19(June) Find maximum volume of a cuboid

19. Find Maximum Volume of a Cuboid

The problem can be found at the following link: Question Link

Problem Description

You are given the perimeter and the area of a cuboid. Your task is to return the maximum volume that can be formed with the given perimeter and surface area. Round off the result to 2 decimal places. For the given parameters, it is guaranteed that an answer always exists.

Examples:

Input:

perimeter = 22, area = 15

Output:

3.02

Explanation: The maximum attainable volume of the cuboid is 3.02.

My Approach

1. Extracting Parameters:

• Calculate the intermediate terms required to derive the dimensions of the cuboid using the provided formulas.

• Part 1: ((\text{{perimeter}} - \sqrt{\text{{perimeter}}^2 - 24 \times \text{{area}}}) / 12)

• Part 2: ((\text{{perimeter}} / 4) - (2 \times \text{{part1}}))

1. Volume Calculation:

• Compute the volume using the formula: (\text{{part1}}^2 \times \text{{part2}}).

1. Return the Result:

• Return the volume rounded to 2 decimal places.

Time and Auxiliary Space Complexity

• Expected Time Complexity: O(1), as we perform a constant number of operations regardless of the input values.

• Expected Auxiliary Space Complexity: O(1), as we only use a constant amount of additional space.

Code

C++:

class Solution {
  public:
    double maxVolume(double p, double a) {
        double part1 = (p - sqrt(pow(p, 2) - (24 * a))) / 12;
        double part2 = (p / 4) - (2 * part1);
        double ans = pow(part1, 2) * part2;
        return round(ans * 100) / 100;
    }
};

Java:

class Solution {
    double maxVolume(double perimeter, double area) {
        double part1 = (perimeter - Math.sqrt(Math.pow(perimeter, 2) - (24 * area))) / 12;
        double part2 = (perimeter / 4) - (2 * part1);
        double ans = Math.pow(part1, 2) * part2;
        return Math.round(ans * 100.0) / 100.0;
    }
}

Python:

class Solution:
    def maxVolume(self, perimeter, area):
        part1 = (perimeter - (perimeter ** 2 - 24 * area) ** 0.5) / 12
        part2 = (perimeter / 4) - (2 * part1)
        ans = part1 ** 2 * part2
        return round(ans, 2)

Contribution and Support

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