How do you solve String Without AAA or BBB on LeetCode?

String Without AAA or BBB (LeetCode #984) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a greedy approach allows us to construct the string without backtracking.

What is the brute force solution for String Without AAA or BBB?

The brute force approach for String Without AAA or BBB is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible strings of length a + b and checking each one for the conditions of having exactly a 'a's and b 'b's, while ensuring no 'aaa' or 'bbb' substrings appear. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve String Without AAA or BBB?

String Without AAA or BBB uses the following concepts: String, Greedy. The recommended patterns to study are: Greedy, Two Pointers.

What are the interview tips for String Without AAA or BBB?

Always explain your thought process clearly when presenting your solution. Consider edge cases and how your solution handles them. Practice writing code on a whiteboard or in an online editor to simulate interview conditions.

What are common mistakes when solving String Without AAA or BBB?

Failing to consider edge cases where one letter count is significantly larger than the other. Not checking for the conditions of 'aaa' or 'bbb' properly in the brute force approach.

#984

String Without AAA or BBB

Medium

StringGreedyGreedyTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a greedy strategy to construct the string directly by ensuring that we never create three consecutive 'a's or 'b's. This is efficient and avoids unnecessary checks.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty result string.
2Step 2: While there are still 'a's or 'b's left, append 'a' or 'b' to the result based on their counts.
3Step 3: If appending 'a' would create 'aaa', append 'b' instead, and vice versa.
4Step 4: Continue until all 'a's and 'b's are used.

solution.py24 lines

1def generate_string(a, b):
2    result = []
3    while a > 0 or b > 0:
4        if a > b:
5            if a > 1:
6                result.append('aa')
7                a -= 2
8            else:
9                result.append('a')
10                a -= 1
11            if b > 0:
12                result.append('b')
13                b -= 1
14        else:
15            if b > 1:
16                result.append('bb')
17                b -= 2
18            else:
19                result.append('b')
20                b -= 1
21            if a > 0:
22                result.append('a')
23                a -= 1
24    return ''.join(result)

ℹ

Complexity note: The time complexity is O(n) since we are constructing the string in a single pass, and the space complexity is O(n) for the resulting string.

1Using a greedy approach allows us to construct the string without backtracking.
2Maintaining the balance between 'a's and 'b's helps avoid invalid substrings.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.