2279. Maximum Bags With Full Capacity of Rocks

Problem Description

In this problem, we're dealing with a set of n bags, each with a certain capacity. The capacity array represents the maximum number of rocks each bag can hold, while the rocks array shows the current number of rocks in each bag. We also have a certain number of additionalRocks that we can distribute across these bags.

The objective is to maximize the number of bags that are filled to their full capacity using the additionalRocks available. A bag is considered to be at full capacity if the number of rocks it contains equals its capacity.

Intuition

The intuition behind the solution is based on the idea that to maximize the number of bags at full capacity, we should fill the bags that need the fewest additional rocks first. This is a greedy approach where we prioritize the bags that are closest to being full because adding rocks to them will quickly increase the count of fully filled bags.

We calculate the difference between the capacity and rocks for each bag, which represents the number of additional rocks needed to fill each bag to capacity. We then sort this list to get the bags that need the least additional rocks at the beginning.

Stepping through this sorted list, we distribute the additionalRocks as long as we have enough to fill a bag. Each time we fill a bag, we decrement the number of additionalRocks by the number used and increment the count of fully-filled bags by one. This process continues until we run out of additionalRocks or fill all the bags. The final count of fully-filled bags is the maximum number we can achieve with the given additionalRocks.

Learn more about Greedy and Sorting patterns.

Solution Approach

The solution implements a simple greedy strategy using Python lists and sorting algorithm. Here's a step-by-step walkthrough of the implementation:

1. Calculate the difference between the capacity and the rocks for each bag to find out how many more rocks are needed to reach full capacity. This is done using a list comprehension:

   1d = [a - b for a, b in zip(capacity, rocks)]

   Here, a represents an element from the capacity array and b represents the corresponding element from the rocks array. The zip function pairs each element from capacity with the corresponding element from rocks.

2. The problem is now reduced to filling the bags with the least difference first. To do this efficiently, sort the list d in non-decreasing order:

   1d.sort()

3. Initialize a variable ans to count the number of bags that can be filled to full capacity:

   1ans = 0

4. Iterate through the sorted list d and try to fill each bag. If the additionalRocks is enough to fill the current bag, increase the ans by 1 and reduce additionalRocks by the amount used:

   1for v in d:
   2    if v <= additionalRocks:
   3        ans += 1
   4        additionalRocks -= v

   • v represents the number of additional rocks needed for the current bag.
   • If additionalRocks is at least v, it means we can fill this bag. Then we update additionalRocks to reflect the rocks used.
   • The loop continues either until there are no more rocks left (additionalRocks is less than the next v in the list) or all bags are checked.

5. Once the loop is complete, ans is the maximum number of bags that can be filled to full capacity, and the function returns this value:

   1return ans

This approach uses the built-in sorting function which typically has a time complexity of O(n log n) where n is the number of elements in the list. The subsequent iteration through the sorted list has a linear time complexity of O(n). As a result, the overall time complexity of this approach is O(n log n). The space complexity is O(n) due to the creation of the list d.

Example Walkthrough

Let's illustrate the solution approach with a small example. Suppose we have the following inputs:

• capacity array: [5, 3, 7]
• rocks array: [4, 2, 5]
• additionalRocks: 3

We want to find out the maximum number of bags that can be filled to their full capacity using these 3 additional rocks.

Following the steps outlined in the solution approach:

1. First, we calculate the difference between capacity and rocks for each bag.

   Using the list comprehension:

   1d = [a - b for a, b in zip(capacity, rocks)]  # d will become [1, 1, 2]

2. We then sort this list to consider the bags that need the fewest additional rocks first:

   1d.sort()  # The sorted list d will remain [1, 1, 2]

3. We initialize ans to keep track of the bags that can be filled to full capacity:

   1ans = 0  # Starts at 0

4. Now, we iterate through the sorted list d and distribute additionalRocks:

   1# Loop through [1, 1, 2] with additionalRocks starting at 3
   2for v in d:
   3    if v <= additionalRocks:
   4        ans += 1            # Increment the count of full bags
   5        additionalRocks -= v  # Decrease the additionalRocks by v

   The first bag needs 1 rock to be full, which we have, so ans becomes 1 and additionalRocks becomes 2.

   The second bag also needs 1 rock, so ans is incremented to 2 and additionalRocks is reduced to 1.

   The third bag needs 2 rocks to be full. However, we only have 1 additionalRock left, so we cannot fill this bag to capacity.

5. The process stops here as we have distributed all additional rocks that we can. The final count of fully-filled bags is the value of ans:

   1return ans  # Returns 2 as the maximum number of full bags

Therefore, we have maximized the number of full bags (2 out of 3) using the 3 additionalRocks provided.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code consists of several parts:

1. The list comprehension d = [a - b for a, b in zip(capacity, rocks)] takes O(n) time, where n is the number of elements in capacity and rocks.
2. Sorting the list d.sort() has a time complexity of O(n log n) because it uses the Timsort algorithm which is Python's standard sorting algorithm.
3. The loop for v in d: iterates through each element of the list d once, giving a time complexity of O(n).

Since sorting the list is the most expensive operation, the overall time complexity of the code is O(n log n).

Space Complexity

The space complexity of the code also involves a few components:

1. The list comprehension generates a new list d of size n, resulting in O(n) space complexity.
2. Sorting the list is done in-place in Python, so it doesn't require additional space other than some constant workspace, hence O(1).
3. The variables ans and additionalRocks use constant space, O(1).

Combining these, the total space complexity of the code is O(n) because the new list d is the dominant factor.

Learn more about how to find time and space complexity quickly using problem constraints.