Leetcode #2071: Maximum Number of Tasks You Can Assign
In this guide, we solve Leetcode #2071 Maximum Number of Tasks You Can Assign 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
You have n tasks and m workers. Each task has a strength requirement stored in a 0-indexed integer array tasks, with the ith task requiring tasks[i] strength to complete.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Greedy, Queue, Array, Two Pointers, Binary Search, Sorting, Monotonic Queue
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
Example
Input: tasks = [3,2,1], workers = [0,3,3], pills = 1, strength = 1
Output: 3
Explanation:
We can assign the magical pill and tasks as follows:
- Give the magical pill to worker 0.
- Assign worker 0 to task 2 (0 + 1 >= 1)
- Assign worker 1 to task 1 (3 >= 2)
- Assign worker 2 to task 0 (3 >= 3)
Python Solution
class Solution:
def maxTaskAssign(
self, tasks: List[int], workers: List[int], pills: int, strength: int
) -> int:
def check(x):
i = 0
q = deque()
p = pills
for j in range(m - x, m):
while i < x and tasks[i] <= workers[j] + strength:
q.append(tasks[i])
i += 1
if not q:
return False
if q[0] <= workers[j]:
q.popleft()
elif p == 0:
return False
else:
p -= 1
q.pop()
return True
n, m = len(tasks), len(workers)
tasks.sort()
workers.sort()
left, right = 0, min(n, m)
while left < right:
mid = (left + right + 1) >> 1
if check(mid):
left = mid
else:
right = mid - 1
return left
Complexity
The time complexity is , and the space complexity is , where is the number of tasks. The space complexity is , where is the number of tasks.
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.