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Making File Names Unique

Given an array of strings names of size n. You will create n folders in your file system such that, at the ith minute, you will create a folder with the name names[i].

Since two files cannot have the same name, if you enter a folder name that was previously used, the system will have a suffix addition to its name in the form of (k), where, k is the smallest positive integer such that the obtained name remains unique.

Return an array of strings of length n where ans[i] is the actual name the system will assign to the ith folder when you create it.

 

Example 1:

Input: names = ["pes","fifa","gta","pes(2019)"]
Output: ["pes","fifa","gta","pes(2019)"]
Explanation: Let's see how the file system creates folder names:
"pes" --> not assigned before, remains "pes"
"fifa" --> not assigned before, remains "fifa"
"gta" --> not assigned before, remains "gta"
"pes(2019)" --> not assigned before, remains "pes(2019)"

Example 2:

Input: names = ["gta","gta(1)","gta","avalon"]
Output: ["gta","gta(1)","gta(2)","avalon"]
Explanation: Let's see how the file system creates folder names:
"gta" --> not assigned before, remains "gta"
"gta(1)" --> not assigned before, remains "gta(1)"
"gta" --> the name is reserved, system adds (k), since "gta(1)" is also reserved, systems put k = 2. it becomes "gta(2)"
"avalon" --> not assigned before, remains "avalon"

Example 3:

Input: names = ["onepiece","onepiece(1)","onepiece(2)","onepiece(3)","onepiece"]
Output: ["onepiece","onepiece(1)","onepiece(2)","onepiece(3)","onepiece(4)"]
Explanation: When the last folder is created, the smallest positive valid k is 4, and it becomes "onepiece(4)".

 

Constraints:

  • 1 <= names.length <= 5 * 104
  • 1 <= names[i].length <= 20
  • names[i] consists of lowercase English letters, digits, and/or round brackets.

Solution Explanation

This problem requires creating unique folder names, adding suffixes like (k) when a name already exists. The solutions use a hash map (dictionary in Python) to efficiently track used names and their suffix counts.

Approach:

  1. Initialization: A hash map (e.g., d in the code) stores folder names as keys and their suffix counts (or 1 if not suffixed) as values.

  2. Iteration: The code iterates through the input names array.

  3. Uniqueness Check: For each name, it checks if the name already exists in the hash map.

  4. Suffix Generation: If a name collision occurs:

    • The current suffix count (k) from the hash map is retrieved.
    • A while loop finds the smallest available k such that name(k) isn't already used.
    • The name is updated with the new suffix (k).
    • The suffix count k+1 is stored back in the hash map.

  5. Hash Map Update: After processing each name (whether it was modified or not), its updated name is added/updated in the hash map with a count of 1 (or updated count if it already existed).

  6. Return: The modified names array is returned.

Time Complexity: O(N*M), where N is the length of the input array names, and M is the maximum length of a name string. In the worst-case scenario, many name collisions could occur, requiring multiple iterations within the while loop to find a unique suffix. However, the average-case complexity is much closer to O(N) because collisions are relatively rare.

Space Complexity: O(N), as the hash map could potentially store all the unique names generated. In the worst-case scenario (all names are the same), the hashmap would only store the original name and its updated suffix versions.

Example Walkthrough (Java):

Let's trace Example 2: names = ["gta","gta(1)","gta","avalon"]

  1. d is initialized as an empty HashMap.
  2. "gta": d doesn't contain "gta", so it's added: d = {"gta": 1}.
  3. "gta(1)": d doesn't contain "gta(1)", so it's added: d = {"gta": 1, "gta(1)": 1}.
  4. "gta": d contains "gta", so k becomes 1. The while loop checks for "gta(1)", which exists, so k becomes 2. "gta(2)" is added: names[2] becomes "gta(2)", and d updates to d = {"gta": 2, "gta(1)": 1, "gta(2)": 1}.
  5. "avalon": d doesn't contain "avalon", so it's added: d = {"gta": 2, "gta(1)": 1, "gta(2)": 1, "avalon": 1}.
  6. The updated names array ["gta","gta(1)","gta(2)","avalon"] is returned.

The other languages (Python, C++, Go, TypeScript) follow the same logic, differing only in syntax and data structure implementation.