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 which is 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)".
Example 4:
Input: names = ["wano","wano","wano","wano"] Output: ["wano","wano(1)","wano(2)","wano(3)"] Explanation: Just increase the value of k each time you create folder "wano".
Example 5:
Input: names = ["kaido","kaido(1)","kaido","kaido(1)"] Output: ["kaido","kaido(1)","kaido(2)","kaido(1)(1)"] Explanation: Please note that system adds the suffix (k) to current name even it contained the same suffix before.
Constraints:
1 <= names.length <= 5 * 10^4
1 <= names[i].length <= 20
names[i]
consists of lower case English letters, digits and/or round brackets.
Solution: Hashtable
Use a hashtable to store the mapping form base_name to its next suffix index.
Time complexity: O(n)
Space complexity: O(n)
C++
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// Author: Huahua class Solution { public: vector<string> getFolderNames(vector<string>& names) { vector<string> ans; unordered_map<string, int> m; // base_name -> next suffix. for (const string& name : names) { string unique_name = name; int j = m[name]; if (j > 0) { while (m.count(unique_name = name + "(" + to_string(j++) + ")")); m[name] = j; } m[unique_name] = 1; ans.push_back(unique_name); } return ans; } }; |
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