267. Palindrome Permutation II

Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form.

Example 1:

Input: "aabb"
Output: ["abba", "baab"]
Example 2:

Input: "abc"
Output: []



Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form.
Example 1:
Input: "aabb"
Output: ["abba", "baab"]
Example 2:
Input: "abc"
Output: []


https://leetcode.com/problems/palindrome-permutation-ii/description/



class Solution {
    public List<String> generatePalindromes(String s) {
        int odd = 0;
        String mid = "";
        List<String> res = new ArrayList<>();
        List<Character> list = new ArrayList<>();
        Map<Character, Integer> map = new HashMap<>();
        
        
        // build freq map from string 
        for(int i = 0; i < s.length(); i++){
            char c = s.charAt(i);
            if(!map.containsKey(c)){
                map.put(c, 1);
            }else{
                map.put(c, map.get(c) + 1);
            }
            odd += map.get(c) % 2 == 0 ? -1 : 1;
        }
        
        // check if can form a palindrom 
        if(odd > 1) return res;
        
        // add half freq of chars to a list 
        for(Map.Entry<Character, Integer> entry : map.entrySet()){
            char key = entry.getKey();
            int val = entry.getValue();
            
            if(val % 2 != 0) mid += key;
            
            for(int i = 0; i < val / 2; i++) list.add(key);
        }
        
        // permutation on the first half, reverse it , becomes the second half , 
        // add mid if necessary 
        boolean[] used = new boolean[list.size()];
        StringBuilder sb = new StringBuilder();
        permutation(res, list, mid, used, sb);
        return res;
    }
    private void permutation(List<String> res, List<Character> list, String mid, boolean[] used, StringBuilder sb){
        
        // base case 
        if(sb.length() == list.size()){
            res.add(sb.toString() + mid + sb.reverse().toString());
            sb.reverse(); // ?
            return;
        }
        
        for(int i = 0; i < list.size(); i++){
            // avoid duplication
            // aaaaabbbbc
            if(i > 0 && list.get(i) == list.get(i - 1) && !used[i - 1]) continue;
            
            if(!used[i]){
                used[i] = true;
                sb.append(list.get(i));
                permutation(res, list, mid, used, sb);
                // backtracking
                used[i] = false;
                sb.deleteCharAt(sb.length() - 1);
            }
        }
    }
}


Input:
"aabbcc"
Output:
["abccba","ccbbcc","baccab","ccaacc","cabbac","bbaabb"]
Expected:
["abccba","acbbca","baccab","bcaacb","cabbac","cbaabc"]