LeetCode in Kotlin
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 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.
Solution
class Solution {
fun getFolderNames(names: Array<String>): Array<String> {
val map = HashMap<String, Int>()
for (i in names.indices) {
var prefix = map.getOrDefault(names[i], 0)
if (prefix != 0) {
val raw = names[i]
while (map.getOrDefault(names[i], 0) != 0) {
names[i] = "$raw($prefix)"
prefix++
}
map[raw] = prefix
}
map[names[i]] = 1
}
return names
}
}