1487. Making File Names Unique 

Description

Given an array of strings names of size n. You will create n folders in your file system such that, at the i^th 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 i^th 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.

Solutions

Solution 1: Hash Table

We can use a hash table $d$ to record the minimum available index for each folder name, where $d[name] = k$ means the minimum available index for the folder $name$ is $k$. Initially, $d$ is empty since there are no folders.

Next, we iterate through the folder names array. For each file name $name$:

If $name$ is already in $d$, it means the folder $name$ already exists, and we need to find a new folder name. We can keep trying $name(k)$, where $k$ starts from $d[name]$, until we find a folder name $name(k)$ that does not exist in $d$. We add $name(k)$ to $d$, update $d[name]$ to $k + 1$, and then update $name$ to $name(k)$.
If $name$ is not in $d$, we can directly add $name$ to $d$ and set $d[name]$ to $1$.
Then, we add $name$ to the answer array and continue to the next file name.

After traversing all file names, we obtain the answer array.

In the code implementation below, we directly modify the $names$ array without using an extra answer array.

The complexity is $O(L)$, and the space complexity is $O(L)$, where $L$ is the sum of the lengths of all file names in the $names$ array.

Python3

class Solution:
    def getFolderNames(self, names: List[str]) -> List[str]:
        d = defaultdict(int)
        for i, name in enumerate(names):
            if name in d:
                k = d[name]
                while f'{name}({k})' in d:
                    k += 1
                d[name] = k + 1
                names[i] = f'{name}({k})'
            d[names[i]] = 1
        return names

Java

class Solution {
    public String[] getFolderNames(String[] names) {
        Map<String, Integer> d = new HashMap<>();
        for (int i = 0; i < names.length; ++i) {
            if (d.containsKey(names[i])) {
                int k = d.get(names[i]);
                while (d.containsKey(names[i] + "(" + k + ")")) {
                    ++k;
                }
                d.put(names[i], k);
                names[i] += "(" + k + ")";
            }
            d.put(names[i], 1);
        }
        return names;
    }
}

C++

class Solution {
public:
    vector<string> getFolderNames(vector<string>& names) {
        unordered_map<string, int> d;
        for (auto& name : names) {
            int k = d[name];
            if (k) {
                while (d[name + "(" + to_string(k) + ")"]) {
                    k++;
                }
                d[name] = k;
                name += "(" + to_string(k) + ")";
            }
            d[name] = 1;
        }
        return names;
    }
};

Go

func getFolderNames(names []string) []string {
 d := map[string]int{}
 for i, name := range names {
  if k, ok := d[name]; ok {
   for {
    newName := fmt.Sprintf("%s(%d)", name, k)
    if d[newName] == 0 {
     d[name] = k + 1
     names[i] = newName
     break
    }
    k++
   }
  }
  d[names[i]] = 1
 }
 return names
}

TypeScript

function getFolderNames(names: string[]): string[] {
    let d: Map<string, number> = new Map();
    for (let i = 0; i < names.length; ++i) {
        if (d.has(names[i])) {
            let k: number = d.get(names[i]) || 0;
            while (d.has(names[i] + '(' + k + ')')) {
                ++k;
            }
            d.set(names[i], k);
            names[i] += '(' + k + ')';
        }
        d.set(names[i], 1);
    }
    return names;
}