Maximum Bags With Full Capacity of Rocks

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Greedy AlgorithmsArrays

You have n bags numbered from 0 to n - 1. You are given two 0-indexed integer arrays capacity and rocks. The i^th bag can hold a maximum of capacity[i] rocks and currently contains rocks[i] rocks. You are also given an integer additionalRocks, the number of additional rocks you can place in any of the bags.

Return the maximum number of bags that could have full capacity after placing the additional rocks in some bags.

Example 1:

Input: capacity = [2,3,4,5], rocks = [1,2,4,4], additionalRocks = 2
Output: 3
Explanation:
Place 1 rock in bag 0 and 1 rock in bag 1.
The number of rocks in each bag are now [2,3,4,4].
Bags 0, 1, and 2 have full capacity.
There are 3 bags at full capacity, so we return 3.
It can be shown that it is not possible to have more than 3 bags at full capacity.
Note that there may be other ways of placing the rocks that result in an answer of 3.

Example 2:

Input: capacity = [10,2,2], rocks = [2,2,0], additionalRocks = 100
Output: 3
Explanation:
Place 8 rocks in bag 0 and 2 rocks in bag 2.
The number of rocks in each bag are now [10,2,2].
Bags 0, 1, and 2 have full capacity.
There are 3 bags at full capacity, so we return 3.
It can be shown that it is not possible to have more than 3 bags at full capacity.
Note that we did not use all of the additional rocks.

Constraints:

n == capacity.length == rocks.length
1 <= n <= 5 * 10⁴
1 <= capacity[i] <= 10⁹
0 <= rocks[i] <= capacity[i]
1 <= additionalRocks <= 10⁹

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 the number of rocks in each bag, and the capacity of each bag? Are they always positive integers?
Can the 'additionalRocks' be zero? What should I return if 'additionalRocks' is zero and no bags can be fully filled?
Is there a limit on the number of bags (the length of the input arrays)?
If there are multiple sets of bags that can be filled to their full capacity using the 'additionalRocks', should I return the maximum possible number of such bags?
Is it possible for a bag's capacity to be less than the number of rocks it already contains? If so, how should I handle it?

Brute Force Solution

Approach

The brute force method for this problem is like trying out every single combination of how to fill the bags with the extra rocks we have. We check each combination to see how many bags we can completely fill. Then, we pick the combination that allows us to fill the most bags.

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

Consider all possible ways to distribute the extra rocks among the bags.
For each way we distribute the rocks, check how many bags are now completely full.
Keep track of the maximum number of completely full bags we have seen so far.
After checking all the possible rock distributions, report the maximum number of completely full bags found.

Code Implementation

def maximum_bags_brute_force(capacity, rocks, additional_rocks):
    max_full_bags = 0
    number_of_bags = len(capacity)

    # Iterate through all possible distributions of additional rocks
    for i in range(2**(number_of_bags)):
        current_rocks = additional_rocks
        full_bags_count = 0
        temp_rocks = rocks[:]

        # Represents a distribution of rocks among bags.
        for bag_index in range(number_of_bags):
            if (i >> bag_index) & 1:
                needed_rocks = capacity[bag_index] - temp_rocks[bag_index]

                # If we can fill this bag, do so.
                if needed_rocks > 0 and current_rocks >= needed_rocks:
                    current_rocks -= needed_rocks
                    temp_rocks[bag_index] += needed_rocks

        # Count how many bags are full.
        for bag_index in range(number_of_bags):
            if temp_rocks[bag_index] == capacity[bag_index]:
                full_bags_count += 1

        # Update the maximum number of full bags.
        max_full_bags = max(max_full_bags, full_bags_count)

    return max_full_bags

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach considers all possible subsets of bags to fill with extra rocks. Since each bag can either be chosen or not chosen to receive rocks, there are 2^n possible combinations to consider, where n is the number of bags. For each of these 2^n combinations, we iterate through the bags to check how many are full. Therefore, the time complexity is dominated by generating and evaluating these 2^n subsets. Thus, the algorithm takes O(2^n) time.

Space Complexity

O(N!) – The brute force approach, as described, involves considering all possible ways to distribute the extra rocks. This conceptually implies generating combinations or permutations. The number of such combinations/permutations of distributing rocks can grow factorially with the number of bags (N). Therefore, storing and iterating through these combinations, even if not explicitly coded in that way, would implicitly require memory proportional to the number of possible distributions, which can reach up to O(N!), where N is the number of bags. This estimation comes from the fact that for each bag you have to find how many rocks can it accomodate, so it's a type of partition problem which can be at most N! combinations. The plain english explanation doesn't describe any optimizations for space, implying that the combination generation dominates the space complexity.

