Problem Description

You need to create n folders in a file system, where you're given an array of strings names containing the desired folder names. The folders are created one by one in order.

The key constraint is that no two folders can have the same name. When you try to create a folder with a name that already exists, the system automatically appends a suffix in the format (k) to make it unique, where k is the smallest positive integer that results in a unique name.

For example:

• If you try to create a folder named "doc" but it already exists, the system will name it "doc(1)"
• If "doc" and "doc(1)" both exist, the next "doc" would become "doc(2)"
• The suffix always uses the smallest available number that makes the name unique

Your task is to return an array showing the actual names assigned by the system to each folder in the order they were created.

The solution uses a hash table to track:

• Which folder names have been used
• For repeated names, what's the next available suffix number to try

The algorithm works by:

1. Maintaining a dictionary d where d[name] stores the next suffix number to try for that base name
2. For each folder name in the input:
   • If the name already exists, find the smallest k such that name(k) is available
   • Update the tracking information for future occurrences
   • Add the final name (with or without suffix) to the result

Intuition

The core challenge is efficiently handling name collisions. When we encounter a duplicate name, we need to find the smallest available suffix number quickly.

The naive approach would be to check name(1), name(2), name(3), and so on until we find an unused name. However, this could be inefficient if we have many duplicates of the same base name.

The key insight is that we need to track two pieces of information:

1. Which exact folder names have been used (including those with suffixes)
2. For each base name, what's the next suffix number we should try

Why do we need both? Consider this scenario: we create "doc", then "doc(1)", then encounter "doc" again. Without tracking that "doc(1)" is already used, we might try to create it again. And without tracking the next available suffix for "doc", we'd have to start checking from (1) every time.

By maintaining d[name] = k, we're essentially keeping a "bookmark" that says: "The next time you see this base name, start checking from suffix k." This optimization prevents us from rechecking smaller numbers that we know are already taken.

When we successfully use a suffix k for a name, we update our bookmark to k + 1 because we know all suffixes up to k are now taken. This way, future occurrences of the same base name can start their search from a more efficient position.

The hash table serves a dual purpose: it tells us whether a name exists (for collision detection) and helps us track the next available suffix (for efficiency). This combination allows us to handle the folder naming process in an optimal manner.

Solution Approach

We use a hash table d to track folder names and their next available suffix indices. The hash table serves dual purposes: checking if a name exists and storing the minimum available suffix for each base name.

Algorithm Steps:

1. Initialize an empty hash table d = defaultdict(int) to store folder names and their suffix tracking.

2. Iterate through each name in the input array names:

   a. Check if the name already exists in d:

   • If name in d, it means this folder name was already used
   • Retrieve the starting suffix number: k = d[name]
   • Keep incrementing k while f'{name}({k})' exists in d
   • Once we find an available suffix, update d[name] = k + 1 for future occurrences
   • Modify the current name to include the suffix: names[i] = f'{name}({k})'

   b. Add the final name to the hash table:

   • Set d[names[i]] = 1 to mark this exact name (with or without suffix) as used
   • The value 1 here indicates that if this exact string appears again, we should start checking from suffix (1)

3. Return the modified array names which now contains all the actual folder names assigned by the system.

Key Implementation Details:

• We modify the input array in-place rather than creating a new array
• The defaultdict(int) returns 0 for missing keys, but we use it to store the next suffix to try
• When a base name like "doc" gets a suffix to become "doc(2)", we store both:
  • d["doc"] = 3 (next suffix to try for base name "doc")
  • d["doc(2)"] = 1 (marking "doc(2)" as used)

This approach ensures O(1) average-case lookup for checking name existence and maintains optimal suffix tracking for repeated base names.

Example Walkthrough

Let's walk through the algorithm with the input: names = ["doc", "doc", "image", "doc(1)", "doc"]

Initial state: d = {} (empty hash table), result = []

Step 1: Process "doc"

• Check if "doc" exists in d: No, it doesn't
• Add to hash table: d["doc"] = 1
• Result: ["doc"]
• Hash table: {"doc": 1}

Step 2: Process second "doc"

• Check if "doc" exists in d: Yes, it does!
• Get starting suffix: k = d["doc"] = 1
• Check if "doc(1)" exists: No, it's available
• Update: d["doc"] = 2 (next time start from 2)
• Modify name to "doc(1)"
• Add to hash table: d["doc(1)"] = 1
• Result: ["doc", "doc(1)"]
• Hash table: {"doc": 2, "doc(1)": 1}

Step 3: Process "image"

