CF 104842C - C and Pascal Strings

Rating: -
Tags: -
Solve time: 37s
Verified: no

Solution

Problem Understanding

We are given a sequence of bytes, each written as a two-digit hexadecimal number, so each value lies in the range from 0 to 255. This sequence represents a raw memory dump, and the task is to decide whether this dump can be interpreted as a valid C-style string, a valid Pascal-style string, both, or neither.

A C-style string here is defined in a slightly simplified way. We scan the byte array from the start. We may see zero or more “text” bytes, where each text byte must lie in the printable ASCII range from 0x20 to 0x7f inclusive. After that, there must be a single zero byte that terminates the string. Everything after this zero byte is irrelevant junk and does not affect validity.

A Pascal-style string instead begins with a length byte l. The next l bytes must all be in the printable ASCII range from 0x20 to 0x7f inclusive. After those l bytes, the remainder of the array is junk and ignored. The length l is between 0 and 255, but it must also fit into the actual array size, meaning there must be at least l bytes after the length byte.

The key difference is that C strings search for a terminating zero, while Pascal strings explicitly encode their length at the start.

The input size is at most 1000 bytes, so any solution up to O(n^2) is already safe, but the structure is simple enough that a linear scan suffices. This strongly suggests we should avoid any brute-force attempts that try all possible split points without preprocessing.

A few subtle cases matter.

A common mistake is to assume that the first zero byte automatically defines a valid C string. That is incorrect because all preceding bytes must be printable. For example, in the input:

01 00

The zero is present, but the prefix byte 01 is not printable, so this is not a valid C string.

For Pascal strings, another subtle case is when the length byte exceeds the remaining number of bytes. For example:

05 41 42

Here l = 5, but only two bytes follow, so it is invalid even if they were printable.

Finally, l = 0 is valid in Pascal strings, meaning the string may have zero content bytes and immediately proceed to junk.

Approaches

A brute-force approach would try every possible position as a potential C terminator and every possible position as a Pascal length interpretation. For each candidate, it would validate the required constraints by scanning segments of the array. In the worst case, each check costs O(n), and there are O(n) choices for both structures, leading to O(n^2) time.

The key observation is that both structures depend only on simple prefix conditions. For C strings, we only need to know whether a prefix contains only printable bytes. For Pascal strings, we only need to validate a single fixed segment determined by the first byte.

This allows preprocessing a boolean array that tracks whether each prefix is fully printable. Once that is available, C validity reduces to checking positions of zero bytes. Pascal validity is a direct single check.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n^2)	O(1)	Too slow
Optimal Prefix Check	O(n)	O(n)	Accepted

Algorithm Walkthrough

We separate the validation into two independent checks, one for C-style interpretation and one for Pascal-style interpretation.

C-string check

Compute a prefix array where each position i indicates whether all bytes from 0 to i are in the printable range.
Scan the array from left to right and consider every position i where the byte is zero.
For each such position i, check whether all bytes before it are printable using the prefix array.
If at least one such position exists, the array can be interpreted as a valid C string.

The reasoning is that a valid C string is completely determined by choosing where the first zero appears, and the prefix constraint ensures no invalid byte occurs before it.

Pascal-string check

Read the first byte as l, the candidate length.
Check whether l is less than or equal to n minus 1, ensuring there are enough remaining bytes.
Verify that all bytes from index 1 to index l are in the printable range.
If both conditions hold, the array is a valid Pascal string.

This works because Pascal encoding fixes the structure rigidly at the start, so there is no ambiguity once l is chosen.

Final decision

Combine the two boolean results and output one of four cases: both valid, only C valid, only Pascal valid, or neither valid.

Why it works

The correctness comes from the fact that both interpretations impose constraints that are local and prefix based. The C condition depends only on whether a valid terminating