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336. Palindrome Pairs

Description

You are given a 0-indexed array of unique strings words.

A palindrome pair is a pair of integers (i, j) such that:

  • 0 <= i, j < words.length,
  • i != j, and
  • words[i] + words[j] (the concatenation of the two strings) is a palindrome.

Return an array of all the palindrome pairs of words.

You must write an algorithm with O(sum of words[i].length) runtime complexity.

 

Example 1:

Input: words = ["abcd","dcba","lls","s","sssll"]
Output: [[0,1],[1,0],[3,2],[2,4]]
Explanation: The palindromes are ["abcddcba","dcbaabcd","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]]
Explanation: The palindromes are ["a","a"]

 

Constraints:

  • 1 <= words.length <= 5000
  • 0 <= words[i].length <= 300
  • words[i] consists of lowercase English letters.

Solutions

Solution 1

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class Solution:
    def palindromePairs(self, words: List[str]) -> List[List[int]]:
        d = {w: i for i, w in enumerate(words)}
        ans = []
        for i, w in enumerate(words):
            for j in range(len(w) + 1):
                a, b = w[:j], w[j:]
                ra, rb = a[::-1], b[::-1]
                if ra in d and d[ra] != i and b == rb:
                    ans.append([i, d[ra]])
                if j and rb in d and d[rb] != i and a == ra:
                    ans.append([d[rb], i])
        return ans

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class Solution {
    private static final int BASE = 131;
    private static final long[] MUL = new long[310];
    private static final int MOD = (int) 1e9 + 7;
    static {
        MUL[0] = 1;
        for (int i = 1; i < MUL.length; ++i) {
            MUL[i] = (MUL[i - 1] * BASE) % MOD;
        }
    }
    public List<List<Integer>> palindromePairs(String[] words) {
        int n = words.length;
        long[] prefix = new long[n];
        long[] suffix = new long[n];
        for (int i = 0; i < n; ++i) {
            String word = words[i];
            int m = word.length();
            for (int j = 0; j < m; ++j) {
                int t = word.charAt(j) - 'a' + 1;
                int s = word.charAt(m - j - 1) - 'a' + 1;
                prefix[i] = (prefix[i] * BASE) % MOD + t;
                suffix[i] = (suffix[i] * BASE) % MOD + s;
            }
        }
        List<List<Integer>> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (check(i, j, words[j].length(), words[i].length(), prefix, suffix)) {
                    ans.add(Arrays.asList(i, j));
                }
                if (check(j, i, words[i].length(), words[j].length(), prefix, suffix)) {
                    ans.add(Arrays.asList(j, i));
                }
            }
        }
        return ans;
    }

    private boolean check(int i, int j, int n, int m, long[] prefix, long[] suffix) {
        long t = ((prefix[i] * MUL[n]) % MOD + prefix[j]) % MOD;
        long s = ((suffix[j] * MUL[m]) % MOD + suffix[i]) % MOD;
        return t == s;
    }
}

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func palindromePairs(words []string) [][]int {
    base := 131
    mod := int(1e9) + 7
    mul := make([]int, 310)
    mul[0] = 1
    for i := 1; i < len(mul); i++ {
        mul[i] = (mul[i-1] * base) % mod
    }
    n := len(words)
    prefix := make([]int, n)
    suffix := make([]int, n)
    for i, word := range words {
        m := len(word)
        for j, c := range word {
            t := int(c-'a') + 1
            s := int(word[m-j-1]-'a') + 1
            prefix[i] = (prefix[i]*base)%mod + t
            suffix[i] = (suffix[i]*base)%mod + s
        }
    }
    check := func(i, j, n, m int) bool {
        t := ((prefix[i]*mul[n])%mod + prefix[j]) % mod
        s := ((suffix[j]*mul[m])%mod + suffix[i]) % mod
        return t == s
    }
    var ans [][]int
    for i := 0; i < n; i++ {
        for j := i + 1; j < n; j++ {
            if check(i, j, len(words[j]), len(words[i])) {
                ans = append(ans, []int{i, j})
            }
            if check(j, i, len(words[i]), len(words[j])) {
                ans = append(ans, []int{j, i})
            }
        }
    }
    return ans
}

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using System.Collections.Generic;
using System.Linq;

public class Solution {
    public IList<IList<int>> PalindromePairs(string[] words) {
        var results = new List<IList<int>>();
        var reverseDict = words.Select((w, i) => new {Word = w, Index = i}).ToDictionary(w => new string(w.Word.Reverse().ToArray()), w => w.Index);

        for (var i = 0; i < words.Length; ++i)
        {
            var word = words[i];
            for (var j = 0; j <= word.Length; ++j)
            {
                if (j > 0 && IsPalindrome(word, 0, j - 1))
                {
                    var suffix = word.Substring(j);
                    int pairIndex;
                    if (reverseDict.TryGetValue(suffix, out pairIndex) && i != pairIndex)
                    {
                        results.Add(new [] { pairIndex, i});
                    }
                }
                if (IsPalindrome(word, j, word.Length - 1))
                {
                    var prefix = word.Substring(0, j);
                    int pairIndex;
                    if (reverseDict.TryGetValue(prefix, out pairIndex) && i != pairIndex)
                    {
                        results.Add(new [] { i, pairIndex});
                    }
                }
            }
        }

        return results;
    }

    private bool IsPalindrome(string word, int startIndex, int endIndex)
    {
        var i = startIndex;
        var j = endIndex;
        while (i < j)
        {
            if (word[i] != word[j]) return false;
            ++i;
            --j;
        }
        return true;
    }
}

Solution 2

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class Trie {
    Trie[] children = new Trie[26];
    Integer v;

    void insert(String w, int i) {
        Trie node = this;
        for (char c : w.toCharArray()) {
            c -= 'a';
            if (node.children[c] == null) {
                node.children[c] = new Trie();
            }
            node = node.children[c];
        }
        node.v = i;
    }

    Integer search(String w, int i, int j) {
        Trie node = this;
        for (int k = j; k >= i; --k) {
            int idx = w.charAt(k) - 'a';
            if (node.children[idx] == null) {
                return null;
            }
            node = node.children[idx];
        }
        return node.v;
    }
}

class Solution {
    public List<List<Integer>> palindromePairs(String[] words) {
        Trie trie = new Trie();
        int n = words.length;
        for (int i = 0; i < n; ++i) {
            trie.insert(words[i], i);
        }
        List<List<Integer>> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            String w = words[i];
            int m = w.length();
            for (int j = 0; j <= m; ++j) {
                if (isPalindrome(w, j, m - 1)) {
                    Integer k = trie.search(w, 0, j - 1);
                    if (k != null && k != i) {
                        ans.add(Arrays.asList(i, k));
                    }
                }
                if (j != 0 && isPalindrome(w, 0, j - 1)) {
                    Integer k = trie.search(w, j, m - 1);
                    if (k != null && k != i) {
                        ans.add(Arrays.asList(k, i));
                    }
                }
            }
        }
        return ans;
    }

    // TLE
    // private boolean isPalindrome(String w, int i, int j) {
    //     for (; i < j; ++i, --j) {
    //         if (w.charAt(i) != w.charAt(j)) {
    //             return false;
    //         }
    //     }
    //     return true;
    // }

    private boolean isPalindrome(String w, int start, int end) {
        int i = start, j = end;
        for (; i < j; ++i, --j) {
            if (w.charAt(i) != w.charAt(j)) {
                return false;
            }
        }
        return true;
    }
}

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