Finding the Users Active Minutes

Finding the Users Active Minutes - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: logs = [[0,5],[1,2],[0,2],[0,5],[1,3]], k = 5

› Output: [0,2,0,0,0]

💡 Note: User 0 has UAM=2 (minutes 2,5), User 1 has UAM=2 (minutes 2,3). So 0 users have UAM=1, 2 users have UAM=2, and 0 users have UAM=3,4,5.

Example 2 — Single User

$ Input: logs = [[1,1],[1,2],[1,3]], k = 4

› Output: [0,0,1,0]

💡 Note: User 1 has UAM=3 (minutes 1,2,3). So 1 user has UAM=3, others are 0.

Example 3 — Duplicate Minutes

$ Input: logs = [[0,1],[0,1],[1,1]], k = 3

› Output: [2,0,0]

💡 Note: User 0 has UAM=1 (minute 1 counted once), User 1 has UAM=1 (minute 1). So 2 users have UAM=1.

Constraints

1 ≤ logs.length ≤ 10⁴
0 ≤ ID_i ≤ 10⁹
1 ≤ time_i ≤ 10⁵
k is in the range [1, logs.length]

Visualization

Tap to expand

Asked in

LC LeetCode 15 A Amazon 8

The key insight is to group logs by user ID first, then count unique minutes per user, and finally count users by UAM values. Best approach is Two-Pass Hash Map with Time: O(n), Space: O(n).

Common Approaches

✓ Union Find

⏱️ Time: N/A Space: N/A

Brute Force with Nested Loops

⏱️ Time: O(n × u) Space: O(u + k)

Find all unique users first. For each user, scan through all logs to collect their unique active minutes. Count how many users have each UAM value.

Two-Pass Hash Map

⏱️ Time: O(n + u) Space: O(n + k)

Use a hash map to collect unique minutes for each user in one pass. Then iterate through the map to count how many users have each UAM value.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

typedef struct {
    int* parent;
    int* rank;
    int size;
} UnionFind;

UnionFind* createUF(int size) {
    UnionFind* uf = (UnionFind*)malloc(sizeof(UnionFind));
    uf->parent = (int*)malloc(size * sizeof(int));
    uf->rank = (int*)calloc(size, sizeof(int));
    uf->size = size;
    for (int i = 0; i < size; i++) {
        uf->parent[i] = i;
    }
    return uf;
}

int find(UnionFind* uf, int x) {
    if (uf->parent[x] != x) {
        uf->parent[x] = find(uf, uf->parent[x]);
    }
    return uf->parent[x];
}

void unite(UnionFind* uf, int x, int y) {
    int px = find(uf, x);
    int py = find(uf, y);
    if (px == py) return;
    
    if (uf->rank[px] < uf->rank[py]) {
        uf->parent[px] = py;
    } else if (uf->rank[px] > uf->rank[py]) {
        uf->parent[py] = px;
    } else {
        uf->parent[py] = px;
        uf->rank[px]++;
    }
}

int gcd(int a, int b) {
    while (b != 0) {
        int temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

int compare(const void* a, const void* b) {
    return (*(int*)a - *(int*)b);
}

bool solution(int* nums, int n) {
    if (n <= 1) return true;
    
    int* sortedNums = (int*)malloc(n * sizeof(int));
    for (int i = 0; i < n; i++) {
        sortedNums[i] = nums[i];
    }
    qsort(sortedNums, n, sizeof(int), compare);
    
    UnionFind* uf = createUF(n);
    
    // Union positions that can be swapped (gcd > 1)
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (gcd(nums[i], nums[j]) > 1) {
                unite(uf, i, j);
            }
        }
    }
    
    // For each position i, check if the value at position i can reach 
    // the position where it should be in the sorted array
    for (int i = 0; i < n; i++) {
        int currentVal = nums[i];
        bool found = false;
        
        // Find a position in sorted array where currentVal should go
        // and check if it's reachable from position i
        for (int j = 0; j < n; j++) {
            if (sortedNums[j] == currentVal) {
                if (find(uf, i) == find(uf, j)) {
                    // Mark this target position as used
                    sortedNums[j] = -1; // Mark as used
                    found = true;
                    break;
                }
            }
        }
        
        if (!found) {
            free(sortedNums);
            free(uf->parent);
            free(uf->rank);
            free(uf);
            return false;
        }
    }
    
    free(sortedNums);
    free(uf->parent);
    free(uf->rank);
    free(uf);
    return true;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    if (strstr(line, "[]") != NULL) {
        printf("true\n");
        return 0;
    }
    
    int nums[1000];
    int n = 0;
    char* token = strtok(line, "[],\n");
    
    while (token != NULL) {
        if (strlen(token) > 0) {
            nums[n++] = atoi(token);
        }
        token = strtok(NULL, "[],\n");
    }
    
    bool result = solution(nums, n);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

⚡ Linearithmic Space

12.0K Views

Medium Frequency

~15 min Avg. Time

450 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen