What is the time complexity of Maximum Palindromes After Operations?

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

What data structures are used to solve Maximum Palindromes After Operations?

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

What are the key insights for solving Maximum Palindromes After Operations?

Insight 1: The core idea is that whether a word can be converted into a palindrome depends on the count of each character. If there are an even number of occurrences for each character, the word is a palindrome. If there is at most one character with an odd number of occurrences, then the word can be made into a palindrome. Insight 2: We can count the occurrences of all characters across *all* words. We then determine the maximum number of 'pairs' of characters that can be formed (each pair contributing to potentially forming palindromes), and greedily allocate these pairs to the shorter words first.

Which companies ask LeetCode 3035. Maximum Palindromes After Operations in interviews?

This problem is frequently asked at Grammarly, MathWorks.

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Maximum Palindromes After Operations - LeetCode 3035 Solution

Q: How to solve Maximum Palindromes After Operations?

The solution first counts the total number of occurrences of each character across all the words. It then calculates the total number of 'pairs' of characters (total_pairs). The lengths of the words are sorted in ascending order. The solution then iterates through the sorted word lengths, trying to make each word a palindrome. For each word, it calculates the number of pairs needed (length // 2). If the total number of pairs available is greater than or equal to the pairs needed, it decrements the total pairs available and increments the number of palindromes. Otherwise, it stops the loop because we don't have enough pairs to construct any more palindromes.

Maximum Palindromes After Operations - Complete Solution Guide

Maximum Palindromes After Operations is LeetCode problem 3035, 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 a 0-indexed string array words having length n and containing 0-indexed strings. You are allowed to perform the following operation any number of times ( including zero ): Choose integers i , j , x , and y such that 0 <= i, j < n , 0 <= x < words[i].length , 0 <= y < words[j].length , and swap the characters words[i][x] and words[j][y] . Return an integer denoting the maximum number of palindromes words can contain, after performing some operations. Note: i and j may be equal durin

Detailed Explanation

The problem asks us to find the maximum number of strings in a given array `words` that can be turned into palindromes after performing any number of swap operations between characters in the words. A swap operation involves choosing two characters from any two words (possibly the same word) and exchanging them. The goal is to maximize the number of palindromes that can be formed in the `words` array.

Solution Approach

The solution first counts the total number of occurrences of each character across all the words. It then calculates the total number of 'pairs' of characters (total_pairs). The lengths of the words are sorted in ascending order. The solution then iterates through the sorted word lengths, trying to make each word a palindrome. For each word, it calculates the number of pairs needed (length // 2). If the total number of pairs available is greater than or equal to the pairs needed, it decrements the total pairs available and increments the number of palindromes. Otherwise, it stops the loop because we don't have enough pairs to construct any more palindromes.

Step-by-Step Algorithm

Step 1: Combine all the words into a single string and count the occurrences of each character using a hash map (or array).
Step 2: Calculate the total number of character pairs available. This is done by summing the integer division of each character count by 2.
Step 3: Sort the lengths of the words in ascending order.
Step 4: Iterate through the sorted word lengths. For each word length, determine the number of character pairs needed to form a palindrome (length // 2).
Step 5: If the number of available character pairs is greater than or equal to the number of pairs needed, decrement the available pairs by the needed pairs and increment the palindrome count.
Step 6: If the number of available pairs is less than the needed pairs, stop the iteration. No more palindromes can be made.
Step 7: Return the final palindrome count.

Key Insights

Insight 1: The core idea is that whether a word can be converted into a palindrome depends on the count of each character. If there are an even number of occurrences for each character, the word is a palindrome. If there is at most one character with an odd number of occurrences, then the word can be made into a palindrome.
Insight 2: We can count the occurrences of all characters across *all* words. We then determine the maximum number of 'pairs' of characters that can be formed (each pair contributing to potentially forming palindromes), and greedily allocate these pairs to the shorter words first.
Insight 3: Sorting the word lengths in ascending order allows us to greedily make the shorter strings palindromes first. If we can form more palindromes, we form a palindrome for shorter strings. This greedy approach yields optimal solution.

Complexity Analysis

Time Complexity: O(NlogN)

Space Complexity: O(1)

Topics

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

Companies

Asked at: Grammarly, MathWorks.

