Leetcode 1417. 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.

Solution:

Approach 1. Use 2 lists and Character.isDigit to separate characters from digits:

class Solution {

 public String reformat(String s) {

  if (s.length() == 0 || s.length() == 1) return s;

  ArrayList < Character > clist = new ArrayList();

  ArrayList < Character > dlist = new ArrayList();

  for (char ch: s.toCharArray())

  {

   if (Character.isDigit(ch))

   {

    dlist.add(ch);

   } else

    clist.add(ch);

  }



  if (Math.abs(dlist.size() - clist.size()) > 1 || dlist.size() == 0 || clist.size() == 0)

   return "";

  StringBuffer sb = new StringBuffer();

  int i = 0;

  if (dlist.size() > clist.size())

  {

   while (i < clist.size())

   {
    sb.append(dlist.get(i));

    sb.append(clist.get(i));

    i++;
   }

   if (i != dlist.size())

    sb.append(dlist.get(i));



  } else

  {
   {

    while (i < dlist.size())

    {
     sb.append(clist.get(i));

     sb.append(dlist.get(i));

     i++;
    }

    if (i != clist.size())

     sb.append(clist.get(i));



   }

  }



  return sb.toString();





 }

}

Approach 2:  Use 2 lists and char comparision for separation:

class Solution { public String reformat(String s) { List<Character> alpha = new ArrayList<>(); List<Character> digit = new ArrayList<>(); for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); if (c >= 'a' && c <= 'z') alpha.add(c); else digit.add(c); } if (Math.abs(alpha.size() - digit.size()) > 1) return ""; StringBuilder ans = new StringBuilder(); if (alpha.size() > digit.size()) { for (int i = 0; i < digit.size(); i++) { ans.append(alpha.get(i)); ans.append(digit.get(i)); } ans.append(alpha.get(alpha.size() - 1)); } else { for (int i = 0; i < alpha.size(); i++) { ans.append(digit.get(i)); ans.append(alpha.get(i)); } if (digit.size() > alpha.size()) ans.append(digit.get(digit.size() - 1)); } return ans.toString(); } }