# [LeetCode] 336. Palindrome Pairs

## Problem

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"]
Example 2:

Input: ["bat","tab","cat"]
Output: [[0,1],[1,0]]
Explanation: The palindromes are ["battab","tabbat"]

## Solution

``````class Solution {
public List<List<Integer>> palindromePairs(String[] words) {
List<List<Integer>> res = new ArrayList<>();
if (words == null || words.length == 0) return res;
Map<String, Integer> map = new HashMap<>();
for (int i = 0; i < words.length; i++) {
map.put(words[i], i);
}

if (map.containsKey("")) {
int blankIndex = map.get("");
for (int i = 0; i < words.length; i++) {
if (isPalindrome(words[i]) && i != blankIndex) {
}
}
}

for (int i = 0; i < words.length; i++) {
String reversed = new StringBuilder(words[i]).reverse().toString();
if (map.containsKey(reversed)) {
int index = map.get(reversed);
if (index != i) {
}
}
}

for (int i = 0; i < words.length; i++) {
String word = words[i];
for (int j = 1; j < word.length(); j++) {
if (isPalindrome(word.substring(0, j))) {
String reversed = new StringBuilder(word.substring(j)).reverse().toString();
if (map.containsKey(reversed)) {
int index = map.get(reversed);
if (index != i) {
}
}
}
if (isPalindrome(word.substring(j))) {
String reversed = new StringBuilder(word.substring(0, j)).reverse().toString();
if (map.containsKey(reversed)) {
int index = map.get(reversed);
if (index != i) {
}
}
}
}
}

return res;
}

private boolean isPalindrome(String word) {
if (word == null || word.length() <= 1) return true;
int i = 0, j = word.length()-1;
while (i < j) {
if (word.charAt(i) != word.charAt(j)) return false;
i++;
j--;
}
return true;
}
}
``````