i],
.videoId = ids[i],
.maxView = views[i],
};
maxPopularity = max(maxPopularity, static_cast<long>(views[i]));
continue;
}
Creator& creator = nameToCreator[creators[i]];
creator.popularity += views[i];
maxPopularity = max(maxPopularity, static_cast<long>(creator.popularity));
if (creator.maxView < views[i] ||
creator.maxView == views[i] && creator.videoId > ids[i]) {
creator.videoId = ids[i];
creator.maxView = views[i];
}
}
for (const auto& [name, creator] : nameToCreator)
if (creator.popularity == maxPopularity)
ans.push_back({name, creator.videoId});
return ans;
}
}; /* code provided by PROGIEZ */
2456. Most Popular Video Creator LeetCode Solution in Java
class Creator {
public long popularity; // the popularity sum
public String videoId; // the video id that has the maximum view
public int maxView; // the maximum view of the creator
public Creator(long popularity, String videoId, int maxView) {
this.popularity = popularity;
this.videoId = videoId;
this.maxView = maxView;
}
}
class Solution {
public List<List<String>> mostPopularCreator(String[] creators, String[] ids, int[] views) {
List<List<String>> ans = new ArrayList<>();
Map<String, Creator> nameToCreator = new HashMap<>();
long maxPopularity = 0;
for (int i = 0; i < creators.length; ++i) {
if (!nameToCreator.containsKey(creators[i])) {
nameToCreator.put(creators[i], new Creator(views[i], ids[i], views[i]));
maxPopularity = Math.max(maxPopularity, (long) views[i]);
continue;
}
Creator creator = nameToCreator.get(creators[i]);
creator.popularity += views[i];
maxPopularity = Math.max(maxPopularity, (long) creator.popularity);
if (creator.maxView < views[i] ||
creator.maxView == views[i] && creator.videoId.compareTo(ids[i]) > 0) {
creator.videoId = ids[i];
creator.maxView = views[i];
}
}
for (Map.Entry<String, Creator> entry : nameToCreator.entrySet())
if (entry.getValue().popularity == maxPopularity)
ans.add(Arrays.asList(entry.getKey(), entry.getValue().videoId));
return ans;
}
} // code provided by PROGIEZ
2456. Most Popular Video Creator LeetCode Solution in Python
class Creator:
def __init__(self, popularity: int, videoId: str, maxView: int):
self.popularity = popularity # the popularity sum
self.videoId = videoId # the video id that has the maximum view
self.maxView = maxView # the maximum view of the creator
class Solution:
def mostPopularCreator(self, creators: list[str],
ids: list[str],
views: list[int]) -> list[list[str]]:
ans = []
maxPopularity = 0
nameToCreator = {}
for name, id, view in zip(creators, ids, views):
if name not in nameToCreator:
nameToCreator[name] = Creator(view, id, view)
maxPopularity = max(maxPopularity, view)
continue
creator = nameToCreator[name]
creator.popularity += view
maxPopularity = max(maxPopularity, creator.popularity)
if (creator.maxView < view or
creator.maxView == view and creator.videoId > id):
creator.videoId = id
creator.maxView = view
for name, creator in nameToCreator.items():
if creator.popularity == maxPopularity:
ans.append([name, creator.videoId])
return ans # code by PROGIEZ
