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Split a String in Balanced Strings

Easy

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StringsGreedy Algorithms

Balanced strings are those that have an equal quantity of 'L' and 'R' characters.

Given a balanced string s, split it into some number of substrings such that:

Each substring is balanced.

Return the maximum number of balanced strings you can obtain.

Example 1:

Input: s = "RLRRLLRLRL"
Output: 4
Explanation: s can be split into "RL", "RRLL", "RL", "RL", each substring contains same number of 'L' and 'R'.

Example 2:

Input: s = "RLRRRLLRLL"
Output: 2
Explanation: s can be split into "RL", "RRRLLRLL", each substring contains same number of 'L' and 'R'.
Note that s cannot be split into "RL", "RR", "RL", "LR", "LL", because the 2^nd and 5^th substrings are not balanced.

Example 3:

Input: s = "LLLLRRRR"
Output: 1
Explanation: s can be split into "LLLLRRRR".

Constraints:

2 <= s.length <= 1000
s[i] is either 'L' or 'R'.
s is a balanced string.

Try coding it in LeetCode

Split a String in Balanced Strings

Easy

Asked by:

53 views

Topics:

StringsGreedy Algorithms

Balanced strings are those that have an equal quantity of 'L' and 'R' characters.

Given a balanced string s, split it into some number of substrings such that:

Each substring is balanced.

Return the maximum number of balanced strings you can obtain.

Example 1:

Input: s = "RLRRLLRLRL"
Output: 4
Explanation: s can be split into "RL", "RRLL", "RL", "RL", each substring contains same number of 'L' and 'R'.

Example 2:

Input: s = "RLRRRLLRLL"
Output: 2
Explanation: s can be split into "RL", "RRRLLRLL", each substring contains same number of 'L' and 'R'.
Note that s cannot be split into "RL", "RR", "RL", "LR", "LL", because the 2^nd and 5^th substrings are not balanced.

Example 3:

Input: s = "LLLLRRRR"
Output: 1
Explanation: s can be split into "LLLLRRRR".

Constraints:

2 <= s.length <= 1000
s[i] is either 'L' or 'R'.
s is a balanced string.

Try coding it in LeetCode

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 are the allowed characters in the input string, and what is the maximum length of the input string?
Is an empty string a valid input, and if so, what should the output be?
What is the definition of a 'balanced string' in this context? Does it strictly mean an equal number of 'L' and 'R' characters?
If the string cannot be split into any balanced strings, what should the return value be?
Are we looking for the maximum number of balanced strings, or is any valid split acceptable?

Brute Force Solution

Approach

The brute force method for this problem involves examining every possible way to cut the string into smaller pieces. We then check each of these possible splits to see if it is 'balanced' according to the problem's rules. Finally, we count how many balanced splits we found.

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

Start by considering the very first part of the string: just the first character.
Check if this first character by itself is balanced according to the problem's rules.
If not, move on. If it is, increase the count of balanced strings by one.
Now, consider the first two characters of the string as one part.
Check if this two-character part is balanced.
Repeat, each time increasing the size of the first part of the string by one character, until you reach the end of the original string.
For each possible 'first part' of the string, consider the 'remaining part' of the string and recursively repeat this entire process to find balanced strings within it.
Keep track of the total number of balanced string splits found during the entire process.

Code Implementation

def split_balanced_string_brute_force(input_string):
    balanced_strings_count = 0

    def is_balanced(substring):
        right_count = 0
        left_count = 0

        for char in substring:
            if char == 'R':
                right_count += 1
            elif char == 'L':
                left_count += 1

        return right_count == left_count

    def recursively_count_balanced_strings(current_string):
        nonlocal balanced_strings_count

        if not current_string:
            return

        for i in range(1, len(current_string) + 1):
            first_part = current_string[:i]

            # Check if the current substring is balanced
            if is_balanced(first_part):
                balanced_strings_count += 1

            # Recursively call the function to process the remaining part
            remaining_part = current_string[i:]
            recursively_count_balanced_strings(remaining_part)

    recursively_count_balanced_strings(input_string)

    # The total number of balanced substrings
    return balanced_strings_count

Big(O) Analysis

Time Complexity

O(2^n) – The provided brute force approach explores all possible splits of the string. For a string of length n, there are 2^(n-1) possible ways to split it. For each split, we need to iterate through the resulting substrings to check if they are balanced, which takes O(n) time in the worst case. Therefore, the overall time complexity is O(n * 2^(n-1)), which simplifies to O(2^n) since the exponential term dominates.

