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2525. Categorize Box According to Criteria
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
10^{4}
.  Or, the volume of the box is greater or equal to
10^{9}
.
 Any of the dimensions of the box is greater or equal to
 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 10^{4}. Its volume = 24500000 <= 10^{9}. 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 10^{4}. Its volume = 8 * 10^{6} <= 10^{9}. 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 <= 10^{5}
1 <= mass <= 10^{3}
Solutions
Solution 1: Simulation
We can simulate according to the problem description.
The time complexity is $O(1)$, and the space complexity is $O(1)$.

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]; } }

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]; } };

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]

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] }

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]; }

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() } }