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Problem solution · Python

Find Minimum Time to Finish All Jobs

Find Minimum Time to Finish All Jobs: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
28 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Find Minimum Time to Finish All Jobs, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 28 lines of Python from the credited upstream file 1723.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeFind Minimum Time to Finish All Jobs · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def minimumTimeRequired(self, jobs: list[int], k: int) -> int:    ans = sum(jobs)    times = [0] * k  # times[i] := accumulate time of workers[i]     # Assign the most time-consuming job first.    jobs.sort(reverse=True)     def dfs(s: int) -> None:      nonlocal ans      if s == len(jobs):        ans = min(ans, max(times))        return      for i in range(k):        # There is no need to explore assigning jobs[s] to workers[i] further as        # it would not yield better results.        if times[i] + jobs[s] >= ans:          continue        times[i] += jobs[s]        dfs(s + 1)        times[i] -= jobs[s]        # It's always non-optimal to have a worker with no jobs.        if times[i] == 0:          return     dfs(0)    return ans

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