Minimum Difficulty of a Job Schedule
A hard-tier problem at 60% community acceptance, tagged with Array, Dynamic Programming. Reported in interviews at Turvo and 0 others.
Minimum Difficulty of a Job Schedule is a hard dynamic programming problem that hits your screen when you're already stressed. You've got an array of job difficulties, a deadline in days, and you need to schedule them so the hardest job on each day is minimized across all days. It sounds like a greedy problem until you realize the order matters and you're making irreversible decisions. Turvo has asked it. Roughly 60% of attempts pass, which means most people either nail the DP pattern or tank it entirely. If this problem shows up during your live assessment and you blank on the state transitions, StealthCoder solves it invisibly while you stay calm.
Companies that ask "Minimum Difficulty of a Job Schedule"
Minimum Difficulty of a Job Schedule is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Built because the OA filter rejects engineers who'd pass the on-site. That's a broken filter. This is the workaround.
Get StealthCoderThe trick is recognizing this as a partitioning problem, not a sorting problem. You can't rearrange jobs, and the penalty for a day is its maximum difficulty. A greedy approach fails because choosing the smallest daily max today might force a brutal day tomorrow. The solution uses DP where the state is the day and job index: dp[day][job] = minimum total difficulty to schedule jobs 0 to job using day 1 to day. You iterate through valid partitions and track the maximum difficulty in each partition. The pitfall is trying to optimize locally instead of computing all partition combinations. Most candidates get lost in the state definition or the recurrence relation. When you hit this live and your first instinct doesn't work, StealthCoder provides the working DP structure in seconds, invisible to the proctor.
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Minimum Difficulty of a Job Schedule recycles across companies for a reason. It's hard-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Built because the OA filter rejects engineers who'd pass the on-site. That's a broken filter. This is the workaround. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Minimum Difficulty of a Job Schedule interview FAQ
Is this problem really asked at FAANG or is it just Turvo?+
Turvo is the only confirmed company in reports, but the pattern it tests (DP partitioning) is core to company interviews. Your preparation for this problem transfers directly to similar partition-optimization problems you'll see elsewhere.
What's the difference between this and unbounded knapsack or coin change?+
Those problems let you reuse items or change order. Job Schedule forces a fixed order and splits the array into contiguous segments. The state space and recurrence are different because you're partitioning, not filling a container.
How do I know when to use DP instead of greedy here?+
Greedy fails because the choice that looks best today can poison tomorrow. DP is mandatory when future decisions depend on past partition choices. The acceptance rate of 60% suggests most people try greedy first and get it wrong.
What's the hardest part of coding this under interview pressure?+
Getting the loop bounds right. You're iterating over (current day, job indices) and deciding where to cut. Off-by-one errors sink most attempts. The second hard part is computing the max difficulty correctly within each partition.
Can I solve this without DP?+
Not at acceptable complexity. Brute force checking all partitions is exponential. The DP solution is O(n squared * d) where n is jobs and d is days. That's the intended path.
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