All Palindromic Permutations
1. You are given a string of length n.Input Format
2. You have to print all the palindromic permutations of the given string.
3. If no palindromic permutation exists for the given string, print "1".
Note > Check out the question video and write the recursive code as it is intended without
changing signature. The judge can't force you but intends you to teach a concept.
A String of length nOutput Format
Check the sample ouput and question video.Question Video
1 <= length of string <= 15Sample Input
aaabbSample Output
ababa
baaab

Editorial
We can observe those cases when the palindrome will exist.
Cases 
Case 1  Frequency of every character is even. (Example  "aabb", "aabbcc" for these inputs, palindrome will exist.)
Case 2  Number of characters with odd frequency is one and rest all the characters have even frequency. (Example  "aabbc", "a","ahbha" have palindromic permutations but "ab","aabc" does not).
We will store the frequency of each character in a hashmap. If the number of the characters with odd frequency is more than 1, we will print 1 else we will generate all the palindromic permutations using recursion and backtracking.
Way to generate palindromic permutation 
We will generate the permutations for half of the frequency of each character and then append the reverse of the permutation generated.
For example  for "aabb", frequency of each character is even therefore, it is possible to generate the palindromic permutations. We will generate the permutations for "ab" and append reverse of the permutation generated.
Permutations for "ab" are "ab" and "ba"
Answer  "ab" + "ba" = "abba"
"ba" + "ab" = "baab"
If the frequency of one of the character is odd, then we will find the permutation of characters with frequency half of the original (for odd character, frequency will be equal to floor(freq[odd]/2)) and then append the odd character and reverse of the permutation generated.
For example  for "aabbccc"  the permutations of "abc" are 
"abc"
"acb"
"bac"
"bca"
"cab"
"cba"
Answer will be 
"abc" +"c" + "cba" = "abccba"
"acb" + "c" + "bca" = "acbcbca"
"bac" + "c" + "cab" = "bacccab"
"bca" + "c" + "acb" = "bcacacb"
"cab" + "c" + "bac" = "cabcbac"
"cba" + "c" + "abc" = "cbacabc"
Signature 
cs  current spot
ts  total number of spots.
fmap  HashMap that stores the half frequency of each character.
oddc character variable that stores the character with odd frequency(can be null)
asf  String that stores the permutation at each level.
Code 
public static void main(String[] args) { Scanner scn = new Scanner(System.in); String str = scn.nextLine(); HashMap
fmap = new HashMap<>(); for (int i = 0; i < str.length(); i++) { char ch = str.charAt(i); fmap.put(ch, fmap.getOrDefault(ch, 0) + 1); } int ofc = 0; Character oddchar = null; int length = 0; for (int i = 0; i < 26; i++) { char ch = (char) ('a' + i); if (fmap.containsKey(ch)) { int freq = fmap.get(ch); if (freq % 2 != 0) { oddchar = ch; ofc++; } if (ofc > 1) { System.out.println("1"); return; } fmap.put(ch, freq / 2); length += (freq / 2); } } generatepw(1, length, fmap, oddchar, ""); } public static void generatepw(int cs, int ts, HashMap
fmap, Character oddc, String asf) { if (cs == ts + 1) { System.out.println(asf + (oddc == null ? "" : oddc) + reverse(asf)); return; } for (char ch : fmap.keySet()) { if (fmap.get(ch) > 0) { fmap.put(ch, fmap.get(ch)  1); generatepw(cs + 1, ts, fmap,oddc, asf + ch); fmap.put(ch, fmap.get(ch) + 1); } } } java false
Code Explained 
In the main function, we store the frequency of each character with the help of hashmap. After that we count how many characters are there with odd frequency and store the character with odd frequency in oddchar. If the characters with the odd frequency is more than 1, then we print 1 else we call the recursive function. We will also store the new len that is equal to the sum of half of frequencies of all the characters.
Initially cs is 1, ts is equal to the new len. Fmap stores the half frequencies of each character.
At each level we will use a FOR loop of size 26 and check whether the character is present in hashmap. If it is present, we will add it to the asf and increase the count of cs. In pre  order, we will reduce the frequency of each character by one and in post order, we will increase it by one.
Base Case 
When cs will be more than 1, it means we have formed permutation of all the characters. If oddc is not null, we will add it to the asf and append the reverse of the asf and print it.
Recursion Tree 
X  Y represents the characters available  answer so far.

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