What is the time complexity of String Without AAA or BBB?

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

What is the space complexity of String Without AAA or BBB?

The space complexity is O(a+b).

What data structures are used to solve String Without AAA or BBB?

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

What are the key insights for solving String Without AAA or BBB?

Insight 1: The core idea is to use a greedy approach. Prioritize appending the character that appears more frequently (either 'a' or 'b') to maximize the utilization of available characters. Insight 2: To prevent "aaa" or "bbb" from appearing, we must check the last two characters appended to the string. If they are the same, we are forced to append the other character to avoid three consecutive identical characters.

Which companies ask LeetCode 984. String Without AAA or BBB in interviews?

This problem is frequently asked at Zalando.

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String Without AAA or BBB - LeetCode 984 Solution

Q: How to solve String Without AAA or BBB?

The solution uses a greedy approach with a check for consecutive characters. The algorithm iterates while either `a` or `b` is greater than 0. In each iteration, it first checks if the last two characters in the result string are the same. If they are, it means appending the same character again would violate the "aaa" or "bbb" constraint. Therefore, it's *forced* to append the other character. If the last two characters are different, or the string length is less than 2, it greedily appends the character with the higher count (`a` vs. `b`). This process continues until both `a` and `b` become 0, and the resulting string is returned.

String Without AAA or BBB - Complete Solution Guide

String Without AAA or BBB is LeetCode problem 984, 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

Given two integers a and b , return any string s such that: s has length a + b and contains exactly a 'a' letters, and exactly b 'b' letters, The substring 'aaa' does not occur in s , and The substring 'bbb' does not occur in s . Example 1: Input: a = 1, b = 2 Output: "abb" Explanation: "abb", "bab" and "bba" are all correct answers. Example 2: Input: a = 4, b = 1 Output: "aabaa" Constraints: 0 <= a, b <= 100 It is guaranteed such an s exists for the given a and b .

Detailed Explanation

The problem asks us to construct a string `s` of length `a + b` containing exactly `a` 'a' characters and `b` 'b' characters, such that the substrings "aaa" and "bbb" do not appear in `s`. We need to find *any* valid string `s` satisfying these conditions, given the integers `a` and `b`. The constraints specify that 0 <= a, b <= 100 and that a solution is guaranteed to exist.

Solution Approach

The solution uses a greedy approach with a check for consecutive characters. The algorithm iterates while either `a` or `b` is greater than 0. In each iteration, it first checks if the last two characters in the result string are the same. If they are, it means appending the same character again would violate the "aaa" or "bbb" constraint. Therefore, it's *forced* to append the other character. If the last two characters are different, or the string length is less than 2, it greedily appends the character with the higher count (`a` vs. `b`). This process continues until both `a` and `b` become 0, and the resulting string is returned.

Step-by-Step Algorithm

Step 1: Initialize an empty string (or list in Python) to store the result.
Step 2: Enter a `while` loop that continues as long as either `a` or `b` is greater than 0.
Step 3: Inside the loop, check if the length of the result string is at least 2, and if the last two characters of the string are the same.
Step 4: If the last two characters are the same, append the opposite character (if they are 'a', append 'b', and vice versa) and decrement the corresponding count (`a` or `b`).
Step 5: If the last two characters are different (or the string length is less than 2), compare `a` and `b`. Append 'a' to the string if `a > b`, and 'b' if `b >= a`. Decrement the corresponding count.
Step 6: Repeat steps 3-5 until both `a` and `b` are 0.
Step 7: Return the resulting string.

Key Insights

Insight 1: The core idea is to use a greedy approach. Prioritize appending the character that appears more frequently (either 'a' or 'b') to maximize the utilization of available characters.
Insight 2: To prevent "aaa" or "bbb" from appearing, we must check the last two characters appended to the string. If they are the same, we are forced to append the other character to avoid three consecutive identical characters.
Insight 3: The problem guarantees that a solution always exists, implying we don't need to worry about scenarios where we run out of a particular character while being forced to append it.

Complexity Analysis

Time Complexity: O(a+b)

Space Complexity: O(a+b)

Topics

This problem involves: String, Greedy.

Companies

Asked at: Zalando.

