Given a string s, return all the palindromic permutations (without duplicates) of it.
You may return the answer in any order. If s has no palindromic permutation, return
an empty list.
Example 1:
Input: s = "aabb" Output: ["abba","baab"]
Example 2:
Input: s = "abc" Output: []
Constraints:
1 <= s.length <= 16sconsists of only lowercase English letters.
Solution:
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/**
* Approach: To construct a palindrome since all the input characters are used, we will generate the prefix permutations
* and append the middle and do a reverse of prefix to get the suffix for the final string.
* Palindrome = prefix + middle + (reverse of prefix)
* Validation Check: If the number of characters with odd occurances > 1 we cannot form a palindrome with it since only one element can be the center .
* So we do backtracking based on the hashmap count of chars.
* Input is broken into half counts for each of the characters to generate prefix permutations.
* Time Complexity: O((n/2 + 1)!)
* Space Complexity: O(n)
**/
class Solution {
public List<String> generatePalindromes(String s) {
Map<Character, Integer> map = new HashMap<>();
for (char ch : s.toCharArray()) {
map.merge(ch, 1, Integer::sum);
}
Character odd = null;
//Detect chars with odd count.
int oddCount = 0;
for (Character ch : map.keySet()) {
if (map.get(ch) % 2 == 1) {
odd = ch;
oddCount++;
}
//Reduce map count size to half for generating only prefix.
map.put(ch, map.get(ch) / 2);
}
List<String> res = new ArrayList<>();
if (oddCount > 1) {
return res;
}
backtrack(res, s, odd, "", map);
return res;
}
void backtrack(List<String> res, String s, Character odd, String cur, Map<Character, Integer> map) {
if (cur.length() == s.length() / 2) {
String middle = odd == null ? "" : odd + "";
res.add(cur + middle + new StringBuilder(cur).reverse().toString());
return;
}
for (Character ch : map.keySet()) {
int count = map.getOrDefault(ch, 0);
if (count != 0) {
map.merge(ch, -1, Integer::sum);
backtrack(res, s, odd, cur + ch, map);
map.merge(ch, 1, Integer::sum);
}
}
}
}