How do you solve Longest Balanced Substring I on LeetCode?

Longest Balanced Substring I (LeetCode #3713) 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 frequencies.

What is the time complexity of Longest Balanced Substring I?

The optimal solution for Longest Balanced Substring I runs in O(n) time and uses O(n) space. Iterate through the string while maintaining counts, leading to linear time complexity.

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

The brute force approach for Longest Balanced Substring I 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 I?

Longest Balanced Substring I uses the following concepts: Hash Table, String, Counting, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Longest Balanced Substring I?

Clarify the definition of balanced substring. Discuss edge cases like single character strings. Explain your thought process clearly.

What are common mistakes when solving Longest Balanced Substring I?

Not checking all character frequencies. Ignoring substrings of different lengths.

#3713

Longest Balanced Substring I

Medium

Hash TableStringCountingEnumerationHash 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 check for balanced substrings efficiently.

⚙️

Algorithm

3 steps

1Step 1: Use a hashmap to count character frequencies as you iterate through the string.
2Step 2: For each unique frequency, check if all characters have that frequency.
3Step 3: Update the maximum length of balanced substrings found.

solution.py10 lines

1def longest_balanced(s):
2    max_len = 0
3    for freq in range(1, len(s) // 2 + 1):
4        count = {}
5        for char in s:
6            count[char] = count.get(char, 0) + 1
7            if count[char] == freq:
8                if len(set(count.values())) == 1:
9                    max_len = max(max_len, sum(count.values()))
10    return max_len

ℹ

Complexity note: Iterate through the string while maintaining counts, leading to linear time complexity.

1Balanced substrings require equal character frequencies.
2Using a hashmap allows efficient frequency counting.

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