What is the time complexity of Longest Palindrome by Concatenating Two Letter Words?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Longest Palindrome by Concatenating Two Letter Words?

This problem uses Array, Hash Table, String, Greedy, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Longest Palindrome by Concatenating Two Letter Words?

Insight 1: Recognize that a palindrome can be constructed by pairing words with their reverses (e.g., "lc" and "cl"). Insight 2: Words that are their own reverses (e.g., "gg", "aa") can be used as the central part of the palindrome, but only one such word can be placed in the very center to maximize length.

Which companies ask LeetCode 2131. Longest Palindrome by Concatenating Two Letter Words in interviews?

This problem is frequently asked at Databricks.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Longest Palindrome by Concatenating Two Letter Words - LeetCode 2131 Solution

Longest Palindrome by Concatenating Two Letter Words - Complete Solution Guide

Longest Palindrome by Concatenating Two Letter Words is LeetCode problem 2131, a Medium 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 an array of strings words . Each element of words consists of two lowercase English letters. Create the longest possible palindrome by selecting some elements from words and concatenating them in any order . Each element can be selected at most once . Return the length of the longest palindrome that you can create . If it is impossible to create any palindrome, return 0 . A palindrome is a string that reads the same forward and backward. Example 1: Input: words = ["lc","cl","gg"] O

Detailed Explanation

The problem asks us to find the length of the longest possible palindrome that can be created by concatenating pairs of two-letter words from a given input list of words. Each word can be used at most once. A palindrome is a string that reads the same forwards and backward. The input consists of strings, each containing exactly two lowercase English letters. The output is an integer representing the maximum length of the palindrome.

Solution Approach

The solution uses a hash map (Counter in Python, HashMap in Java and C++, a custom HashTable in C) to count the occurrences of each word. It then iterates through the hash map, looking for words that are the reverse of other words. For each pair of words and their reverses, it takes the minimum count of the two words and multiplies it by 4 (since each pair contributes 2 words of length 2 each). It also keeps track of whether a word that is its own reverse exists with an odd number of occurrences. If so, it includes a central palindrome word of length 2 in the final palindrome, adding 2 to the total length.

Step-by-Step Algorithm

Step 1: Create a hash map to store the frequency of each word in the input list.
Step 2: Initialize a variable `length` to 0, which will store the length of the longest palindrome, and a boolean variable `center_present` to `false` to indicate whether a self-reversing word with an odd count is present.
Step 3: Iterate through the hash map.
Step 4: For each word in the hash map, check if the word is its own reverse (word[0] == word[1]).
Step 5: If the word is its own reverse, add (count // 2) * 4 to `length`. If count is odd, set `center_present` to `true`.
Step 6: If the word is not its own reverse, and if the first character is less than the second, reverse the word and check if its reverse exists in the hash map.
Step 7: If the reverse exists, add min(count, reverse_count) * 4 to `length`.
Step 8: After iterating through the hash map, if `center_present` is `true`, add 2 to `length` (for the central self-reversing word).
Step 9: Return `length`.

Key Insights

Insight 1: Recognize that a palindrome can be constructed by pairing words with their reverses (e.g., "lc" and "cl").
Insight 2: Words that are their own reverses (e.g., "gg", "aa") can be used as the central part of the palindrome, but only one such word can be placed in the very center to maximize length.
Insight 3: We can use a hash map (or similar data structure) to efficiently count the occurrences of each word.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, String, Greedy, Counting.

Companies

Asked at: Databricks.

