hello-algo/en/docs/chapter_greedy/fractional_knapsack_problem.md

# Fractional knapsack problem

!!! question

    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 the figure below.

![Example data of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_example.png)

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 the figure below, **we can arbitrarily split the items and calculate the corresponding value based on the weight proportion**.

1. For item $i$, its value per unit weight is $val[i-1] / wgt[i-1]$, referred to as the unit value.
2. Suppose we put a part of item $i$ with weight $w$ into the knapsack, then the value added to the knapsack is $w \times val[i-1] / wgt[i-1]$.

![Value per unit weight of the item](fractional_knapsack_problem.assets/fractional_knapsack_unit_value.png)

### Greedy strategy determination

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 the figure below 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)

### 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:

```src
[file]{fractional_knapsack}-[class]{}-[func]{fractional_knapsack}
```

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 $O(n)$**.

### 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 the figure below, 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)