What is the time complexity of Construct the Longest New String?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Construct the Longest New String?

This problem uses Math, Dynamic Programming, Greedy, Brainteaser. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Construct the Longest New String?

Insight 1: To avoid "AAA" and "BBB", we need to alternate "AA" and "BB" as much as possible. The number of "AB" strings just acts as separators that allow us to use up the "AA" and "BB" strings more efficiently. Insight 2: The maximum length is achieved when we use all the "AB" strings available. The core idea is to make the counts of 'AA' and 'BB' as close as possible to maximize usage without creating 'AAA' or 'BBB'.

Which companies ask LeetCode 2745. Construct the Longest New String in interviews?

This problem is frequently asked at Guidewire, Zalando.

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Construct the Longest New String - LeetCode 2745 Solution

Construct the Longest New String - Complete Solution Guide

Construct the Longest New String is LeetCode problem 2745, 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 three integers x , y , and z . You have x strings equal to "AA" , y strings equal to "BB" , and z strings equal to "AB" . You want to choose some (possibly all or none) of these strings and concatenate them in some order to form a new string. This new string must not contain "AAA" or "BBB" as a substring. Return the maximum possible length of the new string . A substring is a contiguous non-empty sequence of characters within a string. Example 1: Input: x = 2, y = 5, z = 1 Output:

Detailed Explanation

The problem requires constructing the longest possible string using a given number of "AA", "BB", and "AB" substrings, such that the resulting string does not contain "AAA" or "BBB" as substrings. The input consists of three integers, x (number of "AA" strings), y (number of "BB" strings), and z (number of "AB" strings). The goal is to return the maximum possible length of the string that meets these constraints.

Solution Approach

The solution is based on the observation that to maximize the length of the string, we want to use as many "AA" and "BB" strings as possible while avoiding the forbidden substrings. We can effectively alternate them. We add the 'AB' strings in between 'AA' and 'BB' substrings. If the number of 'AA' and 'BB' strings is the same, we can use all of them along with all the 'AB' strings. If they are different, we can use twice the minimum of x and y, then potentially one more instance of the more numerous string type (x or y) to connect the end and avoid building up more than two same characters in a row. The logic effectively boils down to using `min(x, y)` pairs of "AA" and "BB", all the "AB" strings and one additional "AA" or "BB" if x!=y to connect to the end.

Step-by-Step Algorithm

Step 1: Find the minimum between x and y using `min(x, y)` function. This gives the maximum number of "AA" and "BB" pairs that can be used without creating "AAA" or "BBB".
Step 2: If x and y are equal, then all x and y can be used. Thus, result = (2 * x + z) * 2. We multiply by two because each string is length 2.
Step 3: If x and y are not equal, then result = (2 * min(x, y) + 1 + z) * 2. Because one of the 'AA' or 'BB' will be more than the other, you can insert a single instance of the more numerous one to avoid having only one left and creating "AAA" or "BBB" at the end. We multiply by two because each string is length 2.
Step 4: Return the final result.

Key Insights

Insight 1: To avoid "AAA" and "BBB", we need to alternate "AA" and "BB" as much as possible. The number of "AB" strings just acts as separators that allow us to use up the "AA" and "BB" strings more efficiently.
Insight 2: The maximum length is achieved when we use all the "AB" strings available. The core idea is to make the counts of 'AA' and 'BB' as close as possible to maximize usage without creating 'AAA' or 'BBB'.
Insight 3: If x and y are equal, we can use all x "AA" strings and all y "BB" strings along with z "AB" strings. If x and y are unequal, we can use all the 'AB' strings, and nearly all the available 'AA's and 'BB's. To ensure the constraint of not having "AAA" or "BBB", we have to consider the case where we can only use min(x,y) of both. The remaining one can be used at most once to avoid having "AAA" or "BBB".

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Math, Dynamic Programming, Greedy, Brainteaser.

Companies

Asked at: Guidewire, Zalando.