String Without AAA or BBB | LeetCode 984

Medium

String Greedy

Time: O(a+b)Space: O(a+b)

Solution Code

Available in:

class Solution {
    public String strWithout3a3b(int a, int b) {
        StringBuilder res = new StringBuilder();
        while (a > 0 || b > 0) {
            // First, check if we are forced to append a character to avoid "aaa" or "bbb"
            if (res.length() >= 2 && res.charAt(res.length() - 1) == res.charAt(res.length() - 2)) {
                if (res.charAt(res.length() - 1) == 'a') {
                    res.append('b');
                    b--;
                } else {  // res[-1] == 'b'
                    res.append('a');
                    a--;
                }
            }
            // Otherwise, make a greedy choice to append the character that is more frequent
            else {
                if (a > b) {
                    res.append('a');
                    a--;
                } else {  // b >= a
                    res.append('b');
                    b--;
                }
            }
        }
        return res.toString();
    }
}

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Problem Info

Description

Problem 984: String Without AAA or BBB - Detailed explanation and solution approaches.

Topics

String Greedy

Companies

Zalando

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:984

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: Initialize an empty string (or list in Python) to store the result.
2
Step 2: Enter a `while` loop that continues as long as either `a` or `b` is greater than 0.
3
Step 3: Inside the loop, check if the length of the result string is at least 2, and if the last two characters of the string are the same.
4
Step 4: If the last two characters are the same, append the opposite character (if they are 'a', append 'b', and vice versa) and decrement the corresponding count (`a` or `b`).
5
Step 5: If the last two characters are different (or the string length is less than 2), compare `a` and `b`. Append 'a' to the string if `a > b`, and 'b' if `b >= a`. Decrement the corresponding count.
6
Step 6: Repeat steps 3-5 until both `a` and `b` are 0.
7
Step 7: Return the resulting string.

💡

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(a+b)

💾

Space Complexity

O(a+b)

📝

Why?

The `while` loop iterates a maximum of `a + b` times because, in each iteration, either `a` or `b` is decremented. Therefore, the time complexity is O(a+b). The resulting string can have a maximum length of `a + b`, so the space complexity to store the string is also O(a+b).

📖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

Mathematical Approach

While the greedy approach is simple and efficient, it's conceptually possible to derive a more mathematical construction. This would likely involve calculating how many pairs of 'aa' or 'bb' you can use, and then distributing the remaining single characters appropriately. However, this would be considerably more complex to implement and understand.

Time: O(a+b) in the best case (pre-calculated string). O(1) if string is pre-calculated.Space: O(a+b)

Trade-offs: Generally not recommended due to increased complexity with no significant performance improvement. The greedy approach is simpler and easier to reason about.

✨

Key Insights

1Insight 1: The core idea is to use a greedy approach. Prioritize appending the character that appears more frequently (either 'a' or 'b') to maximize the utilization of available characters.
2Insight 2: To prevent "aaa" or "bbb" from appearing, we must check the last two characters appended to the string. If they are the same, we are forced to append the other character to avoid three consecutive identical characters.
3Insight 3: The problem guarantees that a solution always exists, implying we don't need to worry about scenarios where we run out of a particular character while being forced to append it.

📚

Key Terms Explained

For Students

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: a = 0, b = n. The solution will produce a string of 'b's, ensuring no "bbb" appears because only 'b's are used. For example, if a=0, b=3 the result is 'bbb', however the prompt guarantees that 'bbb' is not possible in this case, meaning that the values of a and b are constrained such that the greedy approach always finds a solution.
Edge case 2: a = n, b = 0. Similar to the previous case, the solution produces a string of 'a's, ensuring no "aaa" appears because only 'a's are used. Again, this is constrained by the problem statement.
Edge case 3: a = b. The algorithm will produce a balanced string alternating 'a' and 'b', like "ababab...", which obviously satisfies the constraints.
Edge case 4: a close to b (e.g. a = b+1 or b= a+1). Again, the greedy approach, will ensure that no three consecutive 'a's or 'b's occur.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check the last two characters before appending. This leads to the "aaa" or "bbb" substrings.

✗Common mistake 2: Incorrectly handling the case where `a == b`. It's important to handle the tie case correctly so the characters alternate correctly.

✗Common mistake 3: Not handling the base cases where one of `a` or `b` is zero, leading to edge-case failures or infinite loops. The while loop condition `a > 0 or b > 0` already handles this, but it's worth noting.

✅

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: This problem demonstrates a simplified resource allocation problem. Imagine scheduling tasks of two types ('a' and 'b') with constraints on consecutive tasks of the same type. The algorithm helps generate a schedule that avoids overloading the system with one type of task.

🏦Real-world scenario 2: It is applicable in data compression where you may want to alternate symbols to improve compression ratio by preventing long runs of the same symbol. This is a basic demonstration of run-length encoding and the importance of alternating values for entropy.

🎮Real-world scenario 3: Can be used as a basic building block in generating balanced test data for software applications. Ensuring no more than two consecutive values is useful for avoiding edge case bugs triggered by overly similar inputs.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

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

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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: August 24, 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 14

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

@DevExpert92Jul 7

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "String Without AAA or BBB"?

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

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