How does the time complexity O(n) compare to O(n^2)? a) O(n) is faster for large n b) O(n^2) is faster for all n c) O(n) and O(n^2) are always equivalent d) O(n) is faster for smal... How does the time complexity O(n) compare to O(n^2)? a) O(n) is faster for large n b) O(n^2) is faster for all n c) O(n) and O(n^2) are always equivalent d) O(n) is faster for small n, but O(n^2) becomes faster as n grows.
Understand the Problem
The question is asking for a comparison between two different time complexity notations, O(n) and O(n^2), in terms of their performance as n (the size of the input) grows. The options provided suggest different possibilities regarding their speed for varying sizes of n.
Answer
O(n) is faster for large n.
The final answer is O(n) is faster for large n.
Answer for screen readers
The final answer is O(n) is faster for large n.
More Information
Time complexity O(n) is linear, meaning the execution time grows proportionally with input size. O(n^2) is quadratic, meaning the execution time grows quadratically. As n increases, O(n) will outperform O(n^2) because quadratic growth is faster than linear.
Tips
A common mistake is not considering the scale of n. For smaller values, the difference in time isn't as noticeable, but for larger n, linear complexity always becomes more efficient than quadratic.
Sources
- Is O(2^n) better than O(n^2)? - Quora - quora.com
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