What is the time complexity of Palindrome Pairs?

The optimal solution has a time complexity of O(n*k^2).

What data structures are used to solve Palindrome Pairs?

This problem uses Array, Hash Table, String, Trie. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Palindrome Pairs?

Insight 1: A crucial observation is that if `words[i] + words[j]` is a palindrome, then we can potentially split `words[i]` into a prefix and a suffix, or `words[j]` into a prefix and suffix, and analyze whether one part is a palindrome and the other part's reverse exists in the `words` array. Insight 2: Using a hash table (or map) to store each word and its index enables efficient lookups (O(1) on average) for the reversed strings.

Which companies ask LeetCode 336. Palindrome Pairs in interviews?

This problem is frequently asked at Airbnb, Wix, Yandex.

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Palindrome Pairs - LeetCode 336 Solution

Q: How to solve Palindrome Pairs?

The provided solution uses a brute-force approach with optimizations. It iterates through each word in the input array `words`. For each word, it splits it into a prefix and a suffix at every possible position. Then, it checks two cases: first, if the suffix is a palindrome, it searches for the reversed prefix in the `words` array. Second, if the prefix is a palindrome, it searches for the reversed suffix in the `words` array. The use of a hash map allows efficient lookups during the reversed prefix and suffix searches. A set is used to avoid duplicate entries in the final result.

Q: What is the space complexity of Palindrome Pairs?

The space complexity is O(n*k).

Palindrome Pairs - Complete Solution Guide

Palindrome Pairs is LeetCode problem 336, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 ["abcd

Detailed Explanation

The problem asks us to find all pairs of distinct indices (i, j) in a given array of unique strings `words` such that the concatenation of `words[i]` and `words[j]` forms a palindrome. The challenge is to find all such pairs with an efficient algorithm. The input is an array of strings, and the output is a list of lists, where each inner list contains two integers representing the indices of a palindrome pair. Constraints include the size of the `words` array and the length of each string within it.

Solution Approach

The provided solution uses a brute-force approach with optimizations. It iterates through each word in the input array `words`. For each word, it splits it into a prefix and a suffix at every possible position. Then, it checks two cases: first, if the suffix is a palindrome, it searches for the reversed prefix in the `words` array. Second, if the prefix is a palindrome, it searches for the reversed suffix in the `words` array. The use of a hash map allows efficient lookups during the reversed prefix and suffix searches. A set is used to avoid duplicate entries in the final result.

Step-by-Step Algorithm

Step 1: Create a hash map (word_map) to store each word in `words` along with its index. This allows for O(1) lookup of word indices later.
Step 2: Initialize an empty set `result` to store the palindrome pairs as tuples (i, j). Using a set avoids duplicate pairs.
Step 3: Iterate through each word in the `words` array using index `i`.
Step 4: For each word, iterate through all possible split positions `j` (from 0 to length of word).
Step 5: Split the current word into `prefix` (words[i][0:j]) and `suffix` (words[i][j:])
Step 6: Check if the `suffix` is a palindrome. If it is, reverse the `prefix` and check if the reversed `prefix` exists in `word_map` and is not the same word as `words[i]`. If both are true, add the pair (i, index of reversed prefix) to the `result` set.
Step 7: Check if the `prefix` is a palindrome and `j != n` (to avoid double counting when the full word is a palindrome, which is already covered by j=0 in the first case). If it is, reverse the `suffix` and check if the reversed `suffix` exists in `word_map` and is not the same word as `words[i]`. If both are true, add the pair (index of reversed suffix, i) to the `result` set.
Step 8: Convert the `result` set to a list of lists and return it.

Key Insights

Insight 1: A crucial observation is that if `words[i] + words[j]` is a palindrome, then we can potentially split `words[i]` into a prefix and a suffix, or `words[j]` into a prefix and suffix, and analyze whether one part is a palindrome and the other part's reverse exists in the `words` array.
Insight 2: Using a hash table (or map) to store each word and its index enables efficient lookups (O(1) on average) for the reversed strings.
Insight 3: It's important to consider cases where one of the strings is empty or when a string is itself a palindrome. Also, prevent adding duplicate pairs to the result.

Complexity Analysis

Time Complexity: O(n*k^2)

Space Complexity: O(n*k)

Topics

This problem involves: Array, Hash Table, String, Trie.

Companies

Asked at: Airbnb, Wix, Yandex.

Palindrome Pairs | LeetCode 336

Hard

Array Hash Table String Trie

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

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<List<Integer>> palindromePairs(String[] words) {
        Map<String, Integer> wordMap = new HashMap<>();
        for (int i = 0; i < words.length; i++) {
            wordMap.put(words[i], i);
        }

        Set<List<Integer>> result = new HashSet<>();

        for (int i = 0; i < words.length; i++) {
            String word = words[i];
            int n = word.length();
            for (int j = 0; j <= n; j++) {
                String prefix = word.substring(0, j);
                String suffix = word.substring(j);

                // Case 1: words[i] is the first part, find a partner for the second part.
                // word + partner. `word = prefix + suffix`.
                // If suffix is a palindrome, we seek a partner that is the reverse of the prefix.
                if (new StringBuilder(suffix).reverse().toString().equals(suffix)) {
                    String revPrefix = new StringBuilder(prefix).reverse().toString();
                    if (wordMap.containsKey(revPrefix) && wordMap.get(revPrefix) != i) {
                        result.add(Arrays.asList(i, wordMap.get(revPrefix)));
                    }
                }

