Minimize Maximum Pilot Workload
Reported by candidates from Spotnana's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
Spotnana reported this one in July 2026, and the title hides a classic. Split an array into k contiguous groups and minimize the largest group sum. It's binary search on the answer with a greedy check. With up to 100000 flights and durations up to a billion, trying every partition is dead on arrival, so the input size tells you exactly what the OA wants. If you've seen Split Array Largest Sum, you've seen this. If you blank on the night, StealthCoder sits invisibly on your screen as a safety net and hands you the structure.
The problem
You are given an integer array flights, where flights[i] is the duration of the i-th flight, and an integer k representing the number of available pilots. Assign every flight to exactly one pilot while preserving the original flight order. Each pilot must receive one non-empty contiguous sequence of flights. A pilot's workload is the sum of the durations assigned to that pilot. Return the minimum possible value of the maximum workload among all k pilots. Function minimizeMaximumPilotTime(flights: int[], k: int) → long Examples Example 1 flights = [10,20,30,40] k = 2 return = 60 Assign [10,20,30] to one pilot and [40] to the other. Their workloads are 60 and 40. No partition can make the larger workload less than 60. Example 2 flights = [5,1,2,7,3,4] k = 3 return = 8 The groups [5,1,2], [7], and [3,4] have workloads 8, 7, and 7. Constraints 1 <= k <= flights.length <= 100000 1 <= flights[i] <= 1000000000 The answer fits in a signed 64-bit integer.
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick: the answer lives between max(flights) and sum(flights). Binary search that range. For a candidate cap, walk the array greedily, adding flights to the current pilot until the next one would exceed the cap, then start a new pilot. Count pilots. If the count is at most k, the cap works, so search lower. Otherwise search higher. Each check is O(n), total O(n log(sum)). Pitfalls: use 64-bit for sums since 100000 times 1e9 overflows int. Set the lower bound to the max element, not 1, or a single flight won't fit. Also remember each pilot needs a non-empty group, but since k is at most the length, fewer groups than k can always be split further without raising the max. If you freeze live, StealthCoder is the hedge that gives you the binary search skeleton fast.
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Minimize Maximum Pilot Workload FAQ
What's the trick in Minimize Maximum Pilot Workload?+
Binary search on the answer, not on the array. Pick a maximum workload, greedily pack flights into pilots without exceeding it, and count pilots. If you need k or fewer, that cap is feasible. Shrink the range until low meets high. The feasibility check is monotonic, which is why it works.
Why can't I use brute force or plain DP here?+
With n up to 100000, trying all partitions is exponential, and the standard O(n^2 * k) DP is far too slow. Binary search over the sum range takes about 50 iterations of an O(n) scan, which fits comfortably.
What are the binary search bounds?+
Low is the largest single flight, because some pilot must take it. High is the total sum, which is the case of one pilot doing everything. Use 64-bit integers, since the total can reach 100000 times 1e9.
Is this a known LeetCode problem?+
Yes. It's Split Array Largest Sum with flights instead of numbers and pilots instead of subarrays. If you've solved that one, the code is nearly identical. Just rename variables and double check the long return type.
How do I prepare for this in 48 hours?+
Write the greedy feasibility function from memory, then wrap it in a binary search twice. Test on both examples: 60 for the first and 8 for the second. Also learn the sibling problems like capacity to ship packages, since they use the same skeleton and show up often.