LeetCode 1417. Reformat The String

Description

https://leetcode.com/problems/reformat-the-string/

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:

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:

Input: s = "leetcode"
Output: ""
Explanation: "leetcode" has only characters so we cannot separate them by digits.

Example 3:

Input: s = "1229857369"
Output: ""
Explanation: "1229857369" has only digits so we cannot separate them by characters.

Example 4:

Input: s = "covid2019"
Output: "c2o0v1i9d"

Example 5:

Input: s = "ab123"
Output: "1a2b3"

Constraints:

  • 1 <= s.length <= 500
  • s consists of only lowercase English letters and/or digits.

Explanation

Get both lists of characters and digits from the s. If their length difference is greater than 1, we can’t reformat the string. Otherwise, just build a string starting with the longer list first.

Python Solution

class Solution:
    def reformat(self, s: str) -> str:
        
        digits = []
        chars = []
        
        for c in s:
            
            if c.isalpha():
                chars.append(c)
            elif c.isdigit():
                digits.append(c)
                
                
        if abs(len(digits) - len(chars)) > 1:
            return ""
        
        if len(digits) > len(chars):
            longer = digits 
            shorter = chars
        else:
            longer = chars
            shorter = digits
            
        result = ""
        
        for i in range(len(s)):
            if i % 2 == 0:
                result += longer.pop(0)
            else:
                result += shorter.pop(0)
                
                
        return result        
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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