Maximum Price to Fill a Bag

#383 Most AskedMedium

Maximum Price to Fill a Bag #383 Most Asked

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Greedy Algorithms

You are given a bag that can hold at most maxWeight kg, and you have items represented by two 0-indexed integer arrays: price and weight. price[i] and weight[i] represent the price and weight of the i^th item, respectively.

You are allowed to put any number of each item into the bag as long as the cumulative weight does not exceed maxWeight. Return the maximum total price of items you can put into the bag.

Example 1:

Input: price = [1,2,3], weight = [4,5,1], maxWeight = 15
Output: 42
Explanation: The maximum price can be obtained by putting:
- 15 / 1 = 15 of the 3^rd item which costs 15 * 3 = 45.
Therefore the maximum price we can obtain is 45.

Example 2:

Input: price = [1,2,3], weight = [4,5,6], maxWeight = 15
Output: 6
Explanation: The maximum price can be obtained by putting:
- 0 of the 1^st item which costs 0 * 1 = 0.
- 0 of the 2^nd item which costs 0 * 2 = 0.
- 2 of the 3^rd item which costs 2 * 3 = 6.
Therefore the maximum price we can obtain is 6.

Constraints:

1 <= price.length == weight.length <= 10⁵
1 <= price[i] <= 10³
1 <= weight[i] <= 10³
1 <= maxWeight <= 10³

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for prices, weights, and bagWeight? Can they be zero or negative?
If the bagWeight is zero or if all weights are greater than bagWeight, what value should I return?
Can I assume that prices and weights arrays will always have the same length and are not null or empty?
Are prices and weights integers or floating-point numbers?
If multiple combinations achieve the maximum price, is any one acceptable?

Brute Force Solution

Approach

The brute force approach to maximizing the price of items in a bag means we'll explore absolutely every possible combination of items. We check each combination to see if it fits within the bag's weight limit and then calculate the total price for that combination. Finally, we compare all the valid combinations to find the one with the highest total price.

Here's how the algorithm would work step-by-step:

Consider including no items in the bag - this is one possibility to check.
Next, consider including only the first item in the bag. See if its weight is okay and, if so, note its price.
Then, consider including only the second item in the bag. Again, check its weight and note its price.
Do this for every single item individually.
Now, try every possible pair of items: item 1 and item 2, item 1 and item 3, item 2 and item 3, and so on. For each pair, check if their combined weight is acceptable. If it is, calculate and note the combined price.
Continue this pattern, trying all possible groups of three items, then four items, and so on, up to including every single item in the bag together.
For each group of items, always check if their combined weight is within the limit. If it's too heavy, discard that combination.
After considering every possible combination, look back at all the combinations that were within the weight limit. Find the combination among them that has the highest total price.
The combination with the highest price is the answer.

Code Implementation

def maximum_price_to_fill_bag_brute_force(
    item_weights,
    item_prices,
    maximum_weight_of_bag
):
    number_of_items = len(item_weights)
    maximum_price_achieved = 0

    # Iterate through all possible combinations of items
    for i in range(1 << number_of_items):
        current_weight = 0
        current_price = 0

        # Check each item to see if it's included in the current combination
        for item_index in range(number_of_items):
            if (i >> item_index) & 1:
                # Add the item's weight and value to the current total
                current_weight += item_weights[item_index]
                current_price += item_prices[item_index]

        # Check if the current combination is within the weight limit
        if current_weight <= maximum_weight_of_bag:
            # Update the maximum price if the current combination is more valuable
            maximum_price_achieved = max(
                maximum_price_achieved, current_price
            )

    return maximum_price_achieved

def main():
    item_weights = [10, 20, 30]
    item_prices = [60, 100, 120]
    maximum_weight_of_bag = 50
    result = maximum_price_to_fill_bag_brute_force(
        item_weights,
        item_prices,
        maximum_weight_of_bag
    )
    print(f"Maximum price: {result}")

if __name__ == "__main__":
    main()

Big(O) Analysis

Time Complexity

O(2^n) – The described brute force approach examines every possible combination of items. With n items, there are 2^n possible subsets (each item can either be included or excluded). For each subset, we calculate the total weight and price, which takes O(n) time in the worst case to iterate through the items in the subset. Therefore, the overall time complexity is O(n * 2^n). However, as 2^n dominates n, we can approximate the runtime as O(2^n), representing exponential growth.

Space Complexity

O(1) – The brute force approach, as described, only requires a few variables to keep track of the current combination's weight and price, as well as the maximum price found so far. No auxiliary data structures that scale with the number of items (N) are used to store combinations or intermediate results. Therefore, the space complexity remains constant regardless of the input size N, leading to O(1) auxiliary space.

