To determine the best possible upper bound for $T(n)$ using the given recurrence relation

$T(n) \leq 2T\left(\left\lfloor \frac{n}{2} \right\rfloor\right) + cn,$

we analyze it with the Master Theorem for divide and conquer recurrences.

General Form of the Master Theorem:

If a recurrence is of the form

$T(n) = aT\left(\frac{n}{b}\right) + f(n),$

where:

• $a \geq 1$ is the number of subproblems,
• $b > 1$ is the factor by which the size of the problem is divided,
• $f(n)$ is the cost of the work done outside the recursive calls,

then we compare $n^{\log_b a}$ with $f(n)$ to determine $T(n)$:

1. Case 1: If $f(n) \in O(n^{\log_b a - \epsilon})$ for some $\epsilon > 0$, then $T(n) \in \Theta(n^{\log_b a})$.
2. Case 2: If $f(n) \in \Theta(n^{\log_b a})$, then $T(n) \in \Theta(n^{\log_b a} \log n)$.
3. Case 3: If $f(n) \in \Omega(n^{\log_b a + \epsilon})$ for some $\epsilon > 0$, and if $af\left(\frac{n}{b}\right) \leq cf(n)$ for some $c < 1$ and sufficiently large $n$, then $T(n) \in \Theta(f(n))$.

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Applying Master Theorem to the Given Recurrence:

From $T(n) \leq 2T\left(\left\lfloor \frac{n}{2} \right\rfloor\right) + cn$, we identify:

• $a = 2$ (two subproblems),
• $b = 2$ (problem is divided by $2$),
• $f(n) = cn$ (work done outside the recursive calls).

Step 1: Compute $\log_b a = \log_2 2 = 1$.

• The term $n^{\log_b a} = n^1$ is comparable to $f(n) = cn$.

Step 2: Compare $f(n)$ with $n^{\log_b a}$:

• Since $f(n) \in \Theta(n^1)$, this matches Case 2 of the Master Theorem.

Step 3: Conclusion:

For Case 2, the solution to the recurrence is

$T(n) \in \Theta(n \log n).$

Thus, the best possible upper bound for $T(n)$ using Big-O notation is

$T(n) \in O(n \log n).$

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Summary:

The divide-and-conquer algorithm has a time complexity of $O(n \log n)$.

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Do you want further clarifications or deeper insights into divide-and-conquer algorithms? Here are five related questions for you:

1. What are examples of problems solved by divide-and-conquer algorithms with $O(n \log n)$ complexity?
2. How would the solution change if $f(n)$ were $O(n^2)$ instead of $O(n)$?
3. Could the recurrence be solved using an alternative method, like the recursion tree method?
4. How does the Master Theorem handle floors/ceilings in recurrences?
5. What would happen if the constant $c$ were smaller or dependent on $n$?

Tip: When solving recurrences, always compare $n^{\log_b a}$ and $f(n)$ explicitly to ensure you're applying the correct case of the Master Theorem!