Time complexities of three algorithms are given. Which should execute the slowest for large values of N?Select one:O(n^2)O(2N)O(log N)O(N)

1. ### Break Down the Problem

We need to compare the time complexities of four algorithms:

1. $O(n^2)$
2. $O(2^N)$
3. $O(\log N)$
4. $O(N)$

The goal is to determine which time complexity results in the slowest execution time as $N$ becomes large.

2. ### Relevant Concepts

• Big O Notation: It describes the upper limit of time complexity, helping us understand how algorithms will scale with increasing input sizes.
• When comparing time complexities, we look at how each function grows as $N$ approaches infinity.

3. ### Analysis and Detail

• $O(n^2)$ grows quadratically, so it will be slower than linear time complexity for larger values of $N$.
• $O(2^N)$ grows exponentially, which means that even for relatively small values of $N$, this complexity rapidly becomes larger than polynomial or linear complexities.
• $O(\log N)$ grows very slowly; logarithmic time complexity is one of the most efficient complexities.
• $O(N)$ grows linearly and is generally slower than logarithmic but faster than both quadratic and exponential complexities.

4. ### Verify and Summarize

Among the complexities:

• Slowest Growth: $O(\log N)$
• Faster than $O(\log N)$: $O(N)$
• Slower than $O(N)$: $O(n^2)$
• Slowest Overall: $O(2^N)$

Final Answer

The algorithm with time complexity $O(2^N)$ should execute the slowest for large values of $N$.