Optimal Solution

Approach

The goal is to maximize the total price of items we can fit into a bag with a weight limit. The optimal approach prioritizes items that give us the most value for their weight and uses only a fraction of any item if we can't fit the entire item.

Here's how the algorithm would work step-by-step:

Calculate the price-to-weight ratio for each item. This tells us how much 'bang for your buck' each item offers.
Sort all the items by their price-to-weight ratio, with the highest ratio items first. This ensures we consider the most valuable items first.
Start adding items to the bag, beginning with the items that have the highest price-to-weight ratio.
If an item fits completely into the remaining space in the bag, add the whole item.
If an item doesn't fit completely, take as much of the item as possible to fill the remaining space in the bag. Calculate the value of this partial item based on the fraction of the item you took.
Repeat steps 3-5 until the bag is full or you run out of items.
The total price of all the items (or portions of items) in the bag is the maximum possible price you can achieve.

Code Implementation

def maximum_price_to_fill_bag(capacity, items):
    # Calculate the value-to-weight ratio for each item
    for item in items:
        item['ratio'] = item['value'] / item['weight']

    # Sort items based on value-to-weight ratio in descending order
    items.sort(key=lambda item: item['ratio'], reverse=True)

    total_value = 0.0
    current_weight = 0.0

    # Iterate through the sorted items
    for item in items:
        if current_weight + item['weight'] <= capacity:
            # Add the entire item if it fits
            current_weight += item['weight']
            total_value += item['value']

        else:
            # Add a fraction of the item to fill the bag completely
            remaining_capacity = capacity - current_weight
            fraction = remaining_capacity / item['weight']
            total_value += item['value'] * fraction
            current_weight = capacity
            break #Bag is full, no need to check the remaining items

    return total_value

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation impacting the time complexity is the sorting of items based on their price-to-weight ratio. Sorting algorithms like merge sort or quicksort, commonly used in standard library sort functions, have a time complexity of O(n log n), where n is the number of items. Iterating through the sorted items to fill the bag involves a single loop that runs at most n times, contributing O(n). Since O(n log n) grows faster than O(n), the overall time complexity is O(n log n).

Space Complexity

O(N) – The algorithm sorts the items by their price-to-weight ratio. This sorting operation typically requires auxiliary space. In many implementations, like those using merge sort or quicksort with auxiliary space, sorting N items results in O(N) extra space for temporary arrays. Therefore, the dominant space complexity factor is the space used for sorting, which is proportional to the number of items N.

Edge Cases

Case	How to Handle
prices or weights array is null or empty	Return 0 if either array is null or empty since there are no items to put in the bag.
bagWeight is 0	Return 0 if bagWeight is 0 since nothing can be added to the bag.
prices or weights have negative values	Return an error or treat negative prices as 0 and negative weights as invalid and skip them.
prices or weights have zero values	Handle zero weights carefully to avoid division by zero errors, and zero prices should not negatively impact the result.
Very large bagWeight causing integer overflow	Use a data type that can handle larger numbers, such as long, for intermediate calculations to avoid overflow.
weights are all larger than bagWeight	Return 0 since no item can fit in the bag.
prices or weights array contain extremely large numbers	Use appropriate data types (e.g., long) and check for potential overflow or performance issues with large calculations.
prices and weights have mismatched lengths	Throw an IllegalArgumentException or return 0 if lengths are different, since you can't match prices with weights.

prices or weights array is null or empty

How to Handle:

Return 0 if either array is null or empty since there are no items to put in the bag.

bagWeight is 0

How to Handle:

Return 0 if bagWeight is 0 since nothing can be added to the bag.

prices or weights have negative values

How to Handle:

Return an error or treat negative prices as 0 and negative weights as invalid and skip them.

prices or weights have zero values

How to Handle:

Handle zero weights carefully to avoid division by zero errors, and zero prices should not negatively impact the result.

Very large bagWeight causing integer overflow

How to Handle:

Use a data type that can handle larger numbers, such as long, for intermediate calculations to avoid overflow.

weights are all larger than bagWeight

How to Handle:

Return 0 since no item can fit in the bag.

prices or weights array contain extremely large numbers

How to Handle:

Use appropriate data types (e.g., long) and check for potential overflow or performance issues with large calculations.

prices and weights have mismatched lengths

How to Handle:

Throw an IllegalArgumentException or return 0 if lengths are different, since you can't match prices with weights.