#2131

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Longest Palindrome by Concatenating Two Letter Words

Find the maximum-length palindrome by combining two-letter words using array scanning and hash table lookups efficiently.

array hash table string greedy counting

Problem Statement

You are given an array of strings words, where each word consists of exactly two lowercase English letters. Your task is to select some words and concatenate them in any order to form the longest possible palindrome. Each word can be used at most once.

Return the length of the longest palindrome that can be created. If no palindrome can be formed, return 0. Consider that a valid palindrome must mirror around its center, and pairing reversed words or placing a word with identical letters centrally maximizes length.

Examples

Example 1

Input: words = ["lc","cl","gg"]

Output: 6

One longest palindrome is "lc" + "gg" + "cl" = "lcggcl", of length 6. Note that "clgglc" is another longest palindrome that can be created.

Example 2

Input: words = ["ab","ty","yt","lc","cl","ab"]

Output: 8

One longest palindrome is "ty" + "lc" + "cl" + "yt" = "tylcclyt", of length 8. Note that "lcyttycl" is another longest palindrome that can be created.

Example 3

Input: words = ["cc","ll","xx"]

Output: 2

One longest palindrome is "cc", of length 2. Note that "ll" is another longest palindrome that can be created, and so is "xx".

Constraints

1 <= words.length <= 105
words[i].length == 2
words[i] consists of lowercase English letters.

Solution Approach

Count Words Using a Hash Map

Iterate through words and store the frequency of each in a hash map. This allows O(1) lookup for reversed pairs. Identical-letter words require separate tracking for potential center placement.

Pair Reversed Words

For each word, check if its reverse exists in the hash map. Each valid pair contributes twice the word length to the palindrome. Decrement counts to avoid reuse, ensuring proper pairing to maximize mirrored segments.

Add Central Word if Available

After pairing, identify if any word with identical letters remains. Place at most one in the palindrome center to increase length by 2. This step ensures that the palindrome is as long as possible without breaking mirroring.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n) due to single-pass word counting and pairing with hash lookups. Space complexity is O(n) for the hash map storing word frequencies, independent of palindrome construction.

What Interviewers Usually Probe

Check if you can use a hash map to find reversed word pairs efficiently.
Consider how identical-letter words might serve as the palindrome's center.
Be prepared to discuss linear-time pairing versus brute-force concatenation.

Common Pitfalls or Variants

Common pitfalls

Forgetting to decrement counts after pairing leads to overcounting words.
Not handling identical-letter words for central placement reduces palindrome length.
Assuming order of words matters; any order is valid for concatenation.

Follow-up variants

Allow words of length three or more and analyze how center placement generalizes.
Return the actual palindrome string instead of just its length.
Count distinct palindrome arrangements instead of maximum length.

FAQ

What is the primary strategy to solve Longest Palindrome by Concatenating Two Letter Words?

Use array scanning with a hash map to count word frequencies, pair each word with its reverse, and place at most one identical-letter word in the center.

Can a word be used more than once in forming the palindrome?

No, each word can be selected at most once. Pairing and central placement must respect this constraint.

Why are words with identical letters treated differently?

They can serve as the center of the palindrome, allowing maximal length without requiring a mirrored pair.

What is the time complexity of this solution?

The solution runs in O(n) time, scanning all words once and using hash lookups for reversed pairs.

What failure modes should I watch for in this problem?

Common failures include overcounting paired words, ignoring central identical-letter words, and assuming word order affects the palindrome.

terminal

Solution

Solution 1: Greedy + Hash Table

First, we use a hash table $\textit{cnt}$ to count the occurrences of each word.

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class Solution:
    def longestPalindrome(self, words: List[str]) -> int:
        cnt = Counter(words)
        ans = x = 0
        for k, v in cnt.items():
            if k[0] == k[1]:
                x += v & 1
                ans += v // 2 * 2 * 2
            else:
                ans += min(v, cnt[k[::-1]]) * 2
        ans += 2 if x else 0
        return ans

Continue Practicing

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