Given a positive number **S**, find the largest number **n** such that the sum
`1`

is less than or equal to ^{2} + 2^{2} + 3^{2} + ... + n^{2}**S**.

For your information, `n`

^{2} = n * n

**Example:**

Input:S= 35Output:4Explanation:1^{2}+ 2^{2}+ 3^{2}+ 4^{2}= 1 + 4 + 9 + 16 = 30 and 30 <= 35 if we add 5^{2}we end up with 55, which exceeds 35 therefore4is the answer

The core challenge of this problem is to find the largest integer **n** such that the sum of squares from 1 to **n** does not exceed a given number **S**. This problem is significant in various mathematical and computational applications, such as optimization problems and resource allocation.

Potential pitfalls include misunderstanding the sum of squares and not correctly iterating through the numbers to find the largest **n**.

To solve this problem, we can start with a simple iterative approach:

- Initialize a variable to keep track of the total sum of squares.
- Iterate through numbers starting from 1, adding the square of each number to the total sum.
- Stop when adding the next square would exceed
**S**.

This naive approach is straightforward but can be optimized further. However, for this problem, the iterative approach is efficient enough given the constraints.

Here is a step-by-step breakdown of the algorithm:

- Initialize
`totalSum`

to 0 and`n`

to 0. - While the total sum plus the square of the next number is less than or equal to
**S**, increment`n`

and add the square of`n`

to`totalSum`

. - Return
`n`

as the result.

```
public class LargestNumber {
public static int findLargestNumber(int S) {
int totalSum = 0;
int n = 0;
// Iterate while the sum of squares is less than or equal to S
while (totalSum + (n + 1) * (n + 1) <= S) {
n++;
totalSum += n * n;
}
return n;
}
public static void main(String[] args) {
int S = 35;
System.out.println("The largest number n such that the sum of squares is <= " + S + " is: " + findLargestNumber(S));
}
}
```

The time complexity of this approach is **O(n)**, where **n** is the largest number such that the sum of squares is less than or equal to **S**. The space complexity is **O(1)** as we are using a constant amount of extra space.

Potential edge cases include:

**S**being very small (e.g., 1 or 2).**S**being a perfect square sum (e.g., 14, which is 1^{2}+ 2^{2}+ 3^{2}).

These cases are handled by the algorithm as it iterates and checks the sum condition.

To test the solution comprehensively, consider the following test cases:

Input:S = 1Output:1Input:S = 14Output:3Input:S = 50Output:5

Use a testing framework like JUnit to automate and validate these test cases.

When approaching such problems:

- Break down the problem into smaller parts.
- Think about the simplest solution first and then optimize.
- Practice similar problems to improve problem-solving skills.

Understanding and solving this problem helps in developing a strong foundation in iterative algorithms and optimization techniques. Practice and exploration of similar problems can further enhance problem-solving skills.

For further reading and practice: