Three-Resource 0/1 Knapsack

Each item has a value and consumes three independent resources. Select each item at most once without exceeding any of the three capacities, and return the maximum total value.

Function
maximizeKnapsackValue(values: int[], firstCosts: int[], secondCosts: int[], thirdCosts: int[], firstCapacity: int, secondCapacity: int, thirdCapacity: int) → int

Examples
Example 1
values = [10,20]
firstCosts = [1,2]
secondCosts = [1,2]
thirdCosts = [1,2]
firstCapacity = 2
secondCapacity = 2
thirdCapacity = 2
return = 20
Case 1 exercises the documented deterministic contract.
Example 2
values = [6,10,12]
firstCosts = [1,2,3]
secondCosts = [2,1,2]
thirdCosts = [1,2,1]
firstCapacity = 4
secondCapacity = 4
thirdCapacity = 3
return = 18
Case 2 exercises the documented deterministic contract.
Example 3
values = [5]
firstCosts = [0]
secondCosts = [0]
thirdCosts = [0]
firstCapacity = 0
secondCapacity = 0
thirdCapacity = 0
return = 5
Case 3 exercises the documented deterministic contract.

Constraints
1 <= values.length <= 40.
All four item arrays have equal length.
0 <= costs and capacities <= 40.
Values are nonnegative and the answer fits a signed 32-bit integer.

The reduction: dp[a][b][c] is the max value using at most a of resource one, b of resource two, c of resource three. For each item, loop a, b, c downward from the capacities to the item's costs and set dp[a][b][c] = max(dp[a][b][c], dp[a-c1][b-c2][c-c3] + value). Downward iteration is the whole trick, since it stops you reusing an item. Forward loops turn this into unbounded knapsack and you'll get wrong answers. Watch the zero-cost case in Example 3: an item costing nothing must still be counted once, and downward loops handle that if your bounds start at the cost, which is 0. Complexity is 40 items times 41 cubed, about 2.7 million operations, trivial. Initialize everything to 0, not negative infinity, because capacities mean at most. If you freeze on the live OA, StealthCoder is the hedge that reads the problem and hands you this loop structure.

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