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I went with backtracking first and got it mostly right, but the pruning is where things get tricky.
This is a classic load balancing problem that can be solved using binary search on the answer combined with a greedy feasibility check. The key is to recognize that the minimum possible maximum load lies between the maximum job duration and the sum of all job durations, and then binary search for the smallest value for which a valid assignment exists.
Pro tip: Always clarify that jobs are indivisible and each worker's load is the sum of assigned job durations. Mention that while the problem is NP-hard in general, the binary search + greedy approach works because we only need to minimize the maximum, not find an exact partition.
Restate the problem: assign each job to exactly one worker, minimize the maximum total working time. Identify that jobs are indivisible and workers can take any number of jobs.
The answer lies between max(job_durations) and sum(job_durations). For a given candidate maximum load, check if it's possible to assign jobs to at most k workers without exceeding that load using a greedy algorithm.
Binary search over the range [max, sum] to find the smallest feasible maximum load. At each step, run the greedy feasibility check and adjust the bounds accordingly.
Time complexity is O(n log(sum - max)) where n is number of jobs. Space complexity is O(1). Discuss that this is optimal for this problem and mention alternative approaches like DP or heuristics if k is small.
Consider cases like k >= n (each job to a worker), k = 1 (all jobs to one worker), and jobs with large durations. Verify the solution handles these correctly.
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