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Leetcode #1986: Minimum Number of Work Sessions to Finish the Tasks

In this guide, we solve Leetcode #1986 Minimum Number of Work Sessions to Finish the Tasks 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.

Leetcode

Problem Statement

There are n tasks assigned to you. The task times are represented as an integer array tasks of length n, where the ith task takes tasks[i] hours to finish.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Bit Manipulation, Array, Dynamic Programming, Backtracking, Bitmask

Intuition

The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.

A carefully chosen DP state captures exactly what we need to build the final answer.

Approach

Define the DP state and recurrence, then compute states in the correct order.

Optionally compress space once the recurrence is clear.

Steps:

  • Choose a DP state definition.
  • Write the recurrence and base cases.
  • Compute states in the correct order.

Example

Input: tasks = [1,2,3], sessionTime = 3 Output: 2 Explanation: 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.

Python Solution

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]

Complexity

The time complexity is O(n×3n)O(n \times 3^n)O(n×3n), and the space complexity is O(2n)O(2^n)O(2n). The space complexity is O(2n)O(2^n)O(2n).

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.


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