For large values of $n$, Algorithm A (which runs in $O(n \log n)$) is more efficient than Algorithm B (which runs in $O(n^2)$).

Justification:

• Algorithm A's time complexity: $O(n \log n)$:
  The function $n \log n$ grows more slowly than $n^2$ as $n$ increases. The logarithmic function $\log n$ grows very slowly compared to polynomial terms like $n$. Hence, $n \log n$ will always be smaller than $n^2$ for sufficiently large $n$.

• Algorithm B's time complexity: $O(n^2)$:
  The function $n^2$ grows quadratically, which means it increases much faster than a function like $n \log n$ as $n$ becomes large. Quadratic growth is significantly faster compared to logarithmic growth multiplied by a linear factor.

Growth Rate Comparison:

• As $n \to \infty$, the term $n^2$ will vastly outgrow $n \log n$.

  For example, let's compare both functions for large $n$:

  • For $n = 1,000$:

    • $n \log n \approx 1,000 \times 10 = 10,000$
    • $n^2 = 1,000^2 = 1,000,000$

  • For $n = 1,000,000$:

    • $n \log n \approx 1,000,000 \times 20 = 20,000,000$
    • $n^2 = 1,000,000^2 = 1,000,000,000,000$

As $n$ becomes larger, the gap between $n^2$ and $n \log n$ increases even more. Thus, Algorithm A with $O(n \log n)$ is more efficient for large $n$.

Would you like more details or have any questions?

Related questions:

1. Can you provide a comparison for small values of $n$ between the two algorithms?
2. How do logarithmic functions grow compared to polynomial functions in general?
3. What would happen if Algorithm A had a higher constant factor than Algorithm B?
4. Can we visualize the growth of these two functions on a graph?
5. Are there real-world examples where $O(n \log n)$ algorithms are used instead of $O(n^2)$?

Tip: When comparing algorithm efficiency, focus on the growth rates for large $n$, as constant factors and small values of $n$ are less significant in asymptotic analysis.