336. Palindrome Pairs

Description

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”]

Problem URL

---

Solution

给一个word[]，找到数组中所有的i，j使得word[i] + word[j]是回文。

Using Trie tree to solve this proble. Reference

Code

class Solution {
    private static class TrieNode {
        TrieNode[] next;
        int index;
        List<Integer> list;
        
        TrieNode() {
            next = new TrieNode[26];
            index = -1;
            list = new ArrayList<Integer>();
        }
    }
    
    public List<List<Integer>> palindromePairs(String[] words) {
        List<List<Integer>> res = new ArrayList<>();
        if (words == null || words.length == 0){
            return res;
        }
        TrieNode root = new TrieNode();
        
        for (int i = 0; i < words.length; i++){
            addWord(root, words[i], i);
        }
        
        for (int i = 0; i < words.length; i++){
            searchWord(root, words, res, i);
        }
        
        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.list.add(index);
            }
            
            root = root.next[j];
        }
        root.list.add(index);
        root.index = index;
    }
    
    private void searchWord(TrieNode root, String[] words, List<List<Integer>> res, int i){
        for (int j = 0; j < words[i].length(); j++){
            if (root.index >= 0 && root.index != i && isPalindrome(words[i], j, words[i].length() - 1)){
                res.add(Arrays.asList(i, root.index));
            }
            
            root = root.next[words[i].charAt(j) - 'a'];
            if (root == null){
                return;
            }
        }
        
        for (int j : root.list){
            if (i == j) continue;
            res.add(Arrays.asList(i, j));
        }
    }
    
    private boolean isPalindrome(String word, int left, int right){
        while (left < right){
            if (word.charAt(left++) != word.charAt(right--)){
                return false;
            }
        }
        return true;
    }
}

Time Complexity: O(n * k^2)
Space Complexity: O()

---

Review