LeetCode #1817 — MEDIUM

Finding the Users Active Minutes

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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 <= 10⁴
0 <= ID_i <= 10⁹
1 <= time_i <= 10⁵
k is in the range [The maximum UAM for a user, 10⁵].

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[[0,5],[1,2],[0,2],[0,5],[1,3]]
5

Example 2

[[1,1],[2,2],[2,3]]
4

Step 02

Core Insight

What unlocks the optimal approach

Try to find the number of different minutes when action happened for each user.
For each user increase the value of the answer array index which matches the UAM for this user.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1817: Finding the Users Active Minutes
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;
    }
}

// Accepted solution for LeetCode #1817: Finding the Users Active Minutes
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
}

# Accepted solution for LeetCode #1817: Finding the Users Active Minutes
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

// Accepted solution for LeetCode #1817: Finding the Users Active Minutes
struct Solution;

use std::collections::HashMap;
use std::collections::HashSet;

impl Solution {
    fn finding_users_active_minutes(logs: Vec<Vec<i32>>, k: i32) -> Vec<i32> {
        let mut users: HashMap<i32, HashSet<i32>> = HashMap::new();
        let k = k as usize;
        let mut res = vec![0; k];
        for log in logs {
            let id = log[0];
            let time = log[1];
            users.entry(id).or_default().insert(time);
        }
        for times in users.values() {
            let j = times.len() - 1;
            if j < k {
                res[j] += 1;
            }
        }
        res
    }
}

#[test]
fn test() {
    let logs = vec_vec_i32![[0, 5], [1, 2], [0, 2], [0, 5], [1, 3]];
    let k = 5;
    let res = vec![0, 2, 0, 0, 0];
    assert_eq!(Solution::finding_users_active_minutes(logs, k), res);
    let logs = vec_vec_i32![[1, 1], [2, 2], [2, 3]];
    let k = 4;
    let res = vec![1, 1, 0, 0];
    assert_eq!(Solution::finding_users_active_minutes(logs, k), res);
}

// Accepted solution for LeetCode #1817: Finding the Users Active Minutes
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.