Palindrome Pairs solution and explanation
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The problem
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, andwords[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 <= 50000 <= words[i].length <= 300words[i]consists of lowercase English letters.
Approach overview
Problem Overview: You are given a list of unique words. The task is to return all pairs of indices (i, j) such that concatenating words[i] + words[j] forms a palindrome. The challenge is efficiently checking many string combinations without recomputing palindrome checks repeatedly.
Approach 1: Brute Force Check with Reversed Strings (O(n2 * k) time, O(1) extra space)
The straightforward solution checks every ordered pair of words. For each pair (i, j), concatenate the two strings and verify if the result is a palindrome using two pointers from both ends. With n words and average length k, each concatenation check costs O(k), giving overall complexity O(n^2 * k). A small improvement is to precompute reversed versions of each string so you can quickly compare segments instead of building full concatenations. This approach relies mainly on simple array iteration and string comparison. It works well for small input sizes but quickly becomes slow when the number of words grows.
Approach 2: Using Trie Data Structure (O(n * k^2) time, O(n * k) space)
A more scalable approach builds a Trie containing reversed words. While inserting each reversed word, store indices of words whose remaining prefix forms a palindrome. This extra bookkeeping allows efficient matching later. When processing a word, iterate through its characters and walk the Trie simultaneously. If you encounter a Trie node that marks the end of another word and the remaining substring of the current word is a palindrome, you found a valid pair.
After reaching the end of the word in the Trie, any stored indices in that node represent words whose leftover suffix is already palindromic. Those indices also form valid pairs. This technique avoids comparing every pair directly and instead performs structured lookups. The Trie organizes reversed words by prefix, which makes matching efficient. The method combines concepts from Trie, hash table-style indexing, and repeated palindrome checks on substrings.
The key insight is splitting each word at every possible boundary. If the prefix is a palindrome, the reversed suffix may match another word. If the suffix is a palindrome, the reversed prefix may match. The Trie helps locate those matches quickly without scanning all words.
Recommended for interviews: Start with the brute force approach to show understanding of the problem and correctness of palindrome checks. Interviewers typically expect a more optimized design afterward. The Trie-based solution demonstrates stronger algorithmic thinking and knowledge of advanced string indexing techniques, which is why it is the preferred discussion in most senior-level interviews.
Approach 1: Brute Force Check with Reversed Strings
In this method, iterate through all possible pairs of words, concatenate them, and check if the concatenated result is a palindrome. Additionally, use the property that a string and its reverse can help identify palindrome pairs more efficiently.
# Python Solution for Brute Force Check with Reversed Strings
def is_palindrome(s):
return s == s[::-1]
def palindrome_pairs(words):
result = []
for i in range(len(words)):
for j in range(len(words)):
if i != j:
if is_palindrome(words[i] + words[j]):
result.append([i, j])
return result/* Java Solution for Brute Force Check with Reversed Strings */
import java.util.ArrayList;
import java.util.List;
public class Solution {
public boolean isPalindrome(String str) {
int left = 0, right = str.length() - 1;
while (left < right) {
if (str.charAt(left++) != str.charAt(right--)) return false;
}
return true;
}
public List<List<Integer>> palindromePairs(String[] words) {
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < words.length; ++i) {
for (int j = 0; j < words.length; ++j) {
if (i != j && isPalindrome(words[i] + words[j])) {
List<Integer> pair = new ArrayList<>();
pair.add(i);
pair.add(j);
result.add(pair);
}
}
}
return result;
}
}/* C++ Solution for Brute Force Check with Reversed Strings */
#include <iostream>
#include <vector>
#include <string>
using namespace std;
bool isPalindrome(const string& str) {
int left = 0, right = str.size() - 1;
while (left < right) {
if (str[left++] != str[right--]) return false;
}
return true;
}
vector<vector<int>> palindromePairs(vector<string>& words) {
vector<vector<int>> result;
int n = words.size();
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (i != j && isPalindrome(words[i] + words[j])) {
result.push_back({i, j});
}
}
}
return result;
}/* JavaScript Solution for Brute Force Check with Reversed Strings */
function isPalindrome(s) {
return s === s.split('').reverse().join('');
}
function palindromePairs(words) {
const result = [];
for (let i = 0; i < words.length; i++) {
for (let j = 0; j < words.length; j++) {
if (i !== j && isPalindrome(words[i] + words[j])) {
result.push([i, j]);
}
}
}
return result;
