Given $n$ items, the weight of the $i$-th item is $wgt[i-1]$ and its value is $val[i-1]$, and a knapsack with a capacity of $cap$. Each item can be chosen only once, **but a part of the item can be selected, with its value calculated based on the proportion of the weight chosen**, what is the maximum value of the items in the knapsack under the limited capacity? An example is shown in Figure 15-3.
![Example data of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_example.png){ class="animation-figure" }
<palign="center"> Figure 15-3 Example data of the fractional knapsack problem </p>
The fractional knapsack problem is very similar overall to the 0-1 knapsack problem, involving the current item $i$ and capacity $c$, aiming to maximize the value within the limited capacity of the knapsack.
The difference is that, in this problem, only a part of an item can be chosen. As shown in Figure 15-4, **we can arbitrarily split the items and calculate the corresponding value based on the weight proportion**.
Maximizing the total value of the items in the knapsack **essentially means maximizing the value per unit weight**. From this, the greedy strategy shown in Figure 15-5 can be deduced.
1. Sort the items by their unit value from high to low.
2. Iterate over all items, **greedily choosing the item with the highest unit value in each round**.
3. If the remaining capacity of the knapsack is insufficient, use part of the current item to fill the knapsack.
![Greedy strategy of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_greedy_strategy.png){ class="animation-figure" }
<palign="center"> Figure 15-5 Greedy strategy of the fractional knapsack problem </p>
### 2. Code implementation
We have created an `Item` class in order to sort the items by their unit value. We loop and make greedy choices until the knapsack is full, then exit and return the solution:
Apart from sorting, in the worst case, the entire list of items needs to be traversed, **hence the time complexity is $O(n)$**, where $n$ is the number of items.
Since an `Item` object list is initialized, **the space complexity is $O(n)$**.
### 3. Correctness proof
Using proof by contradiction. Suppose item $x$ has the highest unit value, and some algorithm yields a maximum value `res`, but the solution does not include item $x$.
Now remove a unit weight of any item from the knapsack and replace it with a unit weight of item $x$. Since the unit value of item $x$ is the highest, the total value after replacement will definitely be greater than `res`. **This contradicts the assumption that `res` is the optimal solution, proving that the optimal solution must include item $x$**.
For other items in this solution, we can also construct the above contradiction. Overall, **items with greater unit value are always better choices**, proving that the greedy strategy is effective.
As shown in Figure 15-6, if the item weight and unit value are viewed as the horizontal and vertical axes of a two-dimensional chart respectively, the fractional knapsack problem can be transformed into "seeking the largest area enclosed within a limited horizontal axis range". This analogy can help us understand the effectiveness of the greedy strategy from a geometric perspective.
![Geometric representation of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_area_chart.png){ class="animation-figure" }
<palign="center"> Figure 15-6 Geometric representation of the fractional knapsack problem </p>