1487. Making File Names Unique LeetCode Solution

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Problem Statement
Complexity Analysis
Making File Names Unique solution in C++
Making File Names Unique solution in Java
Making File Names Unique solution in Python
Additional Resources

1487. Making File Names Unique LeetCode Solution image

Problem Statement of 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.

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

1487. Making File Names Unique LeetCode Solution in C++

class Solution {
 public:
  vector<string> getFolderNames(vector<string>& names) {
    vector<string> ans;
    unordered_map<string, int> nameToSuffix;

    for (const string& name : names)
      if (const auto it = nameToSuffix.find(name); it != nameToSuffix.cend()) {
        int suffix = it->second;
        string newName = getName(name, ++suffix);
        while (nameToSuffix.contains(newName))
          newName = getName(name, ++suffix);
        nameToSuffix[name] = suffix;
        nameToSuffix[newName] = 0;
        ans.push_back(newName);
      } else {
        nameToSuffix[name] = 0;
        ans.push_back(name);
      }

    return ans;
  }

 private:
  string getName(const string& name, int suffix) {
    return name + "(" + to_string(suffix) + ")";
  }
};
/* code provided by PROGIEZ */

1487. Making File Names Unique LeetCode Solution in Java

class Solution {
  public String[] getFolderNames(String[] names) {
    String[] ans = new String[names.length];
    Map<String, Integer> nameToSuffix = new HashMap<>();

    for (int i = 0; i < names.length; ++i) {
      final String name = names[i];
      if (nameToSuffix.containsKey(name)) {
        int suffix = nameToSuffix.get(name);
        String newName = getName(name, ++suffix);
        while (nameToSuffix.containsKey(newName))
          newName = getName(name, ++suffix);
        nameToSuffix.put(name, suffix);
        nameToSuffix.put(newName, 0);
        ans[i] = newName;
      } else {
        nameToSuffix.put(name, 0);
        ans[i] = name;
      }
    }

    return ans;
  }

  private String getName(final String name, int suffix) {
    return name + "(" + String.valueOf(suffix) + ")";
  }
}
// code provided by PROGIEZ

1487. Making File Names Unique LeetCode Solution in Python

class Solution:
  def getFolderNames(self, names: list[str]) -> list[str]:
    ans = []
    nameToSuffix = {}

    for name in names:
      if name in nameToSuffix:
        suffix = nameToSuffix[name] + 1
        newName = self._getName(name, suffix)
        while newName in nameToSuffix:
          suffix += 1
          newName = self._getName(name, suffix)
        nameToSuffix[name] = suffix
        nameToSuffix[newName] = 0
        ans.append(newName)
      else:
        nameToSuffix[name] = 0
        ans.append(name)

    return ans

  def _getName(self, name: str, suffix: int) -> str:
    return name + '(' + str(suffix) + ')'
 # code by PROGIEZ