Jobs

Interviews

Discussions

Maximum Score After Splitting a String

Easy

Asked by:

More companies

98 views

Topics:

StringsArrays

Given a string s of zeros and ones, return the maximum score after splitting the string into two non-empty substrings (i.e. left substring and right substring).

The score after splitting a string is the number of zeros in the left substring plus the number of ones in the right substring.

Example 1:

Input: s = "011101"
Output: 5 
Explanation: 
All possible ways of splitting s into two non-empty substrings are:
left = "0" and right = "11101", score = 1 + 4 = 5 
left = "01" and right = "1101", score = 1 + 3 = 4 
left = "011" and right = "101", score = 1 + 2 = 3 
left = "0111" and right = "01", score = 1 + 1 = 2 
left = "01110" and right = "1", score = 2 + 1 = 3

Example 2:

Input: s = "00111"
Output: 5
Explanation: When left = "00" and right = "111", we get the maximum score = 2 + 3 = 5

Example 3:

Input: s = "1111"
Output: 3

Constraints:

2 <= s.length <= 500
The string s consists of characters '0' and '1' only.

Try coding it in LeetCode

Maximum Score After Splitting a String

Easy

Asked by:

98 views

Topics:

StringsArrays

Given a string s of zeros and ones, return the maximum score after splitting the string into two non-empty substrings (i.e. left substring and right substring).

The score after splitting a string is the number of zeros in the left substring plus the number of ones in the right substring.

Example 1:

Input: s = "011101"
Output: 5 
Explanation: 
All possible ways of splitting s into two non-empty substrings are:
left = "0" and right = "11101", score = 1 + 4 = 5 
left = "01" and right = "1101", score = 1 + 3 = 4 
left = "011" and right = "101", score = 1 + 2 = 3 
left = "0111" and right = "01", score = 1 + 1 = 2 
left = "01110" and right = "1", score = 2 + 1 = 3

Example 2:

Input: s = "00111"
Output: 5
Explanation: When left = "00" and right = "111", we get the maximum score = 2 + 3 = 5

Example 3:

Input: s = "1111"
Output: 3

Constraints:

2 <= s.length <= 500
The string s consists of characters '0' and '1' only.

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 is the maximum length of the input string `s`?
Does the input string `s` only contain '0's and '1's, or can it contain other characters?
If there are multiple splits that result in the same maximum score, can I return any of them, or is there a specific split I should prioritize?
What should I return if the input string `s` is empty?
Is the goal to minimize space complexity as well, or is the time complexity the primary concern?

Brute Force Solution

Approach

The brute force approach to this problem is like trying every single way to cut the string into two pieces. We calculate a score for each way we cut it. The goal is to find the highest possible score across all the cuts.

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

First, imagine making a cut after the very first character in the string.
Then, count the number of zeros on the left side of the cut and the number of ones on the right side of the cut.
Add these two numbers together to get the score for this particular cut.
Next, imagine moving the cut one position to the right, and repeat the counting and adding process to get a new score.
Keep doing this, moving the cut one position at a time, until you've tried every possible place to make the cut (up to, but not including, cutting after the very last character).
Finally, compare all the scores you calculated, and choose the highest score. That's your answer!

Code Implementation

def max_score_after_splitting_string(string_of_zeros_and_ones):
    max_score = 0
    
    # Iterate through all possible split positions
    for split_position in range(1, len(string_of_zeros_and_ones)): 
        left_substring = string_of_zeros_and_ones[:split_position]
        right_substring = string_of_zeros_and_ones[split_position:]
        
        # Count zeros in the left substring.
        zeros_in_left = left_substring.count('0')

        # Count ones in the right substring
        ones_in_right = right_substring.count('1')

        current_score = zeros_in_left + ones_in_right

        #Update max score if we found a better score
        if current_score > max_score:
            max_score = current_score

    return max_score

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the string of length n, representing each possible split point. For each split point, it counts the zeros on the left and the ones on the right. Counting the zeros and ones requires iterating over the left and right substrings respectively. Since the total string length is n, the counting operations in each iteration together will take O(n) time. Because there are n splits, but the total calculations in a single split equals O(n), the overall complexity can be written as O(n). However, the prompt says the counting happens on either side of the cut. To get the O(n) solution, you would need to do some preprocessing. Considering the given brute force approach, inside the loop which iterates through each element, the number of zeros on the left and ones on the right need to be calculated by traversing these substrings. In the worst-case, where the cut is near the beginning, the right substring is nearly the entire string, causing an n-complexity operation. With n possible cuts, the total runtime complexity is O(n^2). So, the initial O(n) analysis is not correct given the prompt's explanation.

