String Path Compression

Problem Overview

You are given a path-like string where major parts are separated by forward slashes (/) and each major part contains minor parts separated by dots (.). Implement a two-stage compression:

1. The first stage of compression involves replacing each minor part that appears more than once in a major part with a number that represents the number of times that minor part has appeared in the major part up to that point. If a minor part appears only once, it remains unchanged.
2. The second stage of compression involves replacing consecutive sequences of minor parts that are the same with a single instance of that minor part followed by the number of times it was repeated.

Constraints

• The input string will contain only lowercase letters, forward slashes (/), and dots (.).
• The length of the input string will not exceed 1000 characters.

Examples

Example 1

Input: "a.b/c.a.b.a" Output: "a.b/c.a2.b.a"

Explanation:

• In the first stage, a appears twice in the first major part, so it is replaced by a2. The output is "a2.b/c.a.b.a".
• In the second stage, a appears twice in the second major part, so it is replaced by a2. The output is "a2.b/c.a2.b.a".

Example 2

Input: "a.aa.aaa/aaaa" Output: "a3.aa3/aaaa"

Explanation:

• In the first stage, a appears three times in the first major part, so it is replaced by a3. The output is "a3.aa.aaa/aaaa".
• In the second stage, a appears three times in the first minor part of the first major part, so it is replaced by a3. The output is "a3.aa3/aaaa".

Example 3

Input: "a.b/c.a.b.a" Output: "a.b/c.a2.b.a"

Explanation:

• In the first stage, a appears twice in the second major part, so it is replaced by a2. The output is "a.b/c.a2.b.a".
• In the second stage, there are no consecutive sequences of the same minor part, so the output remains "a.b/c.a2.b.a".

Hints

• Use a dictionary to keep track of the counts of minor parts within each major part.
• Iterate through each major part and apply the first stage of compression.
• After the first stage, iterate through each major part again and apply the second stage of compression.

Solution

`python def compress_path(path): parts = path.split('/') compressed_parts = []

for part in parts:
    minor_counts = {}
    compressed_minors = []
    current_count = 1

    for minor in part.split('.'):
        if minor in minor_counts:
            minor_counts[minor] += 1
            current_count += 1
        else:
            if current_count > 1:
                compressed_minors.append(f"{compressed_minors[-1]}{current_count}")
            minor_counts[minor] = 1
            current_count = 1
            compressed_minors.append(minor)

    if current_count > 1:
        compressed_minors.append(f"{compressed_minors[-1]}{current_count}")
    
    compressed_parts.append('.'.join(compressed_minors))

return '/'.join(compressed_parts)

Test cases

print(compress_path("a.b/c.a.b.a")) # Output: "a.b/c.a2.b.a" print(compress_path("a.aa.aaa/aaaa")) # Output: "a3.aa3/aaaa" print(compress_path("a.b/c.a.b.a")) # Output: "a.b/c.a2.b.a" `

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