Here's a freebie. They labeled it as "Hard"

This problem was asked by Google.

Suppose we represent our file system by a string in the following manner:

The string "dir\n\tsubdir1\n\tsubdir2\n\t\tfile.ext" represents:

dir
subdir1
subdir2
file.ext
The directory dir contains an empty sub-directory subdir1 and a sub-directory subdir2 containing a file file.ext.

The string "dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext" represents:

dir
subdir1
file1.ext
subsubdir1
subdir2
subsubdir2
file2.ext
The directory dir contains two sub-directories subdir1 and subdir2. subdir1 contains a file file1.ext and an empty second-level sub-directory subsubdir1. subdir2 contains a second-level sub-directory subsubdir2 containing a file file2.ext.

We are interested in finding the longest (number of characters) absolute path to a file within our file system. For example, in the second example above, the longest absolute path is "dir/subdir2/subsubdir2/file2.ext", and its length is 32 (not including the double quotes).

Given a string representing the file system in the above format, return the length of the longest absolute path to a file in the abstracted file system. If there is no file in the system, return 0.

Note:

The name of a file contains at least a period and an extension.

The name of a directory or sub-directory will not contain a period.

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Octave solution 

@Absinthe

Worst-case space is linear in the number of control characters (newlines plus tabs). I'm not entirely sure on time requirements of the regex but the rest of it is linear in total characters.

function path_length = max_path_length(stringfs)
indent = char(9);
newline = char(10);
if stringfs(1) ~= indent
indent_level = 0;
else
indent_level = find(stringfs ~= indent, 1) - 1; end
path_length = 0;
children_delims = regexp(stringfs, ['\n' repmat('\t', 1, indent_level) '[^\t]']);
for bounds = [1 children_delims + 1; children_delims - 1 numel(stringfs)]
if any(stringfs(bounds(1):bounds(2)) == newline)
cutoff = find(stringfs(bounds(1):bounds(2)) == newline, 1);
this_path_length = cutoff - indent_level + max_path_length(stringfs((bounds(1) + cutoff):bounds(2)));
elseif any(stringfs(bounds(1):bounds(2)) == '.')
this_path_length = bounds(2) - bounds(1) - indent_level + 1;
else
this_path_length = -Inf; end;
path_length = max(path_length, this_path_length); end; end;

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