Space Complexity

O(1) – The algorithm iterates through the string, keeping track of the maximum score found so far. It calculates the score for each split by counting zeros on the left and ones on the right. No auxiliary data structures like arrays, hash maps, or lists are created to store intermediate results or visited states. Therefore, the space used remains constant regardless of the input string's length (N).

Optimal Solution

Approach

The goal is to find the best spot to cut the string so the score is highest. The clever trick is to count the number of zeros on the left and ones on the right without recalculating everything each time, saving a lot of work.

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

First, count all the ones in the entire string. This is a reference point for the right side after the split.
Next, walk through the string one character at a time, imagining a cut at each position.
As you move, keep track of how many zeros you've seen on the left side.
Simultaneously, as you move, subtract from your initial count of ones on the right, based on the actual one or zero value you just parsed.
At each possible cut position, calculate the 'score' by adding the number of zeros on the left and the number of ones on the right.
Remember the highest score you've seen so far as you move through the string.
After checking every possible split position, the highest score you remembered is your answer.

Code Implementation

def max_score_after_split(string_to_split: str) -> int:
    all_ones_count = 0
    for character in string_to_split:
        if character == '1':
            all_ones_count += 1

    left_zeros_count = 0
    maximum_score = 0

    for i in range(len(string_to_split) - 1):
        # Update zeros count for left substring.
        if string_to_split[i] == '0':
            left_zeros_count += 1

        # Update ones count for right substring.
        if string_to_split[i] == '1':
            all_ones_count -= 1

        # Calculate and update maximum score.
        current_score = left_zeros_count + all_ones_count

        # Keep track of the maximum score seen so far
        if current_score > maximum_score:
            maximum_score = current_score

    return maximum_score

Big(O) Analysis

Time Complexity

O(n) – The algorithm first iterates through the string of length n once to count all the ones. Then, it iterates through the string again, performing a constant number of operations (incrementing zero count, decrementing one count, and calculating the score) for each character. Because the algorithm iterates through the string only twice, performing constant-time operations within each iteration, the time complexity is directly proportional to the length of the string, n. Therefore, the time complexity is O(n).

Space Complexity

O(1) – The algorithm uses a constant amount of extra space. It stores the total number of ones, the current count of zeros on the left, and the maximum score seen so far. These variables consume a fixed amount of memory regardless of the input string's length (N), so the auxiliary space is constant.

Edge Cases

Case	How to Handle
Null or empty string input	Return 0 immediately as no split is possible.
String of length 1	Return 0 immediately as no split is possible.
String containing only '0' characters	The maximum score will be length - 1, calculated correctly by counting zeros on the left and ones on the right.
String containing only '1' characters	The maximum score will be 0, calculated correctly as there are no zeros on the left.
String with equal distribution of '0' and '1'	The algorithm correctly iterates to find the optimal split point maximizing the sum of zeros on the left and ones on the right.
Very long string (close to maximum allowed string length)	The linear time complexity (O(n)) of the prefix sum and iteration should scale reasonably well.
String with leading or trailing zeros	The algorithm correctly counts zeros from the left and ones from the right, regardless of their position.
Integer overflow when calculating left and right counts	Using integers for left and right counts is sufficient as string length is constrained by problem description, preventing integer overflow.

Null or empty string input

How to Handle:

Return 0 immediately as no split is possible.

String of length 1

How to Handle:

Return 0 immediately as no split is possible.

String containing only '0' characters

How to Handle:

The maximum score will be length - 1, calculated correctly by counting zeros on the left and ones on the right.

String containing only '1' characters

How to Handle:

The maximum score will be 0, calculated correctly as there are no zeros on the left.

String with equal distribution of '0' and '1'

How to Handle:

The algorithm correctly iterates to find the optimal split point maximizing the sum of zeros on the left and ones on the right.

Very long string (close to maximum allowed string length)

How to Handle:

The linear time complexity (O(n)) of the prefix sum and iteration should scale reasonably well.

String with leading or trailing zeros

How to Handle:

The algorithm correctly counts zeros from the left and ones from the right, regardless of their position.

