LeetCode Problem Workspace
Minimum Number of Work Sessions to Finish the Tasks
Find the minimum number of work sessions needed to finish a set of tasks, considering task durations and session time.
5
Topics
5
Code langs
3
Related
Practice Focus
Medium · State transition dynamic programming
Answer-first summary
Find the minimum number of work sessions needed to finish a set of tasks, considering task durations and session time.
Ace coding interviews with Interview AiBoxInterview AiBox guidance for State transition dynamic programming
To solve the "Minimum Number of Work Sessions to Finish the Tasks" problem, use state transition dynamic programming, a backtracking approach, and bit manipulation. The goal is to minimize work sessions by trying all possible task assignments within the given session time limit. Dynamic programming efficiently handles task combinations and ensures an optimal solution for the minimum number of sessions.
Problem Statement
You are given an array of tasks, where each element represents the time in hours required to complete a specific task. You are also given a session time, which dictates the maximum number of consecutive hours you can work in a single work session. Your goal is to finish all the tasks with the minimum number of work sessions, ensuring each session doesn’t exceed the session time limit.
Return the minimum number of work sessions required to finish all tasks while respecting the session time constraint. This problem can be solved using state transition dynamic programming, where we explore all combinations of task assignments and check if they fit within the session time limit.
Examples
Example 1
Input: tasks = [1,2,3], sessionTime = 3
Output: 2
You can finish the tasks in two work sessions.
- First work session: finish the first and the second tasks in 1 + 2 = 3 hours.
- Second work session: finish the third task in 3 hours.
Example 2
Input: tasks = [3,1,3,1,1], sessionTime = 8
Output: 2
You can finish the tasks in two work sessions.
- First work session: finish all the tasks except the last one in 3 + 1 + 3 + 1 = 8 hours.
- Second work session: finish the last task in 1 hour.
Example 3
Input: tasks = [1,2,3,4,5], sessionTime = 15
Output: 1
You can finish all the tasks in one work session.
Constraints
- n == tasks.length
- 1 <= n <= 14
- 1 <= tasks[i] <= 10
- max(tasks[i]) <= sessionTime <= 15
Solution Approach
State Transition Dynamic Programming
Utilize dynamic programming to represent the state of tasks that have been assigned to a session. By using bitmasking to track the assigned tasks, you can recursively calculate the minimum work sessions required by trying all combinations of task assignments and checking if they fit within the session time limit.
Backtracking with Bitmasking
Leverage backtracking with bitmasking to efficiently explore all task combinations and assign them to work sessions. This method helps prune invalid configurations and avoid redundant calculations while ensuring that the task combinations respect the session time constraint.
Memoization to Optimize Recursion
Apply memoization to store previously computed results for specific task combinations to avoid recalculating the same configuration multiple times. This significantly reduces computation time and makes the solution more efficient.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time complexity depends on the number of subsets of tasks, which is 2^n, where n is the number of tasks. The space complexity depends on the memoization table, which stores results for each subset of tasks, making it O(2^n).
What Interviewers Usually Probe
- Look for understanding of bitmasking and dynamic programming principles.
- Assess the candidate’s ability to optimize recursive solutions using memoization.
- Evaluate how well the candidate can balance brute force and optimization techniques.
Common Pitfalls or Variants
Common pitfalls
- Failing to properly handle all combinations of task assignments within the session time.
- Not using bitmasking effectively to track which tasks have been assigned.
- Overlooking the importance of memoization, leading to redundant calculations and slower performance.
Follow-up variants
- Increase the maximum number of tasks or change the session time limit to test scalability.
- Use this problem as a foundation to solve similar problems involving scheduling and task assignments.
- Modify the problem to include different task durations and session time constraints.
FAQ
How does state transition dynamic programming apply to this problem?
State transition dynamic programming uses a bitmask to represent the set of tasks completed so far, allowing you to efficiently calculate the minimum work sessions needed.
What is bitmasking, and why is it important for this problem?
Bitmasking is a technique used to represent subsets of tasks in a compact form. It helps track which tasks have been assigned and ensures efficient calculation of task combinations.
Can I solve the "Minimum Number of Work Sessions to Finish the Tasks" problem using a greedy approach?
A greedy approach would not guarantee the optimal solution because it may not always make the best choices at each step. Dynamic programming is better suited for this problem.
How do I avoid redundant calculations in this problem?
Use memoization to store the results of previously computed task assignments, preventing redundant recalculations and improving performance.
What is the time complexity of the solution?
The time complexity is O(2^n), where n is the number of tasks, as you need to explore all subsets of tasks to find the optimal solution.
Solution
Solution 1: State Compression Dynamic Programming + Subset Enumeration
We note that $n$ does not exceed $14$, so we can consider using state compression dynamic programming to solve this problem.
class Solution:
def minSessions(self, tasks: List[int], sessionTime: int) -> int:
n = len(tasks)
ok = [False] * (1 << n)
for i in range(1, 1 << n):
t = sum(tasks[j] for j in range(n) if i >> j & 1)
ok[i] = t <= sessionTime
f = [inf] * (1 << n)
f[0] = 0
for i in range(1, 1 << n):
j = i
while j:
if ok[j]:
f[i] = min(f[i], f[i ^ j] + 1)
j = (j - 1) & i
return f[-1]Continue Topic
array
Practice more edge cases under the same topic.
arrow_forwardauto_awesomeContinue Pattern
State transition dynamic programming
Expand the same solving frame across more problems.
arrow_forwardsignal_cellular_altSame Difficulty Track
Medium
Stay on this level to stabilize interview delivery.
arrow_forward