5 Best Ways to Program to Find Maximum Units That Can Be Put on a Truck in Python

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πŸ’‘ Problem Formulation: The task is to develop a Python program that calculates the maximum number of units that can be loaded onto a truck given a limit to the truck’s capacity. Imagine we are given an array boxTypes, where boxTypes[i] = [numberOfBoxes_i, numberOfUnitsPerBox_i], and an integer truckSize, representing the maximum number of boxes the truck can carry. For instance, if the input array is [[1,3],[2,2],[3,1]] and the truck size is 4, the output should be 8 units, by taking one box of type one and two boxes of type two.

Method 1: Sorting and Greedy Approach

This method involves sorting the array of box types based on the number of units per box in descending order. Then, we iterate over the sorted list, adding boxes to the truck until it reaches capacity. The function is designed to ensure that we maximize the unit count by always choosing the box with the highest units available that fits in the truck.

Here’s an example:

def maximumUnits(boxTypes, truckSize):
    boxTypes.sort(key=lambda x: -x[1])
    units = 0
    for box in boxTypes:
        if truckSize >= box[0]:
            units += box[0] * box[1]
            truckSize -= box[0]
        else:
            units += truckSize * box[1]
            break
    return units

# Example usage:
print(maximumUnits([[1,3],[2,2],[3,1]], 4))

The output of this code snippet:

8

This code snippet defines the function maximumUnits() which sorts the input list boxTypes, effectively prioritizing box types with more units per box when picking the boxes to fill the truck. It then iterates over the sorted list, filling the truck while reducing truckSize accordingly, and keeps a running total of the units in units.

Method 2: Heap/Priority Queue

Instead of sorting the entire list, we can use a max-heap or priority queue to fetch the box types with the most units per box efficiently. This way, we always pick the box with maximum units until the truck is at capacity or we run out of boxes.

Here’s an example:

import heapq

def maximumUnits(boxTypes, truckSize):
    # Create a max-heap with the negative number of units per box
    # since heapq in Python is a min-heap
    max_heap = [(-units, boxes) for boxes, units in boxTypes]
    heapq.heapify(max_heap)
    units = 0
    while truckSize > 0 and max_heap:
        units_per_box, boxes = heapq.heappop(max_heap)
        count = min(boxes, truckSize)
        units -= units_per_box * count
        truckSize -= count
    return units

# Example usage:
print(maximumUnits([[1,3],[2,2],[3,1]], 4))

The output of this code snippet:

8

In this code snippet, the maximumUnits() function utilizes a max-heap to keep track of the boxes with the highest units. We use Python’s heapq module which requires us to invert the units for it to work as a max-heap. We then continuously pop the box with the most units and decrement the truck size until no more boxes can fit.

Method 3: Counting Sort for Limited Range of Units

If the number of units per box has a limited range, we can use counting sort which might be more efficient than general sorting. This method counts how many boxes of each type we have and starts filling the truck with the box type that has the highest units per box.

Here’s an example:

def maximumUnits(boxTypes, truckSize):
    # Assuming the maximum number of units per box is known
    max_units_per_box = max(box[1] for box in boxTypes)
    counts = [0] * (max_units_per_box + 1)
    for boxes, units in boxTypes:
        counts[units] += boxes
    units = 0
    for units_per_box in reversed(range(len(counts))):
        if truckSize == 0:
            break
        boxes_to_take = min(counts[units_per_box], truckSize)
        units += boxes_to_take * units_per_box
        truckSize -= boxes_to_take
    return units

# Example usage:
print(maximumUnits([[1,3],[2,2],[3,1]], 4))

The output of this code snippet:

8

This code snippet utilizes the counting sort algorithm to keep a tally of how many boxes exist for each unit value. Starting from the highest value, it calculates how many boxes can be taken at each step and updates both the total units and the remaining truck capacity.

Method 4: Custom Sort with Lambda

If the objective is to maintain the order for boxes, which have the same number of units, we can sort the list using a custom key. We can specify a lambda function that sorts primarily based on units per box, and secondarily on the number of boxes available in case of a tie in units.

Here’s an example:

def maximumUnits(boxTypes, truckSize):
    boxTypes.sort(key=lambda x: (-x[1], x[0]))
    units = 0
    for box in boxTypes:
        count = min(truckSize, box[0])
        units += count * box[1]
        truckSize -= count
        if truckSize == 0:
            break
    return units

# Example usage:
print(maximumUnits([[1,3],[2,2],[3,1]], 4))

The output of this code snippet:

8

By using a custom sort with a lambda function, we can prioritize both the number of units per box and the count of boxes available. The function maximumUnits() then iterates over the sorted list just like in method 1, ensuring an efficient filling of the truck.

Bonus One-Liner Method 5: Using List Comprehension and Sum

For a compact solution, we can use list comprehension to sort the boxes, then sum up the units considering the truck size. This one-liner is less readable, but it constructs the list in sorted order and computes the units as it goes.

Here’s an example:

maximumUnits = lambda bt, ts: sum((box[0] if (ts := ts - box[0]) >= 0 else ts + box[0]) * box[1] for box in sorted(bt, key=lambda x: -x[1]) if ts > 0)

# Example usage:
print(maximumUnits([[1,3],[2,2],[3,1]], 4))

The output of this code snippet:

8

This compact one-liner uses the walrus operator (:=) introduced in Python 3.8 to subtract the number of boxes from truck size inside a list comprehension. It sorts the box types list with a lambda function and computes the sum concurrently, offering a quick and less-verbose solution.

Summary/Discussion

  • Method 1:

    Sorting and Greedy Approach. This method is straightforward and easy to understand. It works well for most case scenarios, but its time complexity is dependent on the sorting algorithm, which is O(n log n).

  • Method 2:

    Heap/Priority Queue. Efficient for large data sets where we don’t need to sort the entire array, as we are constantly fetching the maximum element. However, it can be overkill for smaller or already sorted data sets.

  • Method 3:

    Counting Sort for Limited Range of Units. Extremely efficient if the number of units per box is limited and does not exceed a certain range. Its main weakness is that it’s not suitable for a large range of units per box due to the space complexity.

  • Method 4:

    Custom Sort with Lambda. It’s versatile for scenarios when secondary sorting criteria is needed. While still dependent on sorting time complexity, it provides more control over how the data is sorted.

  • Method 5:

    Bonus One-Liner Method. This method is elegant and concise but sacrifices readability and may be hard to maintain and debug.