What is the time complexity of Maximum Length of a Concatenated String with Unique Characters?

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

What data structures are used to solve Maximum Length of a Concatenated String with Unique Characters?

This problem uses Array, String, Backtracking, Bit Manipulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Length of a Concatenated String with Unique Characters?

Insight 1: Using bit manipulation to represent the presence of characters in a string. Each character 'a' to 'z' can be represented by a bit in an integer, allowing for efficient checking of unique characters and combining strings. Insight 2: Applying backtracking to explore all possible subsequences of the given string array. This exhaustive search allows us to find the subsequence that yields the maximum length string with unique characters.

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Maximum Length of a Concatenated String with Unique Characters - LeetCode 1239 Solution

Q: Which companies ask LeetCode 1239. Maximum Length of a Concatenated String with Unique Characters in interviews?

This problem is frequently asked at Palo Alto Networks.

Maximum Length of a Concatenated String with Unique Characters - Complete Solution Guide

Maximum Length of a Concatenated String with Unique Characters is LeetCode problem 1239, 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 arr . A string s is formed by the concatenation of a subsequence of arr that has unique characters . Return the maximum possible length of s . A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements. Example 1: Input: arr = ["un","iq","ue"] Output: 4 Explanation: All the valid concatenations are: - "" - "un" - "iq" - "ue" - "uniq" ("un" + "iq") - "ique" ("iq" + "ue") Ma

Detailed Explanation

The problem asks us to find the maximum possible length of a string that can be formed by concatenating a subsequence of strings from the given array `arr`, where the resulting concatenated string must have unique characters. A subsequence is obtained by deleting some or no elements from the original array without changing the order of the remaining elements. Essentially, we need to find a combination of strings from `arr` that results in the longest possible string with no repeating characters.

Solution Approach

The solution uses a backtracking approach to explore all possible subsequences of the input array `arr`. It first converts each string in `arr` into a bitmask, where each bit represents the presence of a character from 'a' to 'z'. Strings with duplicate characters are discarded during this initial conversion. The backtracking function recursively explores all combinations of these bitmasks. In each step, it checks if adding a new mask (representing a string) to the current combined mask (representing the concatenated string) would introduce duplicate characters. If not, it adds the mask and recursively explores further combinations. The maximum length of a valid concatenation is tracked and updated throughout the backtracking process. The bit_count function determines the number of unique characters represented by each mask or combined mask.

Step-by-Step Algorithm

Step 1: Convert each string in `arr` into a bitmask. Iterate through each string. For each character in the string, determine its corresponding bit position (e.g., 'a' is bit 0, 'b' is bit 1, and so on). If the bit is already set in the mask, the string has duplicate characters, so skip it. Otherwise, set the bit.
Step 2: Store the valid bitmasks (those representing strings with unique characters) in a list `masks`.
Step 3: Implement a backtracking function `backtrack(start_index, current_mask)` that takes the starting index of the `masks` list and the current combined bitmask as input.
Step 4: In the `backtrack` function, update the `max_len` with the maximum of its current value and the number of set bits in the `current_mask` (which represents the number of unique characters in the current concatenation).
Step 5: Iterate through the remaining masks in the `masks` list, starting from `start_index`. For each `next_mask`, check if it has any characters in common with the `current_mask` using a bitwise AND operation. If `(current_mask & next_mask) == 0`, it means there are no common characters, so add the `next_mask` to the `current_mask` using a bitwise OR operation and recursively call `backtrack` with `i + 1` and the updated mask.
Step 6: After exploring all possibilities, return the `max_len`.

Key Insights

Insight 1: Using bit manipulation to represent the presence of characters in a string. Each character 'a' to 'z' can be represented by a bit in an integer, allowing for efficient checking of unique characters and combining strings.
Insight 2: Applying backtracking to explore all possible subsequences of the given string array. This exhaustive search allows us to find the subsequence that yields the maximum length string with unique characters.
Insight 3: Identifying strings with duplicate characters early on and excluding them from consideration to improve efficiency.

Complexity Analysis

Time Complexity: O(2^n)

Space Complexity: O(n)

Topics

This problem involves: Array, String, Backtracking, Bit Manipulation.

Companies

Asked at: Palo Alto Networks.

Maximum Length of a Concatenated String with Unique Characters | LeetCode 1239

Medium

Array String Backtracking Bit Manipulation

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    private int max_len = 0;

    public int maxLength(List<String> arr) {
        List<Integer> masks = new ArrayList<>();
        for (String s : arr) {
            int mask = 0;
            for (char c : s.toCharArray()) {
                int char_idx = c - 'a';
                if (((mask >> char_idx) & 1) != 0) {
                    mask = 0;
                    break;
                }
                mask |= (1 << char_idx);
            }
            if (mask > 0) {
                masks.add(mask);
            }
        }

        max_len = 0;

        backtrack(0, 0, masks);

        return max_len;
    }

    private void backtrack(int start_index, int current_mask, List<Integer> masks) {
        max_len = Math.max(max_len, Integer.bitCount(current_mask));

        for (int i = start_index; i < masks.size(); i++) {
            int next_mask = masks.get(i);
            if ((current_mask & next_mask) == 0) {
                backtrack(i + 1, current_mask | next_mask, masks);
            }
        }
    }
}

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Problem Info

Description

Problem 1239: Maximum Length of a Concatenated String with Unique Characters - Detailed explanation and solution approaches.

