Jobs

Interviews

Discussions

Special Binary String

Hard

Google

8 views

Topics:

StringsRecursion

You are given a special binary string s. A special binary string has the following two properties:

The number of 0's is equal to the number of 1's.
Every prefix of the binary string has at least as many 1's as 0's.

A move consists of choosing two consecutive, non-empty, special substrings of s, and swapping them. Two strings are consecutive if the last character of the first string is exactly one index before the first character of the second string.

Return the lexicographically largest resulting string possible after applying the mentioned operations on the string.

Example 1:

Input: s = "11011000"
Output: "11100100"
Explanation: The strings "10" [occuring at s[1]] and "1100" [at s[3]] are swapped.
This is the lexicographically largest string possible after some number of swaps.

Example 2:

Input: s = "10"
Output: "10"

Can you provide a well-documented solution in Python?

Special Binary String

Hard

Google

8 views

Topics:

StringsRecursion

You are given a special binary string s. A special binary string has the following two properties:

The number of 0's is equal to the number of 1's.
Every prefix of the binary string has at least as many 1's as 0's.

Return the lexicographically largest resulting string possible after applying the mentioned operations on the string.

Example 1:

Input: s = "11011000"
Output: "11100100"
Explanation: The strings "10" [occuring at s[1]] and "1100" [at s[3]] are swapped.
This is the lexicographically largest string possible after some number of swaps.

Example 2:

Input: s = "10"
Output: "10"

Can you provide a well-documented solution in Python?

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum length of the input special binary string 'S'?
Can the input string 'S' be empty or null?
If multiple lexicographically largest special binary strings are possible, should I return any one of them?
Is the input string 'S' guaranteed to be a valid special binary string or should I validate it first?
What should I return if the input string is already the lexicographically largest special binary string?

Brute Force Solution

Approach

The core idea is to explore every single possible special binary string that could be formed from the input. We will systematically generate all valid combinations and check each one against the rules for special strings.

Here's how the algorithm would work step-by-step:

Start by considering all possible ways to divide the original binary string into smaller pieces.
For each possible piece, check if it follows the rules to be a 'special' binary string (equal number of 0s and 1s, and every prefix has more 1s than 0s).
If the piece is special, great! If not, discard that possibility and move on.
Now, take the special pieces you've found and consider all the different ways you can arrange them together to form a new string.
Check if the new string is also 'special'.
Repeat the process of breaking down and rearranging pieces until you have checked every possible combination.
From all the 'special' strings formed by different combinations, find the one that is largest when compared alphabetically. This will be your final answer.

Code Implementation

def special_binary_string_brute_force(binary_string):
    def is_special(current_string):
        balance = 0
        for character in current_string:
            if character == '1':
                balance += 1
            else:
                balance -= 1
            if balance < 0:
                return False
        return balance == 0

    def generate_special_strings(current_string):
        if not current_string:
            return [""]

        special_substrings = []

        # Find all special substrings of the current string
        for i in range(1, len(current_string) + 1):
            substring = current_string[:i]
            if is_special(substring):
                special_substrings.append(substring)

        all_special_strings = []
        def backtrack(current_combination, remaining_string):
            if not remaining_string:
                if is_special("".join(current_combination)):
                    all_special_strings.append("".join(current_combination))
                return

            #Explore all possible combinations using special substrings
            for special_substring in special_substrings:
                if remaining_string.startswith(special_substring):
                    backtrack(current_combination + [special_substring], remaining_string[len(special_substring):])

        backtrack([], binary_string)
        return all_special_strings

    # Generate all possible special strings
    all_possible_special_strings = generate_special_strings(binary_string)

    # Find the lexicographically largest special string
    if not all_possible_special_strings:
        return ""
    largest_special_string = max(all_possible_special_strings)
    return largest_special_string

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm explores all possible ways to divide the string of length n into substrings. This involves considering every possible start and end index for each substring. The number of possible substrings grows exponentially with the length of the string since each character can either be the start of a substring or not. Additionally, generating all permutations of special substrings, and checking if they are special themselves, contributes further to exponential runtime. Thus, the algorithm has a time complexity of O(2^n).

