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Formatted question description: https://leetcode.ca/all/1817.html

1817. Finding the Users Active Minutes

Level

Medium

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] = [ID_i, time_i] indicates that the user with ID_i performed an action at the minute time_i.

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 <= 10^4
  • 0 <= ID_i <= 10^9
  • 1 <= time_i <= 10^5
  • k is in the range [The maximum UAM for a user, 10^5].

Solution

Use a map to store each user’s active minutes, where the active minutes are stored in a set. Then for each user, obtain the number of active minutes, and update the array answer accordingly. Finally, return answer.

  • class Solution {
    public int[] findingUsersActiveMinutes(int[][] logs, int k) {
        int[] answer = new int[k];
        Map<Integer, Set<Integer>> map = new HashMap<Integer, Set<Integer>>();
        for (int[] log : logs) {
            int id = log[0], time = log[1];
            Set<Integer> set = map.getOrDefault(id, new HashSet<Integer>());
            set.add(time);
            map.put(id, set);
        }
        for (Map.Entry<Integer, Set<Integer>> entry : map.entrySet()) {
            int size = entry.getValue().size();
            answer[size - 1]++;
        }
        return answer;
    }
}

############

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

  • // OJ: https://leetcode.com/problems/finding-the-users-active-minutes/
// Time: O()
// Space: O()
class Solution {
public:
    vector<int> findingUsersActiveMinutes(vector<vector<int>>& A, int k) {
        unordered_map<int, unordered_set<int>> m;
        for (auto &v : A) {
            int id = v[0], time = v[1];
            m[id].insert(time);
        }
        vector<int> ans(k);
        for (auto &[id, s] : m) {
            int cnt = s.size();
            if (cnt <= k) ans[cnt - 1]++;
        }
        return ans;
    }
};

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

############

# 1817. Finding the Users Active Minutes
# https://leetcode.com/problems/finding-the-users-active-minutes/

class Solution:
    def findingUsersActiveMinutes(self, logs: List[List[int]], k: int) -> List[int]:
        res = [0] * k
        mp = collections.defaultdict(set)
        
        for i, time in logs:
            mp[i].add(time)
        
        for v in mp.values():
            res[len(v) - 1] += 1
        
        return res


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

  • function findingUsersActiveMinutes(logs: number[][], k: number): number[] {
    const d: Map<number, Set<number>> = new Map();
    for (const [i, t] of logs) {
        if (!d.has(i)) {
            d.set(i, new Set<number>());
        }
        d.get(i)!.add(t);
    }
    const ans: number[] = Array(k).fill(0);
    for (const [_, ts] of d) {
        ++ans[ts.size - 1];
    }
    return ans;
}

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