Welcome to Subscribe On Youtube
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 * 1041 <= names[i].length <= 20names[i]consists of lowercase English letters, digits, and/or round brackets.
Solutions
-
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; }