Space Complexity

O(N) – The algorithm recursively divides the binary string into smaller pieces. In the worst-case scenario, the depth of the recursion can be proportional to the length of the string, N, where N is the length of the input binary string. Each recursive call potentially creates new substrings and stores them, contributing to auxiliary space. Consequently, the space complexity is O(N) due to the maximum depth of the recursive calls and the storage of substrings.

Optimal Solution

Approach

The core idea is to break down the special binary string into smaller, manageable special substrings. Then, we cleverly re-arrange these smaller special strings to build the largest possible special binary string. Recursion is a key component of this divide-and-conquer strategy.

Here's how the algorithm would work step-by-step:

First, identify all the 'special' parts within the bigger string. A 'special' part starts with '1' and ends with '0', and inside, the number of '1's and '0's are equal.
Think of these 'special' parts as building blocks.
Next, we need to arrange these building blocks. The rule is, a 'special' part is 'bigger' than another if it would come later in a dictionary. So, we sort these blocks from 'biggest' to 'smallest'.
Finally, connect the sorted 'special' blocks together in that order. This gives us the largest possible special binary string.

Code Implementation

def special_binary_string(s): 
    if not s:
        return ""

    special_substrings = []
    balance = 0
    start = 0

    for i, char in enumerate(s):
        if char == '1':
            balance += 1
        else:
            balance -= 1

        # Found a special substring
        if balance == 0:

            # Recursively find special substrings within
            substring = s[start + 1:i]
            special_substrings.append('1' + special_binary_string(substring) + '0')

            start = i + 1

    # Sort substrings lexicographically to maximize the result
    special_substrings.sort(reverse=True)

    # Concatenate the sorted substrings
    return ''.join(special_substrings)

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm recursively decomposes the input string of length n into special substrings. Identifying these substrings involves iterating through the string. In the worst-case scenario, the string might need to be scanned multiple times during the recursive calls, checking and rearranging substrings. The sorting of the special substrings contributes O(k log k) where k is the number of special substrings. Because k can be proportional to n in some cases, and we have nested iterations due to both splitting and sorting, the overall time complexity approximates O(n^2) in the worst case, particularly when many special substrings need sorting.

Space Complexity

O(N) – The dominant space usage stems from the recursion depth. In the worst-case scenario, the special binary string might decompose into nested special substrings, leading to a recursion depth proportional to the length of the string, N. Additionally, the algorithm uses a list to store the 'special' parts, which in the worst case, can also contain substrings that, in total, could equal the length of the original string, N. Therefore, the auxiliary space required is O(N) due to the recursion stack and storage of special substrings. The algorithm creates substrings which can be at most the size of the input string.

Edge Cases

Case	How to Handle
Empty string	Return an empty string as there is no special binary string representation.
Single character string ('0' or '1')	Return the string itself as it is trivially a special binary string of length 1.
String with unbalanced 0s and 1s (more 0s than 1s or vice versa)	Return an empty string because a special binary string must have the same number of 0s and 1s.
String that is already the lexicographically largest special binary string.	The algorithm should correctly identify it as already sorted and potentially return it or its decomposition based on the algorithm's precise goal (e.g. finding a larger one).
String composed of multiple independent special binary strings concatenated together	The algorithm should correctly identify and process each substring independently during decomposition or rearrangement.
String with maximum possible length as defined by constraints (e.g., 50)	Ensure the recursive calls or any other iterative operations do not cause a stack overflow or time limit exceedance; consider memoization or dynamic programming for optimization if needed.
A large number of nested special binary strings causing deep recursion	Ensure the code doesn't exceed recursion depth limits; consider converting the recursive solution to an iterative one with a stack.
String that cannot be decomposed further into smaller valid special binary strings	The algorithm should return the input string itself, indicating that it's the smallest special binary string it can identify.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum length of the input special binary string 'S'?
Can the input string 'S' be empty or null?
If multiple lexicographically largest special binary strings are possible, should I return any one of them?
Is the input string 'S' guaranteed to be a valid special binary string or should I validate it first?
What should I return if the input string is already the lexicographically largest special binary string?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start by considering all possible ways to divide the original binary string into smaller pieces.
For each possible piece, check if it follows the rules to be a 'special' binary string (equal number of 0s and 1s, and every prefix has more 1s than 0s).
If the piece is special, great! If not, discard that possibility and move on.
Now, take the special pieces you've found and consider all the different ways you can arrange them together to form a new string.
Check if the new string is also 'special'.
Repeat the process of breaking down and rearranging pieces until you have checked every possible combination.
From all the 'special' strings formed by different combinations, find the one that is largest when compared alphabetically. This will be your final answer.

