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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 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.

Solutions

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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
  
  

• 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
  }
  

• 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;
  }
  
  

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