                // Case 2: words[i] is the second part, find a partner for the first part.
                // partner + word. `word = prefix + suffix`.
                // If prefix is a palindrome, we seek a partner that is the reverse of the suffix.
                // j != n ensures we don't double-count when the entire word is a palindrome,
                // as the case j=0 (empty prefix) already handles finding the full reverse.
                if (j != n && new StringBuilder(prefix).reverse().toString().equals(prefix)) {
                    String revSuffix = new StringBuilder(suffix).reverse().toString();
                    if (wordMap.containsKey(revSuffix) && wordMap.get(revSuffix) != i) {
                        result.add(Arrays.asList(wordMap.get(revSuffix), i));
                    }
                }
            }
        }

        List<List<Integer>> resultList = new ArrayList<>(result);
        return resultList;
    }
}

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Problem Info

Description

Problem 336: Palindrome Pairs - Detailed explanation and solution approaches.

Topics

Array Hash Table String Trie

Companies

AirbnbWixYandex

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:336

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Create a hash map (word_map) to store each word in `words` along with its index. This allows for O(1) lookup of word indices later.
2
Step 2: Initialize an empty set `result` to store the palindrome pairs as tuples (i, j). Using a set avoids duplicate pairs.
3
Step 3: Iterate through each word in the `words` array using index `i`.
4
Step 4: For each word, iterate through all possible split positions `j` (from 0 to length of word).
5
Step 5: Split the current word into `prefix` (words[i][0:j]) and `suffix` (words[i][j:])
6
Step 6: Check if the `suffix` is a palindrome. If it is, reverse the `prefix` and check if the reversed `prefix` exists in `word_map` and is not the same word as `words[i]`. If both are true, add the pair (i, index of reversed prefix) to the `result` set.
7
Step 7: Check if the `prefix` is a palindrome and `j != n` (to avoid double counting when the full word is a palindrome, which is already covered by j=0 in the first case). If it is, reverse the `suffix` and check if the reversed `suffix` exists in `word_map` and is not the same word as `words[i]`. If both are true, add the pair (index of reversed suffix, i) to the `result` set.
8
Step 8: Convert the `result` set to a list of lists and return it.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*k^2)

💾

Space Complexity

O(n*k)

📝

Why?

The algorithm iterates through `n` words. For each word, it iterates up to `k` times (where `k` is the length of the word) to split it into prefix and suffix. Checking if a substring is a palindrome takes O(k) time. Reversing a string also takes O(k) time. Looking up the reversed string in the hash map takes O(1) time on average. Therefore, the dominant operation is O(k) palindrome check and O(k) string reversal performed `n*k` times giving O(n*k*k) or O(n*k^2). The space complexity is dominated by the `word_map` which stores all the words. The space complexity is O(n*k), where n is the number of words and k is the average length of the words.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Trie-based approach

Build a Trie (prefix tree) from the reversed versions of all the words. Then, for each word, traverse the Trie and search for potential palindrome pairs. This approach leverages the prefix-matching capabilities of Tries.

Time: O(n * k)Space: O(n * k)

Trade-offs: The Trie-based approach can achieve better time complexity in some cases, particularly when there are many words with shared prefixes. However, the implementation can be more complex. The given approach has a simpler implementation at the cost of k factor in the time complexity.

✨

Key Insights

1Insight 1: A crucial observation is that if `words[i] + words[j]` is a palindrome, then we can potentially split `words[i]` into a prefix and a suffix, or `words[j]` into a prefix and suffix, and analyze whether one part is a palindrome and the other part's reverse exists in the `words` array.
2Insight 2: Using a hash table (or map) to store each word and its index enables efficient lookups (O(1) on average) for the reversed strings.
3Insight 3: It's important to consider cases where one of the strings is empty or when a string is itself a palindrome. Also, prevent adding duplicate pairs to the result.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string: If the input contains an empty string, it can form a palindrome pair with any palindrome string in the input. The code handles this correctly by searching for the reverse of the prefix/suffix, which will be the empty string itself when the split occurs at the beginning or end of a word.
Edge case 2: A word is a palindrome itself: The algorithm correctly identifies such words and finds their pairs if the reversed version of the remaining part (empty string or the other part of the split word) also exists in the input.
Edge case 3: Duplicate pairs. The use of a `set` for storing the result pairs ensures that only unique pairs are returned, thus avoiding redundancy.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle edge cases like empty strings or words that are palindromes themselves. The splitting logic must correctly account for prefixes and suffixes of length 0.

✗Common mistake 2: Double-counting palindrome pairs. Using a `set` effectively avoids this pitfall by ensuring uniqueness.

✗Common mistake 3: Incorrectly implementing the palindrome check. Palindrome checks must correctly compare characters from the beginning and end of the string.

✗Common mistake 4: Not checking that the found reversed word isn't the same word you're currently working with, leading to incorrect pairs like (i, i).

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Text editing software: Identifying potential palindrome pairs can be useful in suggesting word completions or corrections in text editors or search engines.

🏦Real-world scenario 2: Bioinformatics: Searching for palindromic sequences in DNA or RNA is a common task, and finding pairs of sequences that form palindromes after concatenation can be relevant in certain biological contexts.

🎮Real-world scenario 3: Data validation: Can be used for validating specific data structures for mirroring consistency.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Palindrome Pairs"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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