What is the time complexity of the following recursive function?

int Dosomething (int n) {

Time left

ut of

if(n≤2)

return 1;

else

return (Dosomething (floor(sqrt)) + n);

}

Let's analyze the time complexity of the given recursive function:

```cpp
int Dosomething(int n) {
    if (n <= 2)
        return 1;
    else
        return (Dosomething(floor(sqrt(n))) + n);
}
```

### Step 1: Understand the Recurrence Relation
The function `Dosomething` operates as follows:
- If `n <= 2`, the function returns `1`.
- Otherwise, it makes a recursive call to `Dosomething(floor(sqrt(n)))` and adds `n` to the result.

This suggests a recurrence relation for the time complexity `T(n)`:

\[
T(n) = T(\sqrt{n}) + n, \quad \text{for } n > 2
\]
\[
T(n) = 1, \quad \text{for } n \leq 2
\]

### Step 2: Solve the Recurrence
To solve this recurrence, let's substitute the value of `n` recursively.

\[
T(n) = n + T(\sqrt{n})
\]
\[
T(n) = n + \sqrt{n} + T(\sqrt{\sqrt{n}})
\]
\[
T(n) = n + \sqrt{n} + \sqrt{\sqrt{n}} + T(\sqrt{\sqrt{\sqrt{n}}}) + \dots
\]

This pattern continues until `n` reduces to 2 or less. Let’s find the number of steps required for `n` to reduce to 2.

### Step 3: Determine the Number of Recursive Calls
In each step, `n` is replaced by its square root. If we let `k` be the number of recursive calls required to reach a base case:

\[
\sqrt{k \text{ times}}{n} = 2
\]

Taking logarithms:

\[
n^{\left(\frac{1}{2}\right)^k} = 2
\]

\[
\left(\frac{1}{2}\right)^k \cdot \log n = \log 2
\]

\[
k = \log(\log n)
\]

### Step 4: Sum the Series
The time complexity `T(n)` involves summing the following series:

\[
T(n) = n + \sqrt{n} + \sqrt{\sqrt{n}} + \dots + 1
\]

This is a geometric series with decreasing terms. The sum of this series is dominated by the first term, `n`. The number of terms is `O(\log(\log n))`, so the series sum can be approximated by `O(n)`.

### Conclusion
Thus, the overall time complexity of the function `Dosomething(int n)` is:

\[
\boxed{O(n)}
\]

This means the time complexity grows linearly with the input size `n`.

---

Would you like to explore how this analysis works or have any further questions?

Here are some related questions:
1. How can you solve other types of recurrence relations similar to this one?
2. What happens if `Dosomething` was called with `n/2` instead of `sqrt(n)`?
3. Can you provide an example of a function with logarithmic time complexity?
4. How does the complexity change if the base condition was different?
5. What is the impact of recursive depth on time complexity?

**Tip:** When analyzing recursive functions, always try to express the recurrence relation clearly and identify how the input size reduces with each call. This is crucial for determining the time complexity.

Explore the time complexity analysis of the recursive function Dosomething(int n). Understand how the function's time complexity is determined using recurrence relations and recursive depth analysis. Suitable for advanced high school and college students interested in algorithms and computational complexity.

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