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The solution itself isn't hard to land on.
Start by framing the problem as a cost minimization with a cardinality constraint, then propose a greedy strategy based on the cost difference between assigning each candidate to NY vs SF. Sort candidates by the absolute difference and assign the N with the largest differences to their cheaper city, filling the remaining slots with the others.
Pro tip: Mention that this greedy approach is optimal and can be proven via an exchange argument, and note that the algorithm runs in O(N log N) time due to sorting, which is efficient for large inputs.
Restate the problem: 2N candidates, each with a cost to fly to NY and a cost to fly to SF. Assign exactly N to each city to minimize total cost.
For each candidate, calculate the difference between their NY cost and SF cost (or vice versa). This represents the penalty for assigning them to the more expensive city.
Sort the candidates in descending order of the absolute cost difference. This prioritizes those with the strongest preference for one city.
Iterate through the sorted list: if a candidate prefers NY (NY cost < SF cost) and NY still has slots, assign them to NY; otherwise assign to SF. If they prefer SF, assign to SF if slots available, else NY. Continue until all are assigned.
Explain why this greedy choice is optimal using an exchange argument: any optimal solution can be transformed to match the greedy assignment without increasing cost. Mention time complexity O(N log N) due to sorting.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.