Maximizing Vessels With Unique Element Sizes In C++

In this guide, we will explore different strategies to optimize containers with distinct element dimensions in C++.

Problem Statement:

Arrange the elements in the array elements which has a size of N. Each element i can be utilized at most elements[i] times. The objective is to construct the highest number of vessels using these elements. Each vessel should contain unique elements and have a different size compared to other vessels.

Example 1:

Input:

elements = {3, 1, 5}, N = 3

Output:

Explanation:

We can use 0 at most 3 times, 1 at most twice, and 2 at most 5 times:

• ✅ Vessel 1: [2]
• ✅ Vessel 2: [0, 2]
• ✅ Vessel 3: [0, 1, 2]

So, the highest possible result is 3 since each container size varies, and each container holds distinct components.

Example 2:

Input:

element = {3, 3, 12, 200}, N = 4

Output:

Explanation:

We can use 0 at most 3 times, 1 at most 3 times, 2 at most 12 times, and 3 at most 200 times.

• ✅ Vessel 1: [3] - Using the element with value 3.
• ✅ Vessel 2: [3, 2] - Using the elements with values 3 and 2.
• ✅ Vessel 3: [3, 2, 1] - Using the elements with values 3, 2, and 1.
• ✅ Vessel 4: [3, 2, 1, 0] - Using the elements with values 3, 2, 1, and 0.

So, the highest possible outcome is 4 due to variations in vessel sizes and the distinct elements in each vessel. These vessels are designed to make the most of the elements available by maximizing the number of vessels created.

Brute Force Approach:

A straightforward method to tackle this issue would be to iterate through all feasible vessel combinations and verify which one meets the specified criteria. Nonetheless, this method becomes unfeasible as the array expands, leading to a rapid proliferation of combinations in proportion to the input's magnitude.

Utilizing a brute force strategy involves employing nested loops to traverse through every conceivable combination, resulting in inefficiency and high computational costs.

C++ Code:

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

// Function to count maximum number of vessels using brute force
int maximumVesselsBruteForce(vector<int> elements, int N)
{
 int maxVessels = 0;

 // Iterate through all possible subsets of vessels
 for (int subset = 0; subset < (1 << N); ++subset)
 {
 int currentLoad = 0;
 int count = 0;

 // Check each vessel in the subset
 for (int i = 0; i < N; ++i)
 {
 if ((subset & (1 << i)) != 0)
 {
 currentLoad += elements[i];
 ++count;
 }
 }

 // Check if the current subset is valid
 if (currentLoad >= (count * (count + 1)) / 2)
 {
 maxVessels = max(maxVessels, count);
 }
 }

 return maxVessels;
}

int main()
{
 int N = 3;
 vector<int> elements = {3, 1,5};

 // Function call
 int res = maximumVesselsBruteForce(elements, N);
 cout << "Maximum number of vessels: " << res << endl;

 return 0;
}

Output:

Maximum number of vessels: 3

Explanation:

1. Header Inclusions:

In this example, the code includes necessary header files like <iostream>, <vector>, and <algorithm> .

1. Namespace:

The using namespace std; statement is used to simplify the code by allowing the use of standard C++ identifiers without the std:: prefix .

1. Function maximumVesselsBruteForce:

• ✅ This function takes a vector of integer elements representing vessels' capacities and an integer N representing the number of vessels.
• ✅ It initializes maxVessels to 0, which will eventually store the maximum number of vessels that can be formed.
• ✅ After that, the function iterates through all possible subsets of vessels using a bitmask (subset).
• ✅ For each subset, it calculates the total load (currentLoad) and the count of vessels in the subset (count).
• ✅ It checks if the current subset is valid based on the condition currentLoad >= (count * (count + 1)) / 2 .
• ✅ If valid, it updates maxVessels with the maximum count.
• ✅ The function returns the maximum number of vessels.

1. main Function:

• ✅ In the main function, an example is given with N = 3 and elements = {3, 1, 5} .
• ✅ The maximumVesselsBruteForce function is called with these parameters, and the result is printed.

1. Example:

• ✅ For the input N = 3 and elements = {3, 1, 5} , the subsets considered are {}, {3}, {1}, {5}, {3, 1}, {3, 5}, {1, 5}, {3, 1, 5}.
• ✅ Among these subsets, {3, 1, 5} is the only valid subset (satisfying the condition currentLoad >= (count * (count + 1)) / 2).
• ✅ Therefore, the output of the program is 3 since the maximum number of vessels in a valid subset is 3.

