Suppose we have an array of n strings called names. We have to make n directories in the file system such that, at the ith minute, we will create a directory with the name names[i]. Two files cannot have the same name, if we enter a duplicate directory name the system will have a suffix addition to its name in the form of (k), here, k is the smallest positive integer such that the obtained name remains unique. We have to find an array of strings of length n where ans[i] is the actual name that will be assign to the ith directory when we create it.
So, if the input is like names = ["my_dir","my_dir(1)","my_new_dir","my_new_dir","abc"], then the output will be ['my_dir', 'my_dir(1)', 'my_new_dir', 'my_new_dir(1)', 'abc'] because "my_new_dir" is already present once, then one (1) is added after the first one.
To solve this, we will follow these steps −
dic := a new map, if some key is not present then return 0
res := a new list
for each name in names, do
if name is not in dic, then
dic[name] := dic[name] + 1
insert name at the end of res
otherwise,
newname := name concatenate '(' concatenate dic[name] concatenate ')'
while newname is present in dic, do
dic[name] := dic[name] + 1
newname := name concatenate '(' concatenate dic[name] concatenate ')'
dic[newname] := 1
insert newname at the end of res
return res
Let us see the following implementation to get better understanding −
Example
from collections import defaultdict
def solve(names):
dic = defaultdict(int)
res = []
for name in names:
if name not in dic:
dic[name] += 1
res.append(name)
else:
newname = name +'(' + str(dic[name]) + ')'
while newname in dic:
dic[name] += 1
newname = name +'(' + str(dic[name]) + ')'
dic[newname] = 1
res.append(newname)
return res
names = ["my_dir","my_dir(1)","my_new_dir","my_new_dir","abc"]
print(solve(names))Input
["my_dir","my_dir(1)","my_new_dir","my_new_dir","abc"]
Output
['my_dir', 'my_dir(1)', 'my_new_dir', 'my_new_dir(1)', 'abc']