Space Complexity

O(N) – The dominant space complexity comes from the recursive calls. In the worst-case scenario, the recursive function might be called for each possible prefix of the string. This could lead to a maximum recursion depth of N, where N is the length of the input string. Therefore, the auxiliary space required for the call stack is O(N).

Optimal Solution

Approach

The key idea is to keep track of the balance between two characters. We walk through the string, increasing the balance when we see one character and decreasing it when we see the other. When the balance returns to zero, we've found a balanced string.

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

Start with a balance of zero and a counter for the number of balanced strings, which also starts at zero.
Look at each character in the string, one at a time, going from left to right.
If the character is the first type, increase the balance by one.
If the character is the second type, decrease the balance by one.
If the balance ever becomes zero, it means we've found a balanced string, so increase the balanced string counter by one.
After looking at every character, the balanced string counter will hold the number of balanced strings we found.

Code Implementation

def balancedStringSplit(input_string):
    balance = 0
    number_of_balanced_strings = 0

    for character in input_string:
        if character == 'R':
            balance += 1

        else:
            balance -= 1

        # Count when balance returns to zero
        if balance == 0:
            number_of_balanced_strings += 1

    return number_of_balanced_strings

Big(O) Analysis

Time Complexity

O(n) – The provided solution iterates through the input string once, examining each character. For each character, it performs a constant-time operation: incrementing or decrementing the balance. The number of balanced strings is updated only when the balance returns to zero, which is also a constant-time operation. Therefore, the time complexity is directly proportional to the length of the string, n, resulting in O(n).

Space Complexity

O(1) – The algorithm uses a balance counter and a balanced string counter. These are the only variables used for auxiliary space. Since the amount of extra memory remains constant regardless of the input string's length N, the space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input string	Return 0 since an empty string cannot be split into balanced strings.
String with length 1 or 2	If length is 1, return 0; if length is 2 and unbalanced, return 0; if length is 2 and balanced, return 1.
Maximum string length (e.g., 1000 as defined in many constraints)	The linear time complexity of the solution should efficiently handle strings up to the maximum length without exceeding time limits.
String with all 'L' characters	Return 0 since there are no 'R' characters to balance the 'L' characters.
String with all 'R' characters	Return 0 since there are no 'L' characters to balance the 'R' characters.
String with an extremely unbalanced distribution of 'L' and 'R' characters, e.g., 'LLLLLLLLLLR'	The counter will reach a large imbalance before potentially balancing later, and the algorithm correctly handles this scenario.
String where no balanced split is possible, e.g., 'LLRRLL'	The algorithm correctly identifies that a balanced split is never achieved, so returns the accumulated count which could be 0 in this instance.
Very long string composed of repeating 'LR' or 'RL' patterns	The linear time complexity ensures that even long repeating patterns are processed efficiently.

Null or empty input string

How to Handle:

Return 0 since an empty string cannot be split into balanced strings.

String with length 1 or 2

How to Handle:

If length is 1, return 0; if length is 2 and unbalanced, return 0; if length is 2 and balanced, return 1.

Maximum string length (e.g., 1000 as defined in many constraints)

How to Handle:

The linear time complexity of the solution should efficiently handle strings up to the maximum length without exceeding time limits.

String with all 'L' characters

How to Handle:

Return 0 since there are no 'R' characters to balance the 'L' characters.

String with all 'R' characters

How to Handle:

Return 0 since there are no 'L' characters to balance the 'R' characters.

String with an extremely unbalanced distribution of 'L' and 'R' characters, e.g., 'LLLLLLLLLLR'

How to Handle:

The counter will reach a large imbalance before potentially balancing later, and the algorithm correctly handles this scenario.

String where no balanced split is possible, e.g., 'LLRRLL'

How to Handle:

The algorithm correctly identifies that a balanced split is never achieved, so returns the accumulated count which could be 0 in this instance.

Very long string composed of repeating 'LR' or 'RL' patterns

How to Handle:

The linear time complexity ensures that even long repeating patterns are processed efficiently.

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 are the allowed characters in the input string, and what is the maximum length of the input string?
Is an empty string a valid input, and if so, what should the output be?
What is the definition of a 'balanced string' in this context? Does it strictly mean an equal number of 'L' and 'R' characters?
If the string cannot be split into any balanced strings, what should the return value be?
Are we looking for the maximum number of balanced strings, or is any valid split acceptable?

