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1817. Finding the Users Active Minutes

中文文档

Description

You are given the logs for users' actions on LeetCode, and an integer k. The logs are represented by a 2D integer array logs where each logs[i] = [IDi, timei] indicates that the user with IDi performed an action at the minute timei.

Multiple users can perform actions simultaneously, and a single user can perform multiple actions in the same minute.

The user active minutes (UAM) for a given user is defined as the number of unique minutes in which the user performed an action on LeetCode. A minute can only be counted once, even if multiple actions occur during it.

You are to calculate a 1-indexed array answer of size k such that, for each j (1 <= j <= k), answer[j] is the number of users whose UAM equals j.

Return the array answer as described above.

 

Example 1:

Input: logs = [[0,5],[1,2],[0,2],[0,5],[1,3]], k = 5
Output: [0,2,0,0,0]
Explanation:
The user with ID=0 performed actions at minutes 5, 2, and 5 again. Hence, they have a UAM of 2 (minute 5 is only counted once).
The user with ID=1 performed actions at minutes 2 and 3. Hence, they have a UAM of 2.
Since both users have a UAM of 2, answer[2] is 2, and the remaining answer[j] values are 0.

Example 2:

Input: logs = [[1,1],[2,2],[2,3]], k = 4
Output: [1,1,0,0]
Explanation:
The user with ID=1 performed a single action at minute 1. Hence, they have a UAM of 1.
The user with ID=2 performed actions at minutes 2 and 3. Hence, they have a UAM of 2.
There is one user with a UAM of 1 and one with a UAM of 2.
Hence, answer[1] = 1, answer[2] = 1, and the remaining values are 0.

 

Constraints:

  • 1 <= logs.length <= 104
  • 0 <= IDi <= 109
  • 1 <= timei <= 105
  • k is in the range [The maximum UAM for a user, 105].

Solutions

Python3

class Solution:
    def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
        d = defaultdict(set)
        for u, t in logs:
            d[u].add(t)
        ans = [0] * k
        for ts in d.values():
            ans[len(ts) - 1] += 1
        return ans

Java

class Solution {
    public int[] findingUsersActiveMinutes(int[][] logs, int k) {
        Map<Integer, Set<Integer>> d = new HashMap<>();
        for (int[] log : logs) {
            int u = log[0], t = log[1];
            d.computeIfAbsent(u, key -> new HashSet<>()).add(t);
        }
        int[] ans = new int[k];
        for (Set<Integer> ts : d.values()) {
            ++ans[ts.size() - 1];
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> findingUsersActiveMinutes(vector<vector<int>>& logs, int k) {
        unordered_map<int, unordered_set<int>> d;
        for (auto& e : logs) d[e[0]].insert(e[1]);
        vector<int> ans(k);
        for (auto& e : d) ++ans[e.second.size() - 1];
        return ans;
    }
};

Go

func findingUsersActiveMinutes(logs [][]int, k int) []int {
	d := map[int]map[int]bool{}
	for _, e := range logs {
		u, t := e[0], e[1]
		if _, ok := d[u]; !ok {
			d[u] = make(map[int]bool)
		}
		d[u][t] = true
	}
	ans := make([]int, k)
	for _, ts := range d {
		ans[len(ts)-1]++
	}
	return ans
}

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