Formatted question description: https://leetcode.ca/all/1487.html

1487. Making File Names Unique (Medium)

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.

Related Topics:
Hash Table, String

Solution 1.

Store the smallest possible index of the name in a map m.

If the name doesn’t exist in m, then we can safely use the name and set m[name] = 1 so that next time you come cross the same name, you can start from name + "(" + 1 + ")".

If the name already exist in m, instead of trying from 1 every time, we start trying from m[name]. The first name + "(" + index + ")" that doesn’t show up in m will be the name we use. Then we set m[name] = index + 1, and m[name + "(" + index + ")"] = 1.

// OJ: https://leetcode.com/problems/making-file-names-unique/

// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<string> getFolderNames(vector<string>& names) {
        int N = names.size();
        vector<string> ans(N);
        unordered_map<string, int> m;
        for (int i = 0; i < N; ++i) {
            auto &name = names[i];
            if (m.count(name) == 0) {
                ans[i] = name;
                m[name] = 1;
            } else {
                int index = m[name];
                while (true) {
                    auto n = name + "(" + to_string(index++) + ")";
                    if (m.count(n) == 0) {
                        ans[i] = n;
                        m[name] = index;
                        m[n] = 1;
                        break;
                    }
                }
            }
        }
        return ans;
    }
};

Java

class Solution {
    public String[] getFolderNames(String[] names) {
        if (names == null || names.length == 0)
            return names;
        int length = names.length;
        String[] actualNames = new String[length];
        Map<String, Integer> map = new HashMap<String, Integer>();
        for (int i = 0; i < length; i++) {
            String name = names[i];
            if (!map.containsKey(name)) {
                actualNames[i] = name;
                map.put(name, 1);
            } else {
                int count = map.get(name);
                while (map.containsKey(name + "(" + count + ")"))
                    count++;
                String actualName = name + "(" + count + ")";
                actualNames[i] = actualName;
                map.put(name, count + 1);
                map.put(actualName, 1);
            }
        }
        return actualNames;
    }
}

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