1. What I learned
a. reversed for loop
To reverse ‘for loop’, write as follows.
for i in range(5,0,-1):
print(i)
#It will print '5,4,3,2,1'
b. Counter class
‘Counter(STRING)’ from ‘collections’ returns a dictionary counting the frequency of each key in STRING.
s = 'Hello'
char_dict = Counter(s)
print(char_dict)
#It will print 'Counter({'l': 2, 'H': 1, 'e': 1, 'o': 1})'
2. How I sloved
I counted each alphabet in ‘S’ using ‘Counter()’. I defined each alphabet as ‘key_list’ to store it and sorted it alphabetically. Two for loops were executed until ‘len(output)’ was equal to ‘len(s)’. The first for loop was the second and third condition of the problem, and the second for loop was the forth and fifth condition.
3. Code
class Solution:
def sortString(self, s: str) -> str:
char_dict = Counter(s)
key_list = sorted(char_dict.keys())
output = ""
while len(output) < len(s):
for e in key_list:
if char_dict[e] > 0:
output += e
char_dict[e] -= 1
for e in reversed(key_list):
if char_dict[e] > 0:
output += e
char_dict[e] -= 1
return output
# my first version, but took too much time
# class Solution:
# def sortString(self, s: str) -> str:
# output = []
# temp = sorted(list(s))
#
# while len(temp)>0:
# output.append(temp[0])
# temp.remove(temp[0])
# for e in temp:
# if e>output[-1]:
# output.append(e)
# temp[temp.index(e)] = ''
# temp = [e for e in temp if e]
#
# if len(temp)==0:
# break
#
# output.append(temp[-1])
# temp.remove(temp[-1])
# for i in range(len(temp)-1,0,-1):
# if temp[i]<output[-1]:
# output.append(temp[i])
# temp[i] = ''
# temp = [e for e in temp if e]
#
# return ''.join(output)
4. Result
Runtime : 64 ms(81.25%), Memory usage : 13.7 MB(86.61%)
(Runtime can be different by a system even if it is a same code.)