336. Palindrome Pairs

Given a list of unique words, find all pairs of distinct indices (i, j) in the given list, so that the concatenation of the two words, i.e. words[i] + words[j] is a palindrome.
Example 1:
Input: ["abcd","dcba","lls","s","sssll"]
Output: [[0,1],[1,0],[3,2],[2,4]] 
Explanation: The palindromes are ["dcbaabcd","abcddcba","slls","llssssll"]






 ["abcd","dcba","lls","s","sssll"]
[[0,1],[1,0],[3,2],[2,4]] 



 ["abcd",      "dcba     " ,"lls"    , "s","     sssll"]
  0                     1。           2。        3。        4。

0， 1 
["abcd",      "dcba  


[1,0]
 dcba   abcd",  



3,2
"s",  ,"lls



2,4
lls   sssll




Solution 1 : dfs (brute force solution ). TLE . Passed 100/ 134 tests 
But not efficient for long strings , probably need to optimize the checkPalindrome function 



class Solution {
    public List<List<Integer>> palindromePairs(String[] words) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> tmp = new ArrayList<>();
        dfs(res, tmp, words);
        return res;
    }
    private void dfs(List<List<Integer>> res, List<Integer> tmp, String[] words){
        // base case 
        if(tmp.size() == 2){
            if(checkPalindrome(tmp, words)){
                res.add(new ArrayList<>(tmp));
            }
            return;
        }
        
        
        // dfs
        if(tmp.size() == 0){
            for(int i = 0; i < words.length; i++){
                tmp.add(i);
                dfs(res, tmp, words);
                // backtrack
                tmp.remove(tmp.size() - 1);
            }
        }else{
            for(int i = 0; i < words.length; i++){
                if(tmp.get(0) != i){
                    tmp.add(i);
                    dfs(res, tmp, words);
                    // backtrack
                    tmp.remove(tmp.size() - 1);
                }
            }
            
        }
        
        
    }
    private boolean checkPalindrome(List<Integer> tmp, String[] words){
        String first = words[tmp.get(0)];
        String second = words[tmp.get(1)];
        String newString = first + second;
        int i = 0;
        int j = newString.length() - 1;
        while(i < j){
            if(newString.charAt(i) != newString.charAt(j)){
                return false;
            }
            i++;
            j--;
        }
        return true;    
    }
}




// bug (idea is same as sol 1 )
class Solution {
    public List<List<Integer>> palindromePairs(String[] words) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> tmp = new ArrayList<>();
        dfs(res, tmp, words);
        return res;
    }
    private void dfs(List<List<Integer>> res, List<Integer> tmp, String[] words){
        // base case 
        if(tmp.size() == 2){
            if(checkPalindrome(tmp, words)){
                res.add(new ArrayList<>(tmp));
            }
            return;
        }
        
        
        // dfs
        if(tmp.size() == 0){
            for(int i = 0; i < words.length; i++){
                tmp.add(i);
                dfs(res, tmp, words);
                // backtrack
                tmp.remove(tmp.size() - 1);
            }
        }else{
            for(int i = 0; i < words.length; i++){
                if(tmp.get(0) != i){
                    tmp.add(i);
                    dfs(res, tmp, words);
                    // backtrack
                    tmp.remove(tmp.size() - 1);
                }
            }
            
        }
        
        
    }
    private boolean checkPalindrome(List<Integer> tmp, String[] words){
        String first = words[tmp.get(0)];
        String second = words[tmp.get(1)];
        int minLen = Math.min(first.length(), second.length());
        for(int i = 0; i < minLen; i++){
            if(first.charAt(i) != second.charAt(i)){
                return false;
            }
        }
        
        String restWord = first.length() < second.length() ? new String(second) : new String(first); // string a = string b, is it okay ?
        int i = 0;
        int j = restWord.length() - minLen - 1;
        while(i < j){
            if(restWord.charAt(i) != restWord.charAt(j)){
                return false;
            }else{
                i++;
                j--;
            }
        }
        return true;
    }
}











// solution 2 :  use trie 
Why can you use trie to make it better ? 

Write down the thins to optimize , from bf to this better solution , 


https://leetcode.com/problems/palindrome-pairs/discuss/79195/O(n-*-k2)-java-solution-with-Trie-structure

Efficient method
It can be done in an efficient manner by using the Trie data structure. The idea is to maintain a Trie of the reverse of all words.
1) Create an empty Trie.
2) Do following for every word:-
    a) Insert reverse of current word.
    b) Also store up to which index it is 
       a palindrome.
3) Traverse list of words again and do following 
   for every word.
    a) If it is available in Trie then return true
    b) If it is partially available
         Check the remaining word is palindrome or not 
         If yes then return true that means a pair
         forms a palindrome.
         Note: Position upto which the word is palindrome
               is stored because of these type of cases.





// solution 3: use HashMap

Why can you use HashMap can make it better ? 

https://blog.csdn.net/qq508618087/article/details/51443809

https://www.jianshu.com/p/fcde442da61b