Topics

Array String Backtracking Bit Manipulation

Companies

Palo Alto Networks

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1239

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

6 steps

1
Step 1: Convert each string in `arr` into a bitmask. Iterate through each string. For each character in the string, determine its corresponding bit position (e.g., 'a' is bit 0, 'b' is bit 1, and so on). If the bit is already set in the mask, the string has duplicate characters, so skip it. Otherwise, set the bit.
2
Step 2: Store the valid bitmasks (those representing strings with unique characters) in a list `masks`.
3
Step 3: Implement a backtracking function `backtrack(start_index, current_mask)` that takes the starting index of the `masks` list and the current combined bitmask as input.
4
Step 4: In the `backtrack` function, update the `max_len` with the maximum of its current value and the number of set bits in the `current_mask` (which represents the number of unique characters in the current concatenation).
5
Step 5: Iterate through the remaining masks in the `masks` list, starting from `start_index`. For each `next_mask`, check if it has any characters in common with the `current_mask` using a bitwise AND operation. If `(current_mask & next_mask) == 0`, it means there are no common characters, so add the `next_mask` to the `current_mask` using a bitwise OR operation and recursively call `backtrack` with `i + 1` and the updated mask.
6
Step 6: After exploring all possibilities, return the `max_len`.

💡

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(2^n)

🐌 Exponential - Very slow! Often needs optimization.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is exponential because we are essentially generating all possible subsets of the input array `arr`. For each element in `arr`, we either include it or exclude it from the subsequence, resulting in 2 choices for each element. Therefore, the total number of subsequences is 2^n. The space complexity is linear because we store the bitmasks in a list of size proportional to the number of valid input strings.

📖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 with Bitmasking

Instead of backtracking, you could potentially use dynamic programming with bitmasking to store the maximum length achieved for each possible mask. This would involve iterating through the masks and updating a DP table.

Time: O(n * 2^26) in the worst case as 26 refers to alphabet. This can be simplified as O(n) since it's constant.Space: O(2^26). Simplified O(1)

Trade-offs: DP might be slightly more efficient in some cases if the same mask is encountered multiple times, avoiding redundant calculations. However, the space complexity can be significantly higher due to the DP table requiring 2^26 entries. Given that the max string length is only 26, a DP implementation could be optimized. The backtracking approach is simpler to implement and often sufficient for the given constraints.

✨

Key Insights

1Insight 1: Using bit manipulation to represent the presence of characters in a string. Each character 'a' to 'z' can be represented by a bit in an integer, allowing for efficient checking of unique characters and combining strings.
2Insight 2: Applying backtracking to explore all possible subsequences of the given string array. This exhaustive search allows us to find the subsequence that yields the maximum length string with unique characters.
3Insight 3: Identifying strings with duplicate characters early on and excluding them from consideration to improve efficiency.

📚

Key Terms Explained

For Students

Array

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

String

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

Edge Cases

Edge case 1: `arr` is empty. The algorithm correctly returns 0 because the loop processing the array will not run and the max length will be initialized as 0.
Edge case 2: `arr` contains an empty string. The empty string does not affect the character count, so including it in a sequence will not violate uniqueness or change the maximum length. The conversion to bitmask will simply result in a zero mask which will then be handled properly during recursion
Edge case 3: `arr` contains a string with length greater than 26. The algorithm implicitly ignores such strings as it's only working with lowercase alphabets 'a' to 'z'. Strings longer than this *must* have repeated characters, which will cause them to be skipped.
Edge case 4: `arr` contains strings with identical characters, e.g., ["aaaa"]. These strings will be skipped during the bitmask creation and will not be included in further calculation.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle duplicate characters within a string. Failing to check for duplicates before constructing the bitmask can lead to incorrect results.

✗Common mistake 2: Not correctly implementing the bitwise AND check for common characters. A wrong implementation of the unique character verification between combined mask and the next mask can corrupt the recursive call.

✗Common mistake 3: Failing to consider all possible subsequences. The backtracking approach needs to ensure that all combinations of strings are explored to find the true maximum length.

✅

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 data storage by identifying unique combinations of data fields that can be stored efficiently without redundancy. This can be seen in database design, where choosing unique key combinations can optimize index space.

🏦Real-world scenario 2: Resource allocation. Consider a scenario where you have a set of resources, each with specific capabilities. This algorithm can be adapted to find the largest combination of resources whose capabilities don't overlap to maximize the total capability without conflicts.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 7, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 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.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

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 Length of a Concatenated String with Unique Characters"?

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 string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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