Find the Longest Balanced Substring of a Binary String

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class Solution:
    def findTheLongestBalancedSubstring(self, s: str) -> int:
        result, balanced = 0, "01"
        while len(balanced) <= len(s):
            if balanced in s: 
                result = len(balanced)
            balanced = '0' + balanced + '1'
        return result

The Python function findTheLongestBalancedSubstring in the Solution class is designed to find the length of the longest balanced substring within the input string s. A balanced string is characterized by having an equal number of ‘0’s and ‘1’s.

Here’s how the function works:

It initiates a variable result to 0 which will eventually hold the length of the longest balanced substring found within s. Another variable balanced is initiated as “01”, representing the smallest possible balanced string.
The function then enters a loop which continues as long as the length of balanced is less than or equal to the length of s. This condition is important because a balanced substring cannot be longer than s itself.
Inside the loop, the function checks if balanced exists as a substring within s. If it does, the function updates result to be the length of balanced.
After this check, the function extends the length of balanced by adding a ‘0’ to its beginning and a ‘1’ to its end. This maintains the balanced nature of the string while increasing its length.
The loop continues with this new balanced string and the process is repeated until the length condition fails.
The function finally returns result, which is now the length of the longest balanced substring found within s.

For the input string s = “10101”, the function will return 4, representing the length of the longest balanced substrings “1010” and “0101”.

The time complexity is O(n^2), where n is the length of the string s. This is due to the worst-case scenario where the function has to check each possible substring of s up to its full length. The space complexity is O(n), as the balanced string can grow up to the length of s.

The findTheLongestBalancedSubstring solution can be broken down into several fundamental building blocks:

String Manipulation: This problem involves working with strings, which includes creating substrings, finding substrings within another string, and concatenating strings.
Looping Structure: The while loop allows the function to repeatedly perform operations on the balanced string until its length exceeds that of s.
Conditional Statements: The if condition checks whether the balanced string exists as a substring within s.
Variable Updating: The function updates variables within the loop. The result variable is updated with the length of the currently found balanced string, and the balanced string is extended by adding ‘0’ to its start and ‘1’ to its end.
In-Memory Data Storage: The function uses in-memory data storage (variables result and balanced) to hold intermediate and final results of computation.
Return Statement: The function uses a return statement to output the final result.

These building blocks, when combined, form the complete solution to find the length of the longest balanced substring within the input string s.

The findTheLongestBalancedSubstring solution uses several high-level coding constructs:

Iterative Construction: The solution iteratively builds balanced strings of increasing length, starting from the smallest possible balanced string, “01”, and extending it by adding ‘0’ to its start and ‘1’ to its end in each iteration. This construct is used to generate all possible balanced strings up to the length of s.
Substring Search: The solution uses the substring search construct to check if a balanced string exists as a substring within s. This is used to find the longest balanced substring in s.
Optimization: The solution uses a greedy approach to continuously update the length of the longest balanced substring found so far. This construct is used to efficiently keep track of the longest balanced substring without needing to store all balanced substrings.
Guard Condition: The solution uses a guard condition to prevent the loop from creating a balanced string longer than s. This is used to ensure the function doesn’t waste time and resources on unnecessary computations.
Result Return: The solution uses a return statement to output the length of the longest balanced substring. This construct is used to provide the final output of the function.

These high-level constructs form the semantic structure of the solution. They show how the solution uses iterative construction, substring search, optimization, a guard condition, and result return to solve the problem.

You can see “Running Longest” coding construct, similar to Running Maximum. There are other high-level coding constructs and strategies at play in the findTheLongestBalancedSubstring solution:

Iterative Generation: The solution continually generates increasingly larger balanced strings, starting from the smallest possible balanced string (“01”). This is an application of iterative generation where the next state depends on the previous state.
Greedy Local Optimization: The solution adopts a greedy approach where it continually updates the longest balanced substring found so far. This allows for locally optimal decisions at each step with the hope of finding a global optimum.
Substring Search: This pattern is evident where the solution checks the existence of each generated balanced string as a substring within the given string s. It’s a form of pattern matching used to solve the problem.
Early Termination: The solution terminates the iterative generation when the length of the generated balanced string exceeds the length of s. This is a form of early termination or a stopping criterion that optimizes the solution by avoiding unnecessary computation.
Result Aggregation: The solution keeps track of the longest length found during the iterations and finally returns this value. This pattern of aggregating or accumulating a result during the iteration and returning it at the end is a common pattern in many problems.

These play a critical role in constructing the solution to this problem.

Q&A

Slice up entire string into units of 0s and 1s. These units should always starts with 0 and ends with 1. In each of these units look for the balanced substring and note down the length of the balanced substring. The result is the maximum length of all the balanced substrings.

Yes, your approach is another effective method to solve the problem, and it can be more efficient than generating all possible balanced strings.

Here’s how it works:

Slicing into Units: The first step is to divide the entire string into units where each unit starts with ‘0’ and ends with ‘1’. This can be done by iterating through the string and creating a new unit whenever we encounter a transition from ‘1’ to ‘0’.
Finding Balanced Substrings: For each unit, we find the longest balanced substring. We can do this by maintaining a balance counter that increments when we encounter ‘0’ and decrements when we encounter ‘1’. When the balance counter is zero, it means we have found a balanced substring.
Updating Result: If the length of the balanced substring is larger than the current maximum length, we update the maximum length.
Returning Result: The result is the maximum length of all the balanced substrings that we have found.

This approach works because a balanced string must start with ‘0’ and end with ‘1’, and the number of ‘0’s must equal the number of ‘1’s. By dividing the string into units that start with ‘0’ and end with ‘1’, we ensure that we consider all possible balanced substrings. By maintaining a balance counter, we can efficiently find balanced substrings within each unit.