关于时间复杂度O(n²)与O(n³)等价性及结果差异的技术问询

Hey there! Let's unpack this question step by step—big O notation can feel counterintuitive at first, but once you grasp what it actually represents, these relationships make perfect sense.

First, What Does Big O Mean? #

Big O notation defines an upper bound on the growth rate of an algorithm's runtime. In plain terms, it tells us the worst-case scenario for how quickly the algorithm's running time will increase as the input size n gets larger. If an algorithm is O(f(n)), that means there exists some constant C and some input size n₀ such that for all n ≥ n₀, the runtime is ≤ C*f(n).

Can an O(n²) Algorithm Be Called O(n³)? #

Absolutely yes. Here's why:

• For large enough n, n² is always smaller than n³ (e.g., when n > 1, n² < n³).
• We can pick a constant C = 1 and n₀ = 1, and the condition n² ≤ 1*n³ holds for all n ≥ 1.
• Since O(n³) describes a looser upper bound than O(n²), any algorithm that fits within the tighter O(n²) bound automatically fits the broader O(n³) bound.

Can an O(n³) Algorithm Be Called O(n²)? #

No, this doesn't hold. To satisfy O(n²), we'd need a constant C where n³ ≤ C*n² for all large n. But if we rearrange that inequality, we get n ≤ C—which can't be true as n grows infinitely large. There's no fixed constant that can keep up with the growth of n here, so O(n³) algorithms can't be categorized as O(n²).

Your Scenario: O(n³) vs O(n²) Results #

When your peer says both O(n³) and O(n²) are valid for the same algorithm, they're technically correct—but with an important caveat:

• Their O(n²) result is a tighter, more precise upper bound that better reflects the algorithm's actual performance.
• Your O(n³) result is a valid but looser upper bound. It's not wrong, but it's less informative because it overestimates the algorithm's growth rate.

Think of it like this: if you know a car's top speed is 100 mph, saying it can't go faster than 150 mph is true, but the 100 mph figure is way more useful. That's exactly what's happening here with O(n²) vs O(n³).

内容的提问来源于stack exchange，提问作者Dana