Sort the dishes by difficulty and compute a prefix maximum of profits to quickly find the best profit for any skill level. Then for each chef, binary search the sorted difficulties to find the highest difficulty they can handle, and add the corresponding maximum profit. This greedy approach works because each chef's assignment is independent and we always pick the best available dish for their skill.
Pro tip: Clarify that multiple chefs can be assigned to the same dish, so there's no need to track dish usage. Also, mention that if a chef cannot cook any dish, they contribute 0 profit, and the algorithm should handle that gracefully.
Restate the problem: each chef must be assigned to exactly one dish they can cook (skill >= difficulty), multiple chefs can share a dish, and we want to maximize total profit. Confirm edge cases like no feasible dish for a chef.
Sort dishes by difficulty. Compute a prefix maximum array where for each dish, we store the maximum profit among all dishes with difficulty <= current difficulty. This allows O(1) retrieval of the best profit for any skill threshold.
For each chef, binary search the sorted difficulties to find the largest difficulty <= chef's skill. If found, add the corresponding prefix maximum profit to the total; otherwise, add 0.
Time complexity: O((n + m) log n) where n is number of dishes and m is number of chefs, due to sorting and binary searches. Space complexity: O(n) for the prefix array. Mention that this is optimal for comparison-based sorting.
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