• Check if "image" exists in d: No, it doesn't
• Add to hash table: d["image"] = 1
• Result: ["doc", "doc(1)", "image"]
• Hash table: {"doc": 2, "doc(1)": 1, "image": 1}

Step 4: Process "doc(1)"

• Check if "doc(1)" exists in d: Yes, it does!
• Get starting suffix: k = d["doc(1)"] = 1
• Check if "doc(1)(1)" exists: No, it's available
• Update: d["doc(1)"] = 2
• Modify name to "doc(1)(1)"
• Add to hash table: d["doc(1)(1)"] = 1
• Result: ["doc", "doc(1)", "image", "doc(1)(1)"]
• Hash table: {"doc": 2, "doc(1)": 2, "image": 1, "doc(1)(1)": 1}

Step 5: Process third "doc"

• Check if "doc" exists in d: Yes, it does!
• Get starting suffix: k = d["doc"] = 2
• Check if "doc(2)" exists: No, it's available
• Update: d["doc"] = 3
• Modify name to "doc(2)"
• Add to hash table: d["doc(2)"] = 1
• Final Result: ["doc", "doc(1)", "image", "doc(1)(1)", "doc(2)"]

The algorithm efficiently handles collisions by tracking the next available suffix for each base name, avoiding redundant checks of already-used suffixes.

Solution Implementation

Time and Space Complexity

The time complexity is O(L) where L is the sum of the lengths of all file names in the names array.

Breaking down the analysis:

• The outer loop iterates through each name once: O(n) iterations where n is the number of names
• For each name, we perform string operations:
  • Checking if a name exists in the dictionary: O(len(name))
  • Creating new strings with suffixes f'{name}({k})': O(len(name))
  • The while loop seems problematic at first glance, but it's amortized O(1) per name because:
    • Each suffix (k) is only generated once across all iterations
    • The counter d[name] tracks the next available suffix, avoiding re-checking previously used suffixes
    • Each unique string is added to the dictionary exactly once
• Dictionary operations (insertion and lookup) with string keys: O(len(key))

Since each character in the input is processed a constant number of times, the total time complexity is O(L).

The space complexity is O(L):

• The dictionary d stores all unique folder names (original and modified)
• In the worst case, all names are unique, so we store each name once
• The modified names array reuses the existing space (in-place modification)
• The total space used is proportional to the total length of all stored strings: O(L)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Not Tracking Both Base Name and Suffixed Names

The Problem: A critical mistake is only tracking the base name without also marking the suffixed versions as occupied. For example, if the input is ["doc", "doc(1)", "doc"], failing to track "doc(1)" as used would cause the third "doc" to incorrectly become "doc(1)" again, creating a duplicate.

Wrong Approach:

# INCORRECT - Only tracks base names
name_counter = {}
for i, name in enumerate(names):
    if name in name_counter:
        suffix = name_counter[name]
        names[i] = f'{name}({suffix})'
        name_counter[name] = suffix + 1
    else:
        name_counter[name] = 1

Correct Solution: The code must track BOTH the base name's next suffix AND mark each created name (including suffixed ones) as used:

name_counter[name] = suffix_num + 1  # Update next suffix for base name
name_counter[names[i]] = 1           # Mark the actual created name as used

Pitfall 2: Inefficient Suffix Search Without Tracking

The Problem: Starting the suffix search from 1 every time a duplicate is encountered leads to O(n²) complexity in worst cases. For input like ["a", "a", "a", ..., "a"] with n occurrences, you'd check suffixes 1, then 1-2, then 1-2-3, etc.

Wrong Approach:

# INEFFICIENT - Always starts from suffix 1
if name in name_counter:
    suffix = 1
    while f'{name}({suffix})' in name_counter:
        suffix += 1
    names[i] = f'{name}({suffix})'

Correct Solution: Store and use the next available suffix for each base name to avoid redundant checks:

suffix_num = name_counter[name]  # Start from the stored next suffix
while f'{name}({suffix_num})' in name_counter:
    suffix_num += 1
name_counter[name] = suffix_num + 1  # Update for next time

Pitfall 3: Confusion About Dictionary Values

The Problem: The dictionary values serve different purposes depending on the key type:

• For base names: stores the next suffix number to try
• For suffixed names: just marks them as used (value is typically 1)

This dual purpose can cause confusion and lead to incorrect implementations.

Example to Illustrate: After processing ["doc", "doc", "doc(2)"]:

• name_counter["doc"] = 3 (next suffix to try is 3)
• name_counter["doc(1)"] = 1 (just marking as used)
• name_counter["doc(2)"] = 1 (just marking as used)

Understanding this distinction is crucial for correct implementation.