}/* C Solution for Brute Force Check with Reversed Strings */
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
bool isPalindrome(char *concatenated) {
int left = 0, right = strlen(concatenated) - 1;
while (left < right) {
if (concatenated[left++] != concatenated[right--]) return false;
}
return true;
}
void palindromePairs(char* words[], int wordsSize, int** result, int* returnSize, int** returnColumnSizes) {
*returnSize = 0;
int bufferSize = wordsSize * wordsSize;
*result = (int*)malloc(bufferSize * 2 * sizeof(int));
*returnColumnSizes = (int*)malloc(bufferSize * sizeof(int));
char buffer[600];
for (int i = 0; i < wordsSize; ++i) {
for (int j = 0; j < wordsSize; ++j) {
if (i != j) {
strcpy(buffer, words[i]);
strcat(buffer, words[j]);
if (isPalindrome(buffer)) {
(*result)[*returnSize * 2] = i;
(*result)[*returnSize * 2 + 1] = j;
(*returnColumnSizes)[*returnSize] = 2;
(*returnSize)++;
}
}
}
}
}/* C# Solution for Brute Force Check with Reversed Strings */
using System;
using System.Collections.Generic;
public class Solution {
public bool IsPalindrome(string str) {
int left = 0, right = str.Length - 1;
while (left < right) {
if (str[left++] != str[right--]) return false;
}
return true;
}
public IList<IList<int>> PalindromePairs(string[] words) {
IList<IList<int>> result = new List<IList<int>>();
for (int i = 0; i < words.Length; ++i) {
for (int j = 0; j < words.Length; ++j) {
if (i != j && IsPalindrome(words[i] + words[j])) {
result.Add(new List<int>{i, j});
}
}
}
return result;
}
}This Python implementation checks all unique pairs of words to find pairs that form palindromes using a simple `is_palindrome` helper function.
Time Complexity: O(n^2 * k)
Space Complexity: O(n^2)
Approach 2: Using Trie Data Structure
Instead of brute force, we can utilize a Trie data structure to store word reversals, which allows faster lookup for potential palindrome content between pairs. This method improves efficiency by focusing on checks only where relevant.
# Trie-based Python solution omitted due to complexity/* Trie-based Java Solution omitted due to complexity *//* Trie-based C++ Solution omitted due to complexity *//* Trie-based JavaScript Solution omitted due to complexity *//* Trie-based C Solution omitted due to complexity *//* Trie-based C# Solution omitted due to complexity */The Python solution constructs a Trie with efficient string reversal and lookup methods employed to minimize unnecessary checks.
Time Complexity: O(n * k^2)
Space Complexity: O(n * k)
Approach 3: Default Approach
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
}Complexity comparison
| Approach | Complexity |
|---|---|
| Brute Force Check with Reversed Strings | Time Complexity: O(n^2 * k) |
| Using Trie Data Structure | Time Complexity: O(n * k^2) |
| Default Approach | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Pair Checking | O(n^2 * k) | O(1) | Good for understanding the problem or when the number of words is small |
| Trie with Reversed Words | O(n * k^2) | O(n * k) | Best for large inputs where checking all pairs is too slow |
Frequently asked questions
Is Palindrome Pairs easy or hard?
Palindrome Pairs is classified as a Hard problem. The difficulty comes from combining multiple concepts: palindrome checking, substring partitioning, and efficient word lookup using structures like Trie.
How to solve Palindrome Pairs in O(n)?
A strict O(n) solution is not achievable because each word must be examined character by character. The closest practical optimization uses Trie or hash-based prefix and suffix checks, resulting in roughly O(n * k^2) time due to palindrome substring checks.
What is the best approach for Palindrome Pairs?
The most efficient practical solution uses a Trie storing reversed words along with palindrome suffix information. This allows quick prefix matching and avoids checking every pair directly. The typical complexity is O(n * k^2) time with O(n * k) space, where n is the number of words and k is the maximum word length.
What data structure is used in Palindrome Pairs?
Efficient solutions rely on a Trie to store reversed words and quickly match prefixes while scanning each word. Some variations also combine hash tables for direct reversed-string lookups and substring palindrome validation.
What is the time complexity of Palindrome Pairs?
The brute force method runs in O(n^2 * k) because it checks every pair of words and verifies whether their concatenation is a palindrome. The optimized Trie-based approach reduces comparisons and typically runs in O(n * k^2) time with O(n * k) memory.
Is Palindrome Pairs asked at Google, Amazon, or Meta?
Palindrome Pairs is considered a classic hard string problem and has appeared in interviews at large tech companies including Google, Amazon, and Meta. It tests knowledge of string manipulation, palindrome properties, and advanced data structures like Trie.
Is there a Python or Java solution for Palindrome Pairs?
Palindrome Pairs can be implemented in Python, Java, C++, and other languages. The logic typically involves building a Trie of reversed words or performing prefix-suffix palindrome checks with a hash map for quick lookups.