Integer overflow when calculating left and right counts

How to Handle:

Using integers for left and right counts is sufficient as string length is constrained by problem description, preventing integer overflow.

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 string `s`?
Does the input string `s` only contain '0's and '1's, or can it contain other characters?
If there are multiple splits that result in the same maximum score, can I return any of them, or is there a specific split I should prioritize?
What should I return if the input string `s` is empty?
Is the goal to minimize space complexity as well, or is the time complexity the primary concern?

Brute Force Solution

Approach

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

First, imagine making a cut after the very first character in the string.
Then, count the number of zeros on the left side of the cut and the number of ones on the right side of the cut.
Add these two numbers together to get the score for this particular cut.
Next, imagine moving the cut one position to the right, and repeat the counting and adding process to get a new score.
Keep doing this, moving the cut one position at a time, until you've tried every possible place to make the cut (up to, but not including, cutting after the very last character).
Finally, compare all the scores you calculated, and choose the highest score. That's your answer!

Code Implementation

def max_score_after_splitting_string(string_of_zeros_and_ones):
    max_score = 0
    
    # Iterate through all possible split positions
    for split_position in range(1, len(string_of_zeros_and_ones)): 
        left_substring = string_of_zeros_and_ones[:split_position]
        right_substring = string_of_zeros_and_ones[split_position:]
        
        # Count zeros in the left substring.
        zeros_in_left = left_substring.count('0')

        # Count ones in the right substring
        ones_in_right = right_substring.count('1')

        current_score = zeros_in_left + ones_in_right

        #Update max score if we found a better score
        if current_score > max_score:
            max_score = current_score

    return max_score

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, count all the ones in the entire string. This is a reference point for the right side after the split.
Next, walk through the string one character at a time, imagining a cut at each position.
As you move, keep track of how many zeros you've seen on the left side.
Simultaneously, as you move, subtract from your initial count of ones on the right, based on the actual one or zero value you just parsed.
At each possible cut position, calculate the 'score' by adding the number of zeros on the left and the number of ones on the right.
Remember the highest score you've seen so far as you move through the string.
After checking every possible split position, the highest score you remembered is your answer.

Code Implementation

def max_score_after_split(string_to_split: str) -> int:
    all_ones_count = 0
    for character in string_to_split:
        if character == '1':
            all_ones_count += 1

    left_zeros_count = 0
    maximum_score = 0

    for i in range(len(string_to_split) - 1):
        # Update zeros count for left substring.
        if string_to_split[i] == '0':
            left_zeros_count += 1

        # Update ones count for right substring.
        if string_to_split[i] == '1':
            all_ones_count -= 1

        # Calculate and update maximum score.
        current_score = left_zeros_count + all_ones_count

        # Keep track of the maximum score seen so far
        if current_score > maximum_score:
            maximum_score = current_score

    return maximum_score

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty string input	Return 0 immediately as no split is possible.
String of length 1	Return 0 immediately as no split is possible.
String containing only '0' characters	The maximum score will be length - 1, calculated correctly by counting zeros on the left and ones on the right.
String containing only '1' characters	The maximum score will be 0, calculated correctly as there are no zeros on the left.
String with equal distribution of '0' and '1'	The algorithm correctly iterates to find the optimal split point maximizing the sum of zeros on the left and ones on the right.
Very long string (close to maximum allowed string length)	The linear time complexity (O(n)) of the prefix sum and iteration should scale reasonably well.
String with leading or trailing zeros	The algorithm correctly counts zeros from the left and ones from the right, regardless of their position.
Integer overflow when calculating left and right counts	Using integers for left and right counts is sufficient as string length is constrained by problem description, preventing integer overflow.

Null or empty string input

How to Handle:

Return 0 immediately as no split is possible.

String of length 1

How to Handle:

Return 0 immediately as no split is possible.

String containing only '0' characters

How to Handle:

The maximum score will be length - 1, calculated correctly by counting zeros on the left and ones on the right.

String containing only '1' characters

How to Handle:

The maximum score will be 0, calculated correctly as there are no zeros on the left.

String with equal distribution of '0' and '1'

How to Handle:

The algorithm correctly iterates to find the optimal split point maximizing the sum of zeros on the left and ones on the right.

Very long string (close to maximum allowed string length)

How to Handle:

The linear time complexity (O(n)) of the prefix sum and iteration should scale reasonably well.

String with leading or trailing zeros

How to Handle:

The algorithm correