Solution For Maximum Units On A Truck

Problem Statement:

You are given two arrays, boxTypes and truckSize, of positive integers where boxTypes[i] = [numberOfBoxes[i], numberOfUnitsPerBox[i]] represents the number of boxes of type i and the number of units in each box of type i, respectively. You are also given an integer truckSize, which represents the maximum number of boxes that can be put on the truck. You can choose any boxes to put on the truck as long as the number of boxes does not exceed truckSize.

Return the maximum total number of units that can be put on the truck.

Solution:

The solution to this problem involves sorting the given boxTypes array in non-increasing order based on the number of units per box. After sorting the array, we can traverse through the array, and for each box type, keep adding boxes to the truck until the truck is full. In each iteration, we keep track of the number of boxes and units added to the truck and update it until the truck is full. The algorithm for the solution is as follows:

1. Sort the boxTypes array in non-increasing order based on the number of units per box.

2. Initialize two variables, total_units and total_boxes, as 0.

3. Traverse through the boxTypes array and for each box type, do the following:

   a. Check if the number of remaining boxes can be added to the truck without exceeding the truck size.

   b. If so, add the remaining boxes and units to the truck and update the total_boxes and total_units variables.

   c. If not, add only the remaining space on the truck and update the total_boxes and total_units variables.

4. Return the value of the total_units variable.

The time complexity of this algorithm is O(nlogn) due to the sorting operation, where n is the length of the boxTypes array. The space complexity is O(1) as we are not using any extra space.

Here is the Python implementation of the solution:

def maximumUnits(boxTypes, truckSize):
# Sort the boxTypes array in non-increasing order based on the number of units per box
boxTypes.sort(key = lambda x: x[1], reverse = True)

# Initialize variables
total_units = 0
total_boxes = 0

# Traverse through the boxTypes array and add boxes to the truck
for boxes, units in boxTypes:
    # Check if the remaining boxes can be added to the truck without exceeding the truck size
    if total_boxes + boxes <= truckSize:
        total_boxes += boxes
        total_units += boxes * units
    # If not, add only the remaining space on the truck
    else:
        remaining_boxes = truckSize - total_boxes
        total_boxes += remaining_boxes
        total_units += remaining_boxes * units
        break

# Return the maximum total number of units that can be put on the truck
return total_units

Example usage

boxTypes = [[1, 3], [2, 2], [3, 1]] truckSize = 4
print(maximumUnits(boxTypes, truckSize)) # Output: 8

Explanation: We can take 1 box of type 1 (3 units per box) and 2 boxes of type 2 (2 units per box) for a total of (1 * 3) + (2 * 2) = 8 units. There is no space for any boxes of type 3.

Step by Step Implementation For Maximum Units On A Truck

/**
 * Given an array of weights and an integer capacity, 
 * return the maximum number of weights that can be 
 * stored in the truck.
 * 
 * @param weights an array of weights
 * @param capacity the maximum weight the truck can hold
 * @return the maximum number of weights that can be stored in the truck
 */
public int maximumUnitsOnATruck(int[] weights, int capacity) {
    // check for invalid inputs
    if (weights == null || weights.length == 0 || capacity <= 0) {
        return 0;
    }
    
    // sort the weights in descending order
    Arrays.sort(weights);
    
    // keep track of the current weight and the number of weights
    int currentWeight = 0;
    int numWeights = 0;
    
    // try to add each weight to the truck
    for (int weight : weights) {
        // if the weight can be added without exceeding the capacity, add it
        if (currentWeight + weight <= capacity) {
            currentWeight += weight;
            numWeights++;
        }
        // otherwise, stop trying to add weights
        else {
            break;
        }
    }
    
    return numWeights;
}

# This problem can be solved using a greedy algorithm. # We start by sorting the weights in descending order. # Then, we take the heaviest item and put it on the truck. # We keep track of the remaining capacity of the truck # and we keep taking the heaviest items until the truck is full. # We continue this process until all items are on the truck. 

def maximumUnits(weights, capacity): 
    
    # sort the weights in descending order 
    weights.sort(reverse = True) 
    
    # initialize the number of units 
    numUnits = 0
    
    # take the heaviest item and put it on the truck 
    for weight in weights: 
        
        # if the weight is greater than the remaining capacity, break 
        if weight > capacity: 
            break 
        
        # otherwise, put the item on the truck 
        else: 
            numUnits += 1 
            capacity -= weight 
    
    return numUnits

var maxUnitsOnATruck = function(weights, maxCapacity) {
  // your code here
};

// Maximum Units on a Truck

// You are given the number of rows n and the number of columns m of a 2D array where each cell contains an integer, the value in that cell. You are also given the number of units that you can move in a single turn k.

// Your goal is to reach the bottom-right corner of the 2D array. You can move up, down, left, or right in each turn. You can start from any cell in the 2D array.

// Return the maximum number of units that you can move.

// Examples:

// Input: n = 3, m = 3, k = 2
// Output: 8
// Explanation:
// [[1,2,3],
//  [4,5,6],
//  [7,8,9]]
// We can reach the bottom-right corner from the starting cell (1, 1) with 8 units of movement: (1, 2), (1, 3), (2, 3), (2, 2), (2, 1), (3, 1), (3, 2), (3, 3).

// Input: n = 3, m = 3, k = 1
// Output: 5
// Explanation:
// [[1,2,3],
//  [4,5,6],
//  [7,8,9]]
// We can reach the bottom-right corner from the starting cell (1, 1) with 5 units of movement: (1, 2), (1, 3), (2, 3), (2, 2), (3, 2).

// Input: n = 3, m = 3, k = 0
// Output: 1
// Explanation:
// [[1,2,3],
//  [4,5,6],
//  [7,8,9]]
// We can reach the bottom-right corner from the starting cell (1, 1) with 1 unit of movement.

// Constraints:

// 1 <= n, m <= 100
// 0 <= k <= 20
// 1 <= array elements <= 100

using System; 

public class Solution { 

// Function to find the maximum units that can be carried 

// on the truck 

static int maxUnits(int numTrucks, int[] load, int[] unitSize) 

{ 

// Initialize result 

int res = 0; 

// Sort loads in decreasing order 

Array.Sort(load); 

Array.Reverse(load); 

// Sort unit sizes in decreasing order 

Array.Sort(unitSize); 

Array.Reverse(unitSize); 

// Find the maximum units that can be carried 

for (int i = 0; i < numTrucks; i++) 

{ 

res += load[i] / unitSize[i]; 

} 

// Return the result 

return res; 

} 

// Driver code 

public static void Main() 

{ 

int numTrucks = 4; 

int[] load = { 10, 12, 8, 20 }; 

int[] unitSize = { 4, 5, 2, 3 }; 

Console.Write(maxUnits(numTrucks, load, unitSize)); 

} 

} 

// This code is contributed by vt_m