1487. Making File Names Unique

Problem

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 * 10^4

• 1 <= names[i].length <= 20

• names[i] consists of lowercase English letters, digits, and/or round brackets.

Solution (Java)

class Solution {
    public String[] getFolderNames(String[] names) {
        HashMap<String, Integer> map = new HashMap<>();
        for (int i = 0; i < names.length; i++) {
            int prefix = map.getOrDefault(names[i], 0);
            if (prefix != 0) {
                String raw = names[i];
                while (map.getOrDefault(names[i], 0) != 0) {
                    names[i] = raw + "(" + prefix + ")";
                    prefix++;
                }
                map.put(raw, prefix);
            }
            map.put(names[i], 1);
        }
        return names;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).