LeetCode-in-Java

336. Palindrome Pairs

Hard

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.

Solution

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class Solution {
    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++) {
            addWord(root, words[i], 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.list.add(index);
            }
            root = root.next[j];
        }
        root.list.add(index);
        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)) {
                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 i, int j) {
        while (i < j) {
            if (word.charAt(i++) != word.charAt(j--)) {
                return false;
            }
        }
        return true;
    }
}

This site is open source. Improve this page.