Code Implementation

def special_binary_string_brute_force(binary_string):
    def is_special(current_string):
        balance = 0
        for character in current_string:
            if character == '1':
                balance += 1
            else:
                balance -= 1
            if balance < 0:
                return False
        return balance == 0

    def generate_special_strings(current_string):
        if not current_string:
            return [""]

        special_substrings = []

        # Find all special substrings of the current string
        for i in range(1, len(current_string) + 1):
            substring = current_string[:i]
            if is_special(substring):
                special_substrings.append(substring)

        all_special_strings = []
        def backtrack(current_combination, remaining_string):
            if not remaining_string:
                if is_special("".join(current_combination)):
                    all_special_strings.append("".join(current_combination))
                return

            #Explore all possible combinations using special substrings
            for special_substring in special_substrings:
                if remaining_string.startswith(special_substring):
                    backtrack(current_combination + [special_substring], remaining_string[len(special_substring):])

        backtrack([], binary_string)
        return all_special_strings

    # Generate all possible special strings
    all_possible_special_strings = generate_special_strings(binary_string)

    # Find the lexicographically largest special string
    if not all_possible_special_strings:
        return ""
    largest_special_string = max(all_possible_special_strings)
    return largest_special_string

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, identify all the 'special' parts within the bigger string. A 'special' part starts with '1' and ends with '0', and inside, the number of '1's and '0's are equal.
Think of these 'special' parts as building blocks.
Next, we need to arrange these building blocks. The rule is, a 'special' part is 'bigger' than another if it would come later in a dictionary. So, we sort these blocks from 'biggest' to 'smallest'.
Finally, connect the sorted 'special' blocks together in that order. This gives us the largest possible special binary string.

Code Implementation

def special_binary_string(s): 
    if not s:
        return ""

    special_substrings = []
    balance = 0
    start = 0

    for i, char in enumerate(s):
        if char == '1':
            balance += 1
        else:
            balance -= 1

        # Found a special substring
        if balance == 0:

            # Recursively find special substrings within
            substring = s[start + 1:i]
            special_substrings.append('1' + special_binary_string(substring) + '0')

            start = i + 1

    # Sort substrings lexicographically to maximize the result
    special_substrings.sort(reverse=True)

    # Concatenate the sorted substrings
    return ''.join(special_substrings)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty string	Return an empty string as there is no special binary string representation.
Single character string ('0' or '1')	Return the string itself as it is trivially a special binary string of length 1.
String with unbalanced 0s and 1s (more 0s than 1s or vice versa)	Return an empty string because a special binary string must have the same number of 0s and 1s.
String that is already the lexicographically largest special binary string.	The algorithm should correctly identify it as already sorted and potentially return it or its decomposition based on the algorithm's precise goal (e.g. finding a larger one).
String composed of multiple independent special binary strings concatenated together	The algorithm should correctly identify and process each substring independently during decomposition or rearrangement.
String with maximum possible length as defined by constraints (e.g., 50)	Ensure the recursive calls or any other iterative operations do not cause a stack overflow or time limit exceedance; consider memoization or dynamic programming for optimization if needed.
A large number of nested special binary strings causing deep recursion	Ensure the code doesn't exceed recursion depth limits; consider converting the recursive solution to an iterative one with a stack.
String that cannot be decomposed further into smaller valid special binary strings	The algorithm should return the input string itself, indicating that it's the smallest special binary string it can identify.