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2525. Categorize Box According to Criteria

SourceBiweekly Contest 95 Q1DifficultyEasyRating1301

Description

Given four integers length, width, height, and mass, representing the dimensions and mass of a box, respectively, return a string representing the category of the box.

  • The box is "Bulky" if:
    • Any of the dimensions of the box is greater or equal to 104.
    • Or, the volume of the box is greater or equal to 109.
  • If the mass of the box is greater or equal to 100, it is "Heavy".
  • If the box is both "Bulky" and "Heavy", then its category is "Both".
  • If the box is neither "Bulky" nor "Heavy", then its category is "Neither".
  • If the box is "Bulky" but not "Heavy", then its category is "Bulky".
  • If the box is "Heavy" but not "Bulky", then its category is "Heavy".

Note that the volume of the box is the product of its length, width and height.

 

Example 1:

Input: length = 1000, width = 35, height = 700, mass = 300
Output: "Heavy"
Explanation: 
None of the dimensions of the box is greater or equal to 104. 
Its volume = 24500000 <= 109. So it cannot be categorized as "Bulky".
However mass >= 100, so the box is "Heavy".
Since the box is not "Bulky" but "Heavy", we return "Heavy".

Example 2:

Input: length = 200, width = 50, height = 800, mass = 50
Output: "Neither"
Explanation: 
None of the dimensions of the box is greater or equal to 104.
Its volume = 8 * 106 <= 109. So it cannot be categorized as "Bulky".
Its mass is also less than 100, so it cannot be categorized as "Heavy" either. 
Since its neither of the two above categories, we return "Neither".

 

Constraints:

  • 1 <= length, width, height <= 105
  • 1 <= mass <= 103

Solutions

Solution 1: Simulation

Thinking

A box is classified by whether any dimension or the volume is bulky and whether the mass is heavy — four labels in all. Each test is a constant-time comparison.

Store \(\textit{bulky}\) and \(\textit{heavy}\) as \(0/1\) and index \([\textit{Neither},\textit{Bulky},\textit{Heavy},\textit{Both}]\) by \(\textit{heavy}\ll 1\mid \textit{bulky}\), avoiding nested branches.

We can simulate according to the problem description.

The time complexity is \(O(1)\), and the space complexity is \(O(1)\).

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class Solution:
    def categorizeBox(self, length: int, width: int, height: int, mass: int) -> str:
        v = length * width * height
        bulky = int(any(x >= 10000 for x in (length, width, height)) or v >= 10**9)
        heavy = int(mass >= 100)
        i = heavy << 1 | bulky
        d = ['Neither', 'Bulky', 'Heavy', 'Both']
        return d[i]

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class Solution {
    public String categorizeBox(int length, int width, int height, int mass) {
        long v = (long) length * width * height;
        int bulky = length >= 10000 || width >= 10000 || height >= 10000 || v >= 1000000000 ? 1 : 0;
        int heavy = mass >= 100 ? 1 : 0;
        String[] d = {"Neither", "Bulky", "Heavy", "Both"};
        int i = heavy << 1 | bulky;
        return d[i];
    }
}

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class Solution {
public:
    string categorizeBox(int length, int width, int height, int mass) {
        long v = (long) length * width * height;
        int bulky = length >= 10000 || width >= 10000 || height >= 10000 || v >= 1000000000 ? 1 : 0;
        int heavy = mass >= 100 ? 1 : 0;
        string d[4] = {"Neither", "Bulky", "Heavy", "Both"};
        int i = heavy << 1 | bulky;
        return d[i];
    }
};

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func categorizeBox(length int, width int, height int, mass int) string {
    v := length * width * height
    i := 0
    if length >= 10000 || width >= 10000 || height >= 10000 || v >= 1000000000 {
        i |= 1
    }
    if mass >= 100 {
        i |= 2
    }
    d := [4]string{"Neither", "Bulky", "Heavy", "Both"}
    return d[i]
}

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function categorizeBox(length: number, width: number, height: number, mass: number): string {
    const v = length * width * height;
    let i = 0;
    if (length >= 10000 || width >= 10000 || height >= 10000 || v >= 1000000000) {
        i |= 1;
    }
    if (mass >= 100) {
        i |= 2;
    }
    return ['Neither', 'Bulky', 'Heavy', 'Both'][i];
}

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impl Solution {
    pub fn categorize_box(length: i32, width: i32, height: i32, mass: i32) -> String {
        let v = (length as i64) * (width as i64) * (height as i64);
        let mut i = 0;

        if length >= 10000 || width >= 10000 || height >= 10000 || v >= 1000000000 {
            i |= 1;
        }

        if mass >= 100 {
            i |= 2;
        }

        let d = vec!["Neither", "Bulky", "Heavy", "Both"];
        d[i].to_string()
    }
}

Solution 2

Thinking

Solution 1 packs the two flags into an index. The same four cases can be written as sequential predicates — both, bulky only, heavy only, neither — with identical results.

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class Solution:
    def categorizeBox(self, length: int, width: int, height: int, mass: int) -> str:
        v = length * width * height
        bulky = any(x >= 10000 for x in (length, width, height)) or v >= 10**9
        heavy = mass >= 100

        if bulky and heavy:
            return "Both"
        if bulky:
            return "Bulky"
        if heavy:
            return "Heavy"

        return "Neither"

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class Solution {
    public String categorizeBox(int length, int width, int height, int mass) {
        long v = (long) length * width * height;
        boolean bulky = length >= 1e4 || width >= 1e4 || height >= 1e4 || v >= 1e9;
        boolean heavy = mass >= 100;

        if (bulky && heavy) {
            return "Both";
        }
        if (bulky) {
            return "Bulky";
        }
        if (heavy) {
            return "Heavy";
        }

        return "Neither";
    }
}

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class Solution {
public:
    string categorizeBox(int length, int width, int height, int mass) {
        long v = (long) length * width * height;
        bool bulky = length >= 1e4 || width >= 1e4 || height >= 1e4 || v >= 1e9;
        bool heavy = mass >= 100;

        if (bulky && heavy) {
            return "Both";
        }
        if (bulky) {
            return "Bulky";
        }
        if (heavy) {
            return "Heavy";
        }

        return "Neither";
    }
};

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func categorizeBox(length int, width int, height int, mass int) string {
    v := length * width * height
    bulky := length >= 1e4 || width >= 1e4 || height >= 1e4 || v >= 1e9
    heavy := mass >= 100
    if bulky && heavy {
        return "Both"
    }
    if bulky {
        return "Bulky"
    }
    if heavy {
        return "Heavy"
    }
    return "Neither"
}

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function categorizeBox(length: number, width: number, height: number, mass: number): string {
    const v = length * width * height;
    const bulky = length >= 1e4 || width >= 1e4 || height >= 1e4 || v >= 1e9;
    const heavy = mass >= 100;
    if (bulky && heavy) {
        return 'Both';
    }
    if (bulky) {
        return 'Bulky';
    }
    if (heavy) {
        return 'Heavy';
    }
    return 'Neither';
}

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impl Solution {
    pub fn categorize_box(length: i32, width: i32, height: i32, mass: i32) -> String {
        let v = length * width * height;
        let bulky = length >= 10000
            || width >= 10000
            || height >= 10000
            || (length as i64) * (width as i64) * (height as i64) >= 1000000000;

        let heavy = mass >= 100;

        if bulky && heavy {
            return "Both".to_string();
        }
        if bulky {
            return "Bulky".to_string();
        }
        if heavy {
            return "Heavy".to_string();
        }

        "Neither".to_string()
    }
}

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