Optimal Solution

Approach

The best way to maximize the number of bags with full capacity is to prioritize filling the bags that require the least amount of additional rocks first. This greedy approach ensures we use our available rocks most efficiently. We determine the 'need' for each bag, then fill the 'neediest' bags first.

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

For each bag, calculate how many more rocks are needed to reach its full capacity.
Arrange the bags in increasing order based on how many more rocks they need (from least to most).
Go through the arranged bags one by one, starting with the bag that needs the fewest rocks.
If you have enough rocks to fill the current bag to capacity, fill it, and reduce the total available rocks by the amount used.
If you don't have enough rocks to fill the current bag, stop the process because you can't fill any more bags completely.
Count how many bags you were able to fill completely. That's your answer.

Code Implementation

def maximum_bags_with_full_capacity(capacity, rocks, additional_rocks):
    number_of_bags = len(capacity)
    rocks_needed = []

    for i in range(number_of_bags):
        rocks_needed.append(capacity[i] - rocks[i])

    # Sort bags by how many more rocks they need.
    bags_with_index = sorted(range(number_of_bags), key=lambda i: rocks_needed[i])

    full_bags_count = 0

    # Iterate through bags in ascending order of rocks needed
    for bag_index in bags_with_index:
        needed_rocks_for_bag = rocks_needed[bag_index]

        # Check if we can fill the current bag completely
        if additional_rocks >= needed_rocks_for_bag:
            additional_rocks -= needed_rocks_for_bag
            full_bags_count += 1

        # If not enough rocks, then stop. Bags are sorted.
        else:
            break

    return full_bags_count

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm iterates through 'n' bags to calculate the remaining capacity (needed rocks) for each. The dominant operation is sorting these 'n' values, which takes O(n log n) time. The subsequent loop iterates through the sorted bags, performing constant-time operations (comparison and subtraction) for each bag. Therefore, the overall time complexity is determined by the sorting step, resulting in O(n log n).

Space Complexity

O(N) – The provided solution calculates the 'need' for each bag and then arranges the bags based on this need. This arrangement requires storing these 'need' values, implying an auxiliary array of size N, where N is the number of bags. The sorting operation might be done in-place, but to definitively establish the order of bags based on 'need', we need to allocate a separate array to store the difference between capacity and rocks. Thus, the space complexity is directly proportional to the number of bags. This results in O(N) space complexity.

Edge Cases

Case	How to Handle
Empty `capacity` or `rocks` array	Return 0 as no bags exist to fill.
`capacity` or `rocks` array is null	Throw an IllegalArgumentException or return 0, depending on requirements.
Different lengths of `capacity` and `rocks` arrays	Throw an IllegalArgumentException as the inputs are invalid.
All bags already have full capacity (rocks[i] == capacity[i] for all i)	Return the length of the capacity array, as all bags are full.
All bags have zero capacity (capacity[i] == 0 for all i)	Return the length of the capacity array, as all bags are already full.
Total available rocks are insufficient to fill any bag completely	Return 0 as no bags can be filled.
Large array sizes causing potential integer overflow when calculating rocks needed	Use long data type for intermediate calculations to prevent overflow and ensure correct sorting/comparison.
`rocks[i]` > `capacity[i]` for any i	Throw IllegalArgumentException, treat rocks[i] as capacity[i] or set rocks[i] to capacity[i] (depending on requirements), and continue.

Empty `capacity` or `rocks` array

How to Handle:

Return 0 as no bags exist to fill.

`capacity` or `rocks` array is null

How to Handle:

Throw an IllegalArgumentException or return 0, depending on requirements.

Different lengths of `capacity` and `rocks` arrays

How to Handle:

Throw an IllegalArgumentException as the inputs are invalid.

All bags already have full capacity (rocks[i] == capacity[i] for all i)

How to Handle:

Return the length of the capacity array, as all bags are full.

All bags have zero capacity (capacity[i] == 0 for all i)

How to Handle:

Return the length of the capacity array, as all bags are already full.

Total available rocks are insufficient to fill any bag completely

How to Handle:

Return 0 as no bags can be filled.

Large array sizes causing potential integer overflow when calculating rocks needed

How to Handle:

Use long data type for intermediate calculations to prevent overflow and ensure correct sorting/comparison.

`rocks[i]` > `capacity[i]` for any i

How to Handle:

Throw IllegalArgumentException, treat rocks[i] as capacity[i] or set rocks[i] to capacity[i] (depending on requirements), and continue.