Construct the Longest New String | LeetCode 2745

Medium

Math Dynamic Programming Greedy Brainteaser

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

Solution Code

Available in:

class Solution {
    public int longestString(int x, int y, int z) {
        if (x == y) {
            return (2 * x + z) * 2;
        } else {
            return (2 * Math.min(x, y) + 1 + z) * 2;
        }
    }
}

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Problem Info

Description

Problem 2745: Construct the Longest New String - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming Greedy Brainteaser

Companies

GuidewireZalando

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2745

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

4 steps

1
Step 1: Find the minimum between x and y using `min(x, y)` function. This gives the maximum number of "AA" and "BB" pairs that can be used without creating "AAA" or "BBB".
2
Step 2: If x and y are equal, then all x and y can be used. Thus, result = (2 * x + z) * 2. We multiply by two because each string is length 2.
3
Step 3: If x and y are not equal, then result = (2 * min(x, y) + 1 + z) * 2. Because one of the 'AA' or 'BB' will be more than the other, you can insert a single instance of the more numerous one to avoid having only one left and creating "AAA" or "BBB" at the end. We multiply by two because each string is length 2.
4
Step 4: Return the final result.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The solution only involves basic arithmetic operations and comparisons. The time complexity is constant because the number of operations does not depend on the input size. Similarly, the space complexity is constant as only a few variables are used, regardless of the values of x, y, and z.

📖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

A dynamic programming approach could be used, but it would be significantly more complex and less efficient than the provided greedy solution. You could define a DP state that tracks the number of 'AA', 'BB', and 'AB' strings used so far, as well as the last two characters added to the string. The transitions would involve adding each type of string and checking for "AAA" and "BBB".

Time: O(x * y * z)Space: O(x * y * z)

Trade-offs: The DP approach is more general and can handle more complex constraints, but it's overkill for this problem. The greedy solution is much simpler and more efficient.

✨

Key Insights

1Insight 1: To avoid "AAA" and "BBB", we need to alternate "AA" and "BB" as much as possible. The number of "AB" strings just acts as separators that allow us to use up the "AA" and "BB" strings more efficiently.
2Insight 2: The maximum length is achieved when we use all the "AB" strings available. The core idea is to make the counts of 'AA' and 'BB' as close as possible to maximize usage without creating 'AAA' or 'BBB'.
3Insight 3: If x and y are equal, we can use all x "AA" strings and all y "BB" strings along with z "AB" strings. If x and y are unequal, we can use all the 'AB' strings, and nearly all the available 'AA's and 'BB's. To ensure the constraint of not having "AAA" or "BBB", we have to consider the case where we can only use min(x,y) of both. The remaining one can be used at most once to avoid having "AAA" or "BBB".

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: x = 0, y = 0, z = 0 (though constraints say x, y, z >= 1). The code handles this gracefully as the min function will still return 0, and the calculation will correctly lead to 0. Although it's not possible based on constraints.
Edge case 2: x is much larger than y, or y is much larger than x. The min(x, y) function ensures that the smaller value is used to create the alternating sequence, preventing the formation of "AAA" or "BBB". The `+ 1` takes care of using the one remaining instance of the more numerous string to avoid 'AAA' or 'BBB' at the end.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the case when x and y are significantly different, leading to "AAA" or "BBB" being formed if the larger value is used without proper handling.

✗Common mistake 2: Forgetting to multiply the result by 2 because each string fragment ("AA", "BB", "AB") has a length of 2.

✅

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: Resource allocation where you have limited resources of two types and need to combine them in a way that maximizes the outcome while avoiding certain combinations (e.g., chemical mixing where certain combinations are unstable).

🏦Real-world scenario 2: Scheduling tasks on a machine where certain task sequences are undesirable due to performance or stability reasons (e.g., alternating between CPU-intensive and I/O-intensive tasks).

💡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

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 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: July 16, 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 13

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

@AlgoMasterJul 6

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

@InterviewProJun 29

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 "Construct the Longest New String"?

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

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