Note: This approach has exponential time complexity and is not scalable for larger input sizes.

Time and Space Complexities

Time Complexity:

• ✅ The main contributor to the time complexity is the nested loop
• ✅ The outer loop runs in O(2^N) time, where N is the number of vessels, as it iterates through all possible subsets.
• ✅ The inner loop runs in O(N) time, where N is the number of vessels, as it iterates through the elements in each subset.

As a result, the total time complexity amounts to O(N * 2^N).

Space Complexity:

• ✅ The space complexity is determined by the additional variables used in the function ( currentLoad and count ), which occupy constant space.
• ✅ The recursion stack also contributes to space complexity.
• ✅ Hence, the overall space complexity is O(N) due to the recursion stack.

Optimal Solution:

To discover the most efficient solution, we must create a plan that effectively identifies the highest quantity of containers that meet the specified criteria.

The most efficient method arranges items according to both their frequency and initial position by employing a vector of pairs. It effectively monitors utilized items through a set to guarantee each container contains unique elements. This strategy is adaptable and offers a more effective alternative to the straightforward technique, especially with extensive input arrays. Leveraging sorting and a set leads to a quicker algorithm for recognizing and optimizing containers with different element quantities. Retaining the original indices ensures the output maintains the sequence of elements from the initial array.

C++ Code:

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

// Function to count the maximum number of vessels that can be formed
int countMaximumVessels(vector<int> vesselCapacities, int numberOfVessels)
{
 // Sort the vessel capacities in ascending order
 sort(vesselCapacities.begin(), vesselCapacities.end());

 long long currentLoad = 0, maximumVessels = 0;

 // Iterate through each vessel capacity
 for (int capacity : vesselCapacities) {
 currentLoad += capacity;

 // Check if the current load is enough to accommodate the next vessel
 if (currentLoad >= ((maximumVessels + 1) * (maximumVessels + 2)) / 2) {
 maximumVessels++;
 }
 }

 // Return the maximum number of vessels that can be formed
 return maximumVessels;
}

int main()
{
 // Example usage
 int numberOfVessels = 4;
 vector<int> vesselCapacities = {3, 3, 12, 200};

 // Function call
 int result = countMaximumVessels(vesselCapacities, numberOfVessels);

 // Display the result
 cout << "Maximum number of vessels: " << result << endl;

 return 0;
}

Output:

Maximum number of vessels: 4

Explanation:

1. Function countMaximumVessels:

• ✅ This function takes a vector of integers vesselCapacities representing the capacities of vessels and an integer numberOfVessels representing the total number of vessels.
• ✅ It sorts the vesselCapacities vector in ascending order.

1. Variables:

• ✅ currentLoad: It keeps track of the current total load of vessels.
• ✅ maximumVessels: It keeps track of the maximum number of vessels that can be formed.

1. Iteration through Vessel Capacities:

• ✅ The function iterates through each vessel capacity in the sorted vector.
• ✅ For each capacity, it adds the capacity to the currentLoad
• ✅ After that, it checks if the current load is enough to accommodate the next vessel based on the condition currentLoad >= ((maximumVessels + 1) * (maximumVessels + 2)) / 2.
• ✅ If the condition is true, it increments maximumVessels because it means the current load can accommodate an additional vessel.

1. Return:

• ✅ The function returns the calculated maximumVessels , representing the maximum number of vessels that can be formed.

1. Main function:

• ✅ In the main function, an example is given with numberOfVessels = 4 and vesselCapacities = {3, 3, 12, 200}.
• ✅ The countMaximumVessels function is called with these parameters, and the result is stored in the variable result.

Time and Space Complexities

Time Complexity:

• ✅ The primary time-consuming operation is sorting the vesselCapacities vector, which has a time complexity of O(N log N), where N is the number of elements in the vector.
• ✅ The subsequent iteration through the sorted vector has an O(N) linear time complexity.
• ✅ Therefore, the overall time complexity is O(N log N + N) , which simplifies to O(N log N).

Space Complexity:

• ✅ The space complexity is determined by the additional variables used in the function ( currentLoad and maximumVessels ), which occupy constant space.
• ✅ Sorting is usually done in-place in C++, so it doesn't contribute to the space complexity.
• ✅ Hence, the overall space complexity is O(1) or constant.