Design Task Manager - Practice Coding Problems

Design Task Manager - Problem

Design and implement a Task Manager system that efficiently handles task operations across multiple users.

Each task has a taskId, belongs to a userId, and has a priority. The system must support:

Adding new tasks
Editing task priorities
Removing tasks
Executing the highest priority task (highest taskId for ties)

Implement the TaskManager class with the following methods:

TaskManager(tasks) - Initialize with list of [userId, taskId, priority] triples
add(userId, taskId, priority) - Add new task (guaranteed unique taskId)
edit(taskId, newPriority) - Update existing task priority
rmv(taskId) - Remove existing task
execTop() - Execute and remove highest priority task, return its userId (return -1 if no tasks)

Input & Output

Example 1 — Basic Operations

$ Input: [["TaskManager", [[1,101,2],[1,102,5]]], ["add", 2,103,1], ["execTop"], ["execTop"]]

› Output: [null, null, 1, 1]

💡 Note: Initialize with 2 tasks. Add task 103. execTop() returns user 1 (task 102, priority 5). execTop() returns user 1 (task 101, priority 2).

Example 2 — Edit and Remove

$ Input: [["TaskManager", [[2,201,3],[2,202,1]]], ["edit", 202,5], ["execTop"], ["rmv", 201], ["execTop"]]

› Output: [null, null, 2, null, -1]

💡 Note: Edit task 202 priority to 5. execTop() returns user 2 (task 202). Remove task 201. execTop() returns -1 (no tasks).

Example 3 — Priority Ties

$ Input: [["TaskManager", [[1,301,2],[2,302,2]]], ["execTop"]]

› Output: [null, 2]

💡 Note: Both tasks have priority 2, but task 302 has higher taskId, so execTop() returns user 2.

Constraints

1 ≤ tasks.length ≤ 10⁵
1 ≤ userId, taskId ≤ 10⁵
1 ≤ priority ≤ 10⁶
At most 5 × 10⁴ calls to add, edit, rmv, and execTop

Visualization

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Understanding the Visualization

Input

Operations list with TaskManager initialization and method calls

Process

Execute operations using hash map + priority queue structure

Output

Return results array with null for void methods and userIds for execTop

Key Takeaway

🎯 Key Insight: Combine hash map's O(1) lookups with priority queue's O(log n) ordering for optimal performance

Asked in

G Google 45 a Amazon 38 f Meta 32 m Microsoft 28

The key insight is combining a hash map for O(1) task lookups with a priority queue for O(log n) priority operations. The optimal approach uses lazy deletion to handle edit/remove operations efficiently. Time: O(log n) per operation, Space: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Simple List Storage	O(n)	O(n)	Store tasks in a simple list and scan through all tasks for each operation
Hash Map + Priority Queue	O(log n)	O(n)	Use hash map for O(1) lookups and priority queue for O(log n) priority operations

Simple List Storage — Algorithm Steps

Step 1: Store tasks in a simple array/list
Step 2: For execTop(), iterate through all tasks to find max priority
Step 3: For edit/remove, linear search to find target task

Visualization

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Step-by-Step Walkthrough

Store Tasks

Keep all tasks in a simple list

Linear Search

Scan every task to find maximum priority

Remove Found

Remove the selected task from list

Code -

solution.c — C

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

#define MAX_TASKS 1000

typedef struct {
    int tasks[MAX_TASKS][3]; // [userId, taskId, priority]
    int size;
} TaskManager;

TaskManager* taskManagerCreate(int** tasks, int tasksSize) {
    TaskManager* tm = (TaskManager*)malloc(sizeof(TaskManager));
    tm->size = tasksSize;
    for (int i = 0; i < tasksSize; i++) {
        tm->tasks[i][0] = tasks[i][0];
        tm->tasks[i][1] = tasks[i][1];
        tm->tasks[i][2] = tasks[i][2];
    }
    return tm;
}

void taskManagerAdd(TaskManager* tm, int userId, int taskId, int priority) {
    tm->tasks[tm->size][0] = userId;
    tm->tasks[tm->size][1] = taskId;
    tm->tasks[tm->size][2] = priority;
    tm->size++;
}

void taskManagerEdit(TaskManager* tm, int taskId, int newPriority) {
    for (int i = 0; i < tm->size; i++) {
        if (tm->tasks[i][1] == taskId) {
            tm->tasks[i][2] = newPriority;
            break;
        }
    }
}

void taskManagerRmv(TaskManager* tm, int taskId) {
    for (int i = 0; i < tm->size; i++) {
        if (tm->tasks[i][1] == taskId) {
            for (int j = i; j < tm->size - 1; j++) {
                tm->tasks[j][0] = tm->tasks[j + 1][0];
                tm->tasks[j][1] = tm->tasks[j + 1][1];
                tm->tasks[j][2] = tm->tasks[j + 1][2];
            }
            tm->size--;
            break;
        }
    }
}

int taskManagerExecTop(TaskManager* tm) {
    if (tm->size == 0) return -1;
    
    int maxPriority = -1;
    int maxTaskId = -1;
    int maxIndex = -1;
    
    for (int i = 0; i < tm->size; i++) {
        int userId = tm->tasks[i][0], taskId = tm->tasks[i][1], priority = tm->tasks[i][2];
        if (priority > maxPriority || (priority == maxPriority && taskId > maxTaskId)) {
            maxPriority = priority;
            maxTaskId = taskId;
            maxIndex = i;
        }
    }
    
    int resultUser = tm->tasks[maxIndex][0];
    taskManagerRmv(tm, tm->tasks[maxIndex][1]);
    return resultUser;
}

int main() {
    printf("[null,null,null,1,2]\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Each execTop() operation requires scanning all n tasks to find maximum

✓ Linear Growth

Space Complexity

O(n)

Storage for n tasks in the list

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Simple List Storage — Algorithm Steps

Visualization

Code -

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