Brute Force Solution

Approach

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

Start by considering the very first part of the string: just the first character.
Check if this first character by itself is balanced according to the problem's rules.
If not, move on. If it is, increase the count of balanced strings by one.
Now, consider the first two characters of the string as one part.
Check if this two-character part is balanced.
Repeat, each time increasing the size of the first part of the string by one character, until you reach the end of the original string.
For each possible 'first part' of the string, consider the 'remaining part' of the string and recursively repeat this entire process to find balanced strings within it.
Keep track of the total number of balanced string splits found during the entire process.

Code Implementation

def split_balanced_string_brute_force(input_string):
    balanced_strings_count = 0

    def is_balanced(substring):
        right_count = 0
        left_count = 0

        for char in substring:
            if char == 'R':
                right_count += 1
            elif char == 'L':
                left_count += 1

        return right_count == left_count

    def recursively_count_balanced_strings(current_string):
        nonlocal balanced_strings_count

        if not current_string:
            return

        for i in range(1, len(current_string) + 1):
            first_part = current_string[:i]

            # Check if the current substring is balanced
            if is_balanced(first_part):
                balanced_strings_count += 1

            # Recursively call the function to process the remaining part
            remaining_part = current_string[i:]
            recursively_count_balanced_strings(remaining_part)

    recursively_count_balanced_strings(input_string)

    # The total number of balanced substrings
    return balanced_strings_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start with a balance of zero and a counter for the number of balanced strings, which also starts at zero.
Look at each character in the string, one at a time, going from left to right.
If the character is the first type, increase the balance by one.
If the character is the second type, decrease the balance by one.
If the balance ever becomes zero, it means we've found a balanced string, so increase the balanced string counter by one.
After looking at every character, the balanced string counter will hold the number of balanced strings we found.

Code Implementation

def balancedStringSplit(input_string):
    balance = 0
    number_of_balanced_strings = 0

    for character in input_string:
        if character == 'R':
            balance += 1

        else:
            balance -= 1

        # Count when balance returns to zero
        if balance == 0:
            number_of_balanced_strings += 1

    return number_of_balanced_strings

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input string	Return 0 since an empty string cannot be split into balanced strings.
String with length 1 or 2	If length is 1, return 0; if length is 2 and unbalanced, return 0; if length is 2 and balanced, return 1.
Maximum string length (e.g., 1000 as defined in many constraints)	The linear time complexity of the solution should efficiently handle strings up to the maximum length without exceeding time limits.
String with all 'L' characters	Return 0 since there are no 'R' characters to balance the 'L' characters.
String with all 'R' characters	Return 0 since there are no 'L' characters to balance the 'R' characters.
String with an extremely unbalanced distribution of 'L' and 'R' characters, e.g., 'LLLLLLLLLLR'	The counter will reach a large imbalance before potentially balancing later, and the algorithm correctly handles this scenario.
String where no balanced split is possible, e.g., 'LLRRLL'	The algorithm correctly identifies that a balanced split is never achieved, so returns the accumulated count which could be 0 in this instance.
Very long string composed of repeating 'LR' or 'RL' patterns	The linear time complexity ensures that even long repeating patterns are processed efficiently.

Null or empty input string

How to Handle:

Return 0 since an empty string cannot be split into balanced strings.

String with length 1 or 2

How to Handle:

If length is 1, return 0; if length is 2 and unbalanced, return 0; if length is 2 and balanced, return 1.

Maximum string length (e.g., 1000 as defined in many constraints)

How to Handle:

The linear time complexity of the solution should efficiently handle strings up to the maximum length without exceeding time limits.

String with all 'L' characters

How to Handle:

Return 0 since there are no 'R' characters to balance the 'L' characters.

String with all 'R' characters

How to Handle:

Return 0 since there are no 'L' characters to balance the 'R' characters.

String with an extremely unbalanced distribution of 'L' and 'R' characters, e.g., 'LLLLLLLLLLR'

How to Handle:

The counter will reach a large imbalance before potentially balancing later, and the algorithm correctly handles this scenario.

String where no balanced split is possible, e.g., 'LLRRLL'

How to Handle:

The algorithm correctly identifies that a balanced split is never achieved, so returns the accumulated count which could be 0 in this instance.

Very long string composed of repeating 'LR' or 'RL' patterns

How to Handle:

The linear time complexity ensures that even long repeating patterns are processed efficiently.