Longest Palindrome by Concatenating Two Letter Words | LeetCode 2131

Medium

Array Hash Table String Greedy Counting

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int longestPalindrome(String[] words) {
        Map<String, Integer> counts = new HashMap<>();
        for (String word : words) {
            counts.put(word, counts.getOrDefault(word, 0) + 1);
        }

        int length = 0;
        boolean centerPresent = false;

        for (Map.Entry<String, Integer> entry : counts.entrySet()) {
            String word = entry.getKey();
            int count = entry.getValue();

            if (word.charAt(0) == word.charAt(1)) {
                length += (count / 2) * 4;
                if (count % 2 == 1) {
                    centerPresent = true;
                }
            } else if (word.charAt(0) < word.charAt(1)) {
                String reverseWord = new StringBuilder(word).reverse().toString();
                if (counts.containsKey(reverseWord)) {
                    int pairs = Math.min(count, counts.get(reverseWord));
                    length += pairs * 4;
                }
            }
        }

        if (centerPresent) {
            length += 2;
        }

        return length;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2131: Longest Palindrome by Concatenating Two Letter Words - Detailed explanation and solution approaches.

Topics

Array Hash Table String Greedy Counting

Companies

Databricks

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2131

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

9 steps

1
Step 1: Create a hash map to store the frequency of each word in the input list.
2
Step 2: Initialize a variable `length` to 0, which will store the length of the longest palindrome, and a boolean variable `center_present` to `false` to indicate whether a self-reversing word with an odd count is present.
3
Step 3: Iterate through the hash map.
4
Step 4: For each word in the hash map, check if the word is its own reverse (word[0] == word[1]).
5
Step 5: If the word is its own reverse, add (count // 2) * 4 to `length`. If count is odd, set `center_present` to `true`.
6
Step 6: If the word is not its own reverse, and if the first character is less than the second, reverse the word and check if its reverse exists in the hash map.
7
Step 7: If the reverse exists, add min(count, reverse_count) * 4 to `length`.
8
Step 8: After iterating through the hash map, if `center_present` is `true`, add 2 to `length` (for the central self-reversing word).
9
Step 9: Return `length`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the two iterations: one to build the hash map and the second to calculate the length of the palindrome. Both iterations are proportional to the number of words in the input. The space complexity is determined by the size of the hash map, which can hold up to n distinct words in the worst case.

📖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

Brute-force approach

Try all possible combinations of words and check if the resulting string is a palindrome, keeping track of the longest palindrome found. This approach would involve generating all permutations or combinations, which can be very slow.

Time: O((2^n) * (2n)), where n is the number of words. The 2^n comes from all the combinations, and the 2n comes from the length of the combined word and therefore the time to check if it's a palindromeSpace: O(2n) - space used to build palindrome for verification.

Trade-offs: The brute-force approach is extremely inefficient and impractical for any input of significant size. It avoids extra space complexity but has terrible performance.

✨

Key Insights

1Insight 1: Recognize that a palindrome can be constructed by pairing words with their reverses (e.g., "lc" and "cl").
2Insight 2: Words that are their own reverses (e.g., "gg", "aa") can be used as the central part of the palindrome, but only one such word can be placed in the very center to maximize length.
3Insight 3: We can use a hash map (or similar data structure) to efficiently count the occurrences of each word.

📚

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!

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

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

Edge Cases

Edge case 1: Empty input list. The algorithm correctly returns 0 in this case, as the hash map will be empty, and no calculations will be performed.
Edge case 2: Input list containing only words that are their own reverses (e.g., ["gg", "aa", "ll"]). The algorithm correctly handles this by using even counts of self-reversing words to form palindromes and using only one self-reversing word with an odd count (if any) in the center.
Edge case 3: Input list with no palindromes possible (e.g., ["ab", "cd", "ef"]). The algorithm correctly returns 0 in this case, as no reverse pairs will be found.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering words that are their own reverses as potential central parts of the palindrome.

✗Common mistake 2: Counting each pair of reversed words twice, once for each word in the pair. The `word[0] < word[1]` check ensures that each pair is only counted once.

✗Common mistake 3: Not taking the minimum of counts of the word and its reverse when combining them which would violate the 'at most once' constraint.

✅

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: Bioinformatics - Finding palindromic sequences in DNA, which are important for gene regulation and other biological processes.

🏦Real-world scenario 2: Data compression - Palindrome detection can be used in data compression algorithms by identifying and compressing repeating patterns.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: October 30, 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 13

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

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Longest Palindrome by Concatenating Two Letter Words"?

For this medium-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.

🤖 Need help with this problem?