How do you solve Longest Balanced Substring II on LeetCode?

Longest Balanced Substring II (LeetCode #3714) 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: Balanced substrings require equal character frequency.

What is the time complexity of Longest Balanced Substring II?

The optimal solution for Longest Balanced Substring II runs in O(n) time and uses O(n) space. The algorithm efficiently checks character frequencies in linear time, using a hashmap for storage.

What is the brute force solution for Longest Balanced Substring II?

The brute force approach for Longest Balanced Substring II is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all possible substrings and count character frequencies. If all characters have the same frequency, it's balanced. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Balanced Substring II?

Longest Balanced Substring II uses the following concepts: Hash Table, String, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Longest Balanced Substring II?

Clarify the definition of balanced substring. Discuss edge cases upfront. Explain your thought process clearly.

What are common mistakes when solving Longest Balanced Substring II?

Overlooking character frequency checks. Not handling edge cases with single characters.

#3714

Longest Balanced Substring II

Medium

Hash TableStringPrefix SumHash MapArray

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)

Use a hashmap to track character frequencies and a sliding window to find the longest balanced substring efficiently.

⚙️

Algorithm

3 steps

1Step 1: Use a hashmap to maintain character counts as you expand the window.
2Step 2: Check if all characters in the hashmap have the same frequency.
3Step 3: Adjust the window size based on character counts to maintain balance.

solution.py13 lines

1def longest_balanced_substring(s):
2    max_length = 0
3    for length in range(1, len(s) + 1):
4        freq = {}
5        for i in range(len(s)):
6            if i >= length:
7                freq[s[i - length]] -= 1
8                if freq[s[i - length]] == 0:
9                    del freq[s[i - length]]
10            freq[s[i]] = freq.get(s[i], 0) + 1
11            if len(freq) > 0 and len(set(freq.values())) == 1:
12                max_length = max(max_length, length)
13    return max_length

ℹ

Complexity note: The algorithm efficiently checks character frequencies in linear time, using a hashmap for storage.

1Balanced substrings require equal character frequency.
2Sliding window helps manage character counts efficiently.

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