Leetcode - Increasing Decreasing String Solution
Given a string s
. You should re-order the string using the following algorithm:
- Pick the smallest character from
s
and append it to the result. - Pick the smallest character from
s
which is greater than the last appended character to the result and append it. - Repeat step 2 until you cannot pick more characters.
- Pick the largest character from
s
and append it to the result. - Pick the largest character from
s
which is smaller than the last appended character to the result and append it. - Repeat step 5 until you cannot pick more characters.
- Repeat the steps from 1 to 6 until you pick all characters from
s
.
In each step, If the smallest or the largest character appears more than once you can choose any occurrence and append it to the result.
Return the result string after sorting s
with this algorithm.
Example 1:
Input: s = "aaaabbbbcccc"
Output: "abccbaabccba"
Explanation: After steps 1, 2 and 3 of the first iteration, result = "abc"
After steps 4, 5 and 6 of the first iteration, result = "abccba"
First iteration is done. Now s = "aabbcc" and we go back to step 1
After steps 1, 2 and 3 of the second iteration, result = "abccbaabc"
After steps 4, 5 and 6 of the second iteration, result = "abccbaabccba"
Example 2:
Input: s = "rat"
Output: "art"
Explanation: The word "rat" becomes "art" after re-ordering it with the mentioned algorithm.
Example 3:
Input: s = "leetcode"
Output: "cdelotee"
Example 4:
Input: s = "ggggggg"
Output: "ggggggg"
Example 5:
Input: s = "spo"
Output: "ops"
Constraints:
1 <= s.length <= 500
s
contains only lower-case English letters.
Solution in Python
from collections import Counter
class Solution:
def sortString(self, s: str) -> str:
s = dict(Counter(s).items())
result = ""
while any(s.values()):
for letter in sorted(s.keys()):
if s[letter]:
result+=letter
s[letter]-=1
for letter in reversed(sorted(s.keys())):
if s[letter]:
result+=letter
s[letter]-=1
return result