Leetcode - Reformat The String Solution
Poor Man
- Published
- Updated
- Reading time
- 2 min read
- Length
- 228 words
Given alphanumeric string s. (Alphanumeric string is a string consisting of lowercase English letters and digits).
You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same type.
Return the reformatted string or return an empty string if it is impossible to reformat the string.
Example 1:
Text
Input: s = "a0b1c2"
Output: "0a1b2c"
Explanation: No two adjacent characters have the same type in "0a1b2c". "a0b1c2", "0a1b2c", "0c2a1b" are also valid permutations.Example 2:
Text
Input: s = "leetcode"
Output: ""
Explanation: "leetcode" has only characters so we cannot separate them by digits.Example 3:
Text
Input: s = "1229857369"
Output: ""
Explanation: "1229857369" has only digits so we cannot separate them by characters.Example 4:
Text
Input: s = "covid2019"
Output: "c2o0v1i9d"Example 5:
Text
Input: s = "ab123"
Output: "1a2b3"Constraints:
1 <= s.length <= 500sconsists of only lowercase English letters and/or digits.
Solution in Python#
Python
class Solution:
def reformat(self, s: str) -> str:
chars = []
nums = []
for i in s:
if i.isalpha():
chars.append(i)
else:
nums.append(i)
if(abs(len(chars)-len(nums)))>1:
return ""
else:
chars.extend([""])
nums.extend([""])
if len(chars)>len(nums):
return "".join(map(lambda x:x[0]+x[1],zip(chars,nums)))
else:
return "".join(map(lambda x:x[0]+x[1],zip(nums,chars)))