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You’re given a binary string ( s ) (a string consisting only of the characters ‘0’ and ‘1’). A binary string is considered balanced if the number of ‘0’s is equal to the number of ‘1’s. Your task is to find the length of the longest balanced substring within the given string ( s ).

Clarifying Questions

1. Can the input string be empty, and if so, what should be the output?
2. Are there any constraints on the length of the input string?
3. Should overlapping balanced substrings be considered?
4. Are nested balanced substrings possible, and if so, should we consider the largest possible substrate?

Example

Consider the binary string s = "00111011".

• The balanced substrings are “0011” and “110011”.
• The longest balanced substring is “0011” which has a length of 4.

Strategy

1. We can use a two-pointer approach, where we maintain a running balance while we traverse through the string.
2. A dictionary can be used to keep track of the first occurrence of each balance. This helps in calculating the length of the substring when we encounter the same balance again.
3. Iterate through the string, updating the balance accordingly:
   • Increment balance for ‘1’.
   • Decrement balance for ‘0’.
4. If the balance is zero at any index, it means we’ve seen a balanced substring from the start to this index.
5. If the balance has been seen before at some previous index, it means the substring between these indices is balanced.

This approach uses the concept of prefix sum to track balances and exploit the properties of balanced segments.

Code

def find_the_longest_balanced_substring(s: str) -> int:
    balance = 0
    balance_map = {0: -1}
    max_length = 0
    
    for i, char in enumerate(s):
        if char == '1':
            balance += 1
        else:
            balance -= 1
        
        if balance in balance_map:
            max_length = max(max_length, i - balance_map[balance])
        else:
            balance_map[balance] = i
    
    return max_length

# Example usage
print(find_the_longest_balanced_substring("00111011")) # Output: 4

Time Complexity

• Time Complexity: ( O(n) ), where ( n ) is the length of the binary string. We only traverse the string once.
• Space Complexity: ( O(n) ), for storing the balance map which in the worst case might grow linearly with respect to the input string if all balances are unique.

The algorithm ensures we efficiently find the longest balanced substring by leveraging the properties of balanced prefixes and their occurrences.

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