Leetcode #2590: Design a Todo List
In this guide, we solve Leetcode #2590 Design a Todo List in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Design a Todo List Where users can add tasks, mark them as complete, or get a list of pending tasks. Users can also add tags to tasks and can filter the tasks by certain tags.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Design, Array, Hash Table, String, Sorting
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input
["TodoList", "addTask", "addTask", "getAllTasks", "getAllTasks", "addTask", "getTasksForTag", "completeTask", "completeTask", "getTasksForTag", "getAllTasks"]
[[], [1, "Task1", 50, []], [1, "Task2", 100, ["P1"]], [1], [5], [1, "Task3", 30, ["P1"]], [1, "P1"], [5, 1], [1, 2], [1, "P1"], [1]]
Output
[null, 1, 2, ["Task1", "Task2"], [], 3, ["Task3", "Task2"], null, null, ["Task3"], ["Task3", "Task1"]]
Explanation
TodoList todoList = new TodoList();
todoList.addTask(1, "Task1", 50, []); // return 1. This adds a new task for the user with id 1.
todoList.addTask(1, "Task2", 100, ["P1"]); // return 2. This adds another task for the user with id 1.
todoList.getAllTasks(1); // return ["Task1", "Task2"]. User 1 has two uncompleted tasks so far.
todoList.getAllTasks(5); // return []. User 5 does not have any tasks so far.
todoList.addTask(1, "Task3", 30, ["P1"]); // return 3. This adds another task for the user with id 1.
todoList.getTasksForTag(1, "P1"); // return ["Task3", "Task2"]. This returns the uncompleted tasks that have the tag "P1" for the user with id 1.
todoList.completeTask(5, 1); // This does nothing, since task 1 does not belong to user 5.
todoList.completeTask(1, 2); // This marks task 2 as completed.
todoList.getTasksForTag(1, "P1"); // return ["Task3"]. This returns the uncompleted tasks that have the tag "P1" for the user with id 1.
// Notice that we did not include "Task2" because it is completed now.
todoList.getAllTasks(1); // return ["Task3", "Task1"]. User 1 now has 2 uncompleted tasks.
Python Solution
class TodoList:
def __init__(self):
self.i = 1
self.tasks = defaultdict(SortedList)
def addTask(
self, userId: int, taskDescription: str, dueDate: int, tags: List[str]
) -> int:
taskId = self.i
self.i += 1
self.tasks[userId].add([dueDate, taskDescription, set(tags), taskId, False])
return taskId
def getAllTasks(self, userId: int) -> List[str]:
return [x[1] for x in self.tasks[userId] if not x[4]]
def getTasksForTag(self, userId: int, tag: str) -> List[str]:
return [x[1] for x in self.tasks[userId] if not x[4] and tag in x[2]]
def completeTask(self, userId: int, taskId: int) -> None:
for task in self.tasks[userId]:
if task[3] == taskId:
task[4] = True
break
# Your TodoList object will be instantiated and called as such:
# obj = TodoList()
# param_1 = obj.addTask(userId,taskDescription,dueDate,tags)
# param_2 = obj.getAllTasks(userId)
# param_3 = obj.getTasksForTag(userId,tag)
# obj.completeTask(userId,taskId)
Complexity
The time complexity is O(n). The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.