Maximum Palindromes After Operations | LeetCode 3035

Medium

Array Hash Table String Greedy Sorting Counting

Time: O(NlogN)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public int maxPalindromesAfterOperations(String[] words) {
        Map<Character, Integer> charCounts = new HashMap<>();
        StringBuilder combined = new StringBuilder();
        for (String word : words) {
            combined.append(word);
        }
        String combinedString = combined.toString();

        for (int i = 0; i < combinedString.length(); i++) {
            char c = combinedString.charAt(i);
            charCounts.put(c, charCounts.getOrDefault(c, 0) + 1);
        }

        int totalPairs = 0;
        for (int count : charCounts.values()) {
            totalPairs += count / 2;
        }

        int[] wordLengths = new int[words.length];
        for (int i = 0; i < words.length; i++) {
            wordLengths[i] = words[i].length();
        }
        Arrays.sort(wordLengths);

        int palindromesCount = 0;
        for (int length : wordLengths) {
            int pairsNeeded = length / 2;
            if (totalPairs >= pairsNeeded) {
                totalPairs -= pairsNeeded;
                palindromesCount++;
            } else {
                break;
            }
        }

        return palindromesCount;
    }
}

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Problem Info

Description

Problem 3035: Maximum Palindromes After Operations - Detailed explanation and solution approaches.

Topics

Array Hash Table String Greedy Sorting Counting

Companies

GrammarlyMathWorks

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:3035

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

7 steps

1
Step 1: Combine all the words into a single string and count the occurrences of each character using a hash map (or array).
2
Step 2: Calculate the total number of character pairs available. This is done by summing the integer division of each character count by 2.
3
Step 3: Sort the lengths of the words in ascending order.
4
Step 4: Iterate through the sorted word lengths. For each word length, determine the number of character pairs needed to form a palindrome (length // 2).
5
Step 5: If the number of available character pairs is greater than or equal to the number of pairs needed, decrement the available pairs by the needed pairs and increment the palindrome count.
6
Step 6: If the number of available pairs is less than the needed pairs, stop the iteration. No more palindromes can be made.
7
Step 7: Return the final palindrome count.

💡

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(NlogN)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the sorting of word lengths. The character counting and the iteration through the sorted lengths are both linear in the number of characters or the number of words, respectively. The space complexity is constant because the size of the character count map/array does not depend on the input size significantly.

📖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

Dynamic Programming

While DP is not naturally suited to this problem, one could theoretically formulate a DP solution based on distributing available characters to form palindromes. This will likely be significantly more complex and less efficient due to its overhead.

Time: Likely O(N^2) or worse, depending on the specific formulationSpace: Likely O(N^2) or worse, depending on the specific formulation

Trade-offs: DP is not a good alternative here. The greedy solution is simpler, more efficient, and correct. A DP approach would be significantly harder to implement and reason about.

✨

Key Insights

1Insight 1: The core idea is that whether a word can be converted into a palindrome depends on the count of each character. If there are an even number of occurrences for each character, the word is a palindrome. If there is at most one character with an odd number of occurrences, then the word can be made into a palindrome.
2Insight 2: We can count the occurrences of all characters across *all* words. We then determine the maximum number of 'pairs' of characters that can be formed (each pair contributing to potentially forming palindromes), and greedily allocate these pairs to the shorter words first.
3Insight 3: Sorting the word lengths in ascending order allows us to greedily make the shorter strings palindromes first. If we can form more palindromes, we form a palindrome for shorter strings. This greedy approach yields optimal solution.

📚

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!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: An empty input array of words. The code implicitly handles this, returning 0 because there are no words to make into palindromes. The `char_counts` and `word_lengths` will be empty, and the loops will not execute.
Edge case 2: All words are already palindromes. The algorithm will still work correctly, as it will correctly count character pairs and assign them to each word in order.
Edge case 3: A single very long word vs. many very short words. The algorithm correctly assigns the pairs to short words before trying to make the longer word a palindrome, prioritizing smaller palindromes which results in an optimal number of palindromes.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the word lengths. Without sorting, the algorithm may not produce the maximum number of palindromes, because it will assign pairs to longer strings when they could be used more efficiently to make multiple shorter strings into palindromes.

✗Common mistake 2: Incorrectly calculating the number of pairs. It's important to do integer division `count // 2` to get the number of pairs for each character and sum them correctly.

✗Common mistake 3: Assuming that any word can be made a palindrome. A word can only be made a palindrome if the total count of characters can be used to form pairs exceeding half the number of characters in a target word

✅

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: Optimizing resource allocation. Imagine having a collection of tasks with varying requirements. This problem's greedy approach is relevant to allocate resources optimally from the pool to complete as many tasks as possible, where some tasks are more important than others.

🏦Real-world scenario 2: Data compression. In data compression, you might want to rearrange the data to form more repeating sequences to improve compression ratios. The underlying principle of grouping similar items is similar to arranging characters to maximize palindromes.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 3, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

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

@AlgoMasterJul 7

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

@InterviewProJun 30

❓

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 "Maximum Palindromes After Operations"?

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?