Big O Notation: Does f(n) = O(g(n)) Imply g(n) = O(f(n))?

Hey guys! Let’s dive into a common question that pops up when we’re talking about Big O notation: If f(n) = O(g(n)) , does that automatically mean g(n) = O(f(n)) ? The short answer is, usually, no. But let’s break it down so we really get why.

Understanding Big O Notation

First, let’s quickly recap what Big O notation is all about. Big O notation is a way to describe the upper bound of an algorithm’s growth rate as the input size ( n ) increases. Essentially, it tells us how the runtime or space requirements of an algorithm scale as the input gets larger and larger. It focuses on the worst-case scenario and ignores constant factors and lower-order terms.

When we say f(n) = O(g(n)) , it means that f(n) grows no faster than g(n) as n approaches infinity. More formally, there exist positive constants c and n₀ such that for all n > n₀ , f(n) <= c * g(n) . In simpler terms, after a certain point ( n₀ ), f(n) will always be less than or equal to some constant multiple ( c ) of g(n) . This doesn’t mean they grow at the same rate; it just means g(n) is at least as fast as f(n) . Think of it like saying “I weigh less than 2000 pounds.” It doesn’t mean I weigh exactly 2000 pounds, just that my weight doesn’t exceed that limit. It’s all about setting a limit from above.

Why f(n) = O(g(n)) Doesn’t Necessarily Imply g(n) = O(f(n))

Okay, now to the heart of the matter. The reason why f(n) = O(g(n)) doesn’t automatically mean g(n) = O(f(n)) is because Big O notation only gives us an upper bound . It tells us that g(n) grows at least as fast as f(n) , but it doesn’t tell us anything about whether f(n) grows at least as fast as g(n) . g(n) could grow much faster than f(n) . Imagine f(n) is a tiny snail crawling along, and g(n) is a rocket ship blasting off. The snail’s speed is certainly O of the rocket ship’s speed, but the rocket ship’s speed isn’t O of the snail’s speed!

Let’s look at some concrete examples to make this crystal clear:

• Example 1: Polynomial vs. Exponential

  • Let f(n) = n^2 (a polynomial function) and g(n) = 2^n (an exponential function).
  • We know that n^2 = O(2^n) because exponential functions grow much faster than polynomial functions. No matter how large n gets, 2^n will eventually overtake n^2 . Formally, you can always find a c and n₀ to satisfy the Big O definition.
  • However, 2^n is not O(n^2) . There’s no constant c that you can multiply n^2 by to always be greater than or equal to 2^n for all sufficiently large n . The exponential function will always win out in the end. It’s like saying a rocket ship is way faster than a bicycle.

• Example 2: Linear vs. Quadratic

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  • Let f(n) = n (a linear function) and g(n) = n^2 (a quadratic function).
  • Here, n = O(n^2) because n^2 grows faster than n . You can easily find constants c and n₀ to show this (e.g., c = 1 , n₀ = 1 ).
  • But n^2 is not O(n) . You can’t multiply n by a constant and have it always be greater than or equal to n^2 for large n . The quadratic function will eventually dominate. The gap between the two gets progressively bigger.

• Example 3: Constant vs. Logarithmic

  • Let f(n) = 1 (a constant function) and g(n) = log(n) (a logarithmic function).
  • 1 = O(log(n)) because the logarithm grows (albeit very slowly) as n increases. A constant value will always be bounded by a growing logarithmic function, eventually.
  • log(n) is not O(1) . A logarithm, however slowly, will always exceed a constant as n approaches infinity. There’s no constant you can pick that will stay above log(n) forever. Logarithmic functions, even though they grow incredibly slowly, do grow.

When is g(n) = O(f(n))?

So, when is it true that if f(n) = O(g(n)) , then g(n) = O(f(n)) ? This is true when f(n) and g(n) grow at the same rate . In other words, when f(n) = Θ(g(n)) . The Big Theta notation (Θ) describes a tight bound . It means that f(n) and g(n) are both upper and lower bounds of each other. There exist positive constants c1 , c2 , and n₀ such that for all n > n₀ , c1 * g(n) <= f(n) <= c2 * g(n) . This implies that f(n) and g(n) have the same growth rate, just potentially with different constant factors. This is the key!

For example:

• If f(n) = 2n and g(n) = n , then f(n) = O(g(n)) and g(n) = O(f(n)) . They both grow linearly, just with a different constant multiple. In this case, f(n) = Θ(g(n)) . Big Theta really shows these two grow at similar rates.
• If f(n) = 100n^2 + 5n and g(n) = n^2 , then f(n) = O(g(n)) and g(n) = O(f(n)) . Both are quadratic, and we ignore lower-order terms and constant factors in Big O. Here, f(n) = Θ(g(n)) . With Big Theta, the lower order terms just don’t matter.

Big O, Big Omega, and Big Theta

To summarize:

• Big O (O): Represents the upper bound or the worst-case scenario. f(n) = O(g(n)) means f(n) grows no faster than g(n) .
• Big Omega (Ω): Represents the lower bound or the best-case scenario. f(n) = Ω(g(n)) means f(n) grows at least as fast as g(n) . This is the opposite of Big O.
• Big Theta (Θ): Represents the tight bound . f(n) = Θ(g(n)) means f(n) and g(n) grow at the same rate. It’s both O and Ω.

Think of it like this:

• Big O: “At worst, my code will take this long.”
• Big Omega: “At best, my code will take this long.”
• Big Theta: “My code will always take roughly this long.”

Practical Implications

Understanding these nuances of Big O notation is crucial when analyzing and comparing algorithms. It helps you make informed decisions about which algorithm is most efficient for a given task. For instance, if you have two algorithms, one with O(n log n) and another with O(n^2) , you know that the O(n log n) algorithm will generally perform better for large input sizes, even though it might have a higher constant factor.

Also, it’s really useful when talking about code in interviews. Being able to explain the complexities and nuances of Big O can set you apart.

Conclusion

So, to reiterate, just because f(n) = O(g(n)) doesn’t automatically mean g(n) = O(f(n)) . It only means g(n) grows at least as fast as f(n) . The reverse is only true when f(n) and g(n) grow at the same rate , meaning f(n) = Θ(g(n)) . Keep these distinctions in mind when you’re analyzing algorithm performance. Happy coding, everyone!