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 lengthn where ans[i] is the actual name the system will assign to the ith folder when you create it.

Examples

Example 1

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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

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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

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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

Method 1 – Hash Map for Name Counting

Intuition

To ensure each file name is unique, we use a hash map to track the next available suffix for each base name. When a duplicate is found, we increment the suffix until a unique name is found.

Approach

  1. Use a hash map to store the next available integer suffix for each name.
  2. For each name in the input:
    • If the name is not in the map, add it to the answer and set its suffix to 1.
    • If the name is already used, increment the suffix until a unique name is found, then add it to the answer and update the map.
  3. Return the answer array.

Code

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class Solution {
public:
    vector<string> getFolderNames(vector<string>& names) {
        unordered_map<string, int> mp;
        vector<string> ans;
        for (auto& name : names) {
            if (!mp.count(name)) {
                ans.push_back(name);
                mp[name] = 1;
            } else {
                int k = mp[name];
                string newName;
                while (true) {
                    newName = name + "(" + to_string(k) + ")";
                    if (!mp.count(newName)) break;
                    k++;
                }
                ans.push_back(newName);
                mp[name] = k + 1;
                mp[newName] = 1;
            }
        }
        return ans;
    }
};
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func getFolderNames(names []string) []string {
    mp := make(map[string]int)
    ans := make([]string, 0, len(names))
    for _, name := range names {
        if _, ok := mp[name]; !ok {
            ans = append(ans, name)
            mp[name] = 1
        } else {
            k := mp[name]
            var newName string
            for {
                newName = name + "(" + strconv.Itoa(k) + ")"
                if _, ok := mp[newName]; !ok {
                    break
                }
                k++
            }
            ans = append(ans, newName)
            mp[name] = k + 1
            mp[newName] = 1
        }
    }
    return ans
}
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class Solution {
    public String[] getFolderNames(String[] names) {
        Map<String, Integer> mp = new HashMap<>();
        String[] ans = new String[names.length];
        for (int i = 0; i < names.length; i++) {
            String name = names[i];
            if (!mp.containsKey(name)) {
                ans[i] = name;
                mp.put(name, 1);
            } else {
                int k = mp.get(name);
                String newName;
                while (true) {
                    newName = name + "(" + k + ")";
                    if (!mp.containsKey(newName)) break;
                    k++;
                }
                ans[i] = newName;
                mp.put(name, k + 1);
                mp.put(newName, 1);
            }
        }
        return ans;
    }
}
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class Solution {
    fun getFolderNames(names: Array<String>): Array<String> {
        val mp = mutableMapOf<String, Int>()
        val ans = Array(names.size) { "" }
        for (i in names.indices) {
            val name = names[i]
            if (name !in mp) {
                ans[i] = name
                mp[name] = 1
            } else {
                var k = mp[name]!!
                var newName: String
                while (true) {
                    newName = "$name($k)"
                    if (newName !in mp) break
                    k++
                }
                ans[i] = newName
                mp[name] = k + 1
                mp[newName] = 1
            }
        }
        return ans
    }
}
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class Solution:
    def getFolderNames(self, names: list[str]) -> list[str]:
        mp: dict[str, int] = {}
        ans: list[str] = []
        for name in names:
            if name not in mp:
                ans.append(name)
                mp[name] = 1
            else:
                k = mp[name]
                while True:
                    new_name = f"{name}({k})"
                    if new_name not in mp:
                        break
                    k += 1
                ans.append(new_name)
                mp[name] = k + 1
                mp[new_name] = 1
        return ans
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impl Solution {
    pub fn get_folder_names(names: Vec<String>) -> Vec<String> {
        use std::collections::HashMap;
        let mut mp = HashMap::new();
        let mut ans = Vec::with_capacity(names.len());
        for name in names.iter() {
            if !mp.contains_key(name) {
                ans.push(name.clone());
                mp.insert(name.clone(), 1);
            } else {
                let mut k = *mp.get(name).unwrap();
                loop {
                    let new_name = format!("{}({})", name, k);
                    if !mp.contains_key(&new_name) {
                        ans.push(new_name.clone());
                        mp.insert(name.clone(), k + 1);
                        mp.insert(new_name, 1);
                        break;
                    }
                    k += 1;
                }
            }
        }
        ans
    }
}
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class Solution {
    getFolderNames(names: string[]): string[] {
        const mp = new Map<string, number>();
        const ans: string[] = [];
        for (const name of names) {
            if (!mp.has(name)) {
                ans.push(name);
                mp.set(name, 1);
            } else {
                let k = mp.get(name)!;
                let newName = '';
                while (true) {
                    newName = `${name}(${k})`;
                    if (!mp.has(newName)) break;
                    k++;
                }
                ans.push(newName);
                mp.set(name, k + 1);
                mp.set(newName, 1);
            }
        }
        return ans;
    }
}

Complexity

  • ⏰ Time complexity: O(n * m), where n is the number of names and m is the average length of the names, since each name and its suffixes are checked for uniqueness.
  • 🧺 Space complexity: O(n * m), for storing all unique names in the hash map.