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# Palindrome Pairs LeetCode Solution

## Problem – Palindrome Pairs

Given a list of unique words, return all the pairs of the distinct indices `(i, j)` in the given list, so that the concatenation of the two words `words[i] + words[j]` is a palindrome.

Example 1:

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

Example 3:

``````Input: words = ["a",""]
Output: [[0,1],[1,0]]
``````

Constraints:

• `1 <= words.length <= 5000`
• `0 <= words[i].length <= 300`
• `words[i]` consists of lower-case English letters.

### Palindrome Pairs LeetCode Solution in Java

``````private static class TrieNode {
TrieNode[] next;
int index;
List<Integer> list;

TrieNode() {
next = new TrieNode[26];
index = -1;
list = new ArrayList<>();
}
}

public List<List<Integer>> palindromePairs(String[] words) {
List<List<Integer>> res = new ArrayList<>();

TrieNode root = new TrieNode();

for (int i = 0; i < words.length; i++) {
}

for (int i = 0; i < words.length; i++) {
search(words, i, root, res);
}

return res;
}

private void addWord(TrieNode root, String word, int index) {
for (int i = word.length() - 1; i >= 0; i--) {
int j = word.charAt(i) - 'a';

if (root.next[j] == null) {
root.next[j] = new TrieNode();
}

if (isPalindrome(word, 0, i)) {
}

root = root.next[j];
}

root.index = index;
}

private void search(String[] words, int i, TrieNode root, List<List<Integer>> res) {
for (int j = 0; j < words[i].length(); j++) {
if (root.index >= 0 && root.index != i && isPalindrome(words[i], j, words[i].length() - 1)) {
}

root = root.next[words[i].charAt(j) - 'a'];
if (root == null) return;
}

for (int j : root.list) {
if (i == j) continue;
}
}

private boolean isPalindrome(String word, int i, int j) {
while (i < j) {
if (word.charAt(i++) != word.charAt(j--)) return false;
}

return true;
}
``````

### Palindrome Pairs LeetCode Solution in Python

``````    def is_palindrome(check):
return check == check[::-1]

words = {word: i for i, word in enumerate(words)}
valid_pals = []
for word, k in words.iteritems():
n = len(word)
for j in range(n+1):
pref = word[:j]
suf = word[j:]
if is_palindrome(pref):
back = suf[::-1]
if back != word and back in words:
valid_pals.append([words[back],  k])
if j != n and is_palindrome(suf):
back = pref[::-1]
if back != word and back in words:
valid_pals.append([k, words[back]])
return valid_pals
``````

### Palindrome Pairs LeetCode Solution in C++

``````class Solution {
public:
vector<vector<int>> palindromePairs(vector<string>& words) {
unordered_map<string, int> dict;
vector<vector<int>> ans;
// build dictionary
for(int i = 0; i < words.size(); i++) {
string key = words[i];
reverse(key.begin(), key.end());
dict[key] = i;
}
// edge case: if empty string "" exists, find all palindromes to become pairs ("", self)
if(dict.find("")!=dict.end()){
for(int i = 0; i < words.size(); i++){
if(i == dict[""]) continue;
if(isPalindrome(words[i])) ans.push_back({dict[""], i}); // 1) if self is palindrome, here ans covers concatenate("", self)
}
}

for(int i = 0; i < words.size(); i++) {
for(int j = 0; j < words[i].size(); j++) {
string left = words[i].substr(0, j);
string right = words[i].substr(j, words[i].size() - j);

if(dict.find(left) != dict.end() && isPalindrome(right) && dict[left] != i) {
ans.push_back({i, dict[left]});     // 2) when j = 0, left = "", right = self, so here covers concatenate(self, "")
}

if(dict.find(right) != dict.end() && isPalindrome(left) && dict[right] != i) {
ans.push_back({dict[right], i});
}
}
}

return ans;
}

bool isPalindrome(string str){
int i = 0;
int j = str.size() - 1;

while(i < j) {
if(str[i++] != str[j--]) return false;
}

return true;
}

};
``````
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