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 <= 50000 <= words[i].length <= 300words[i]consists of lower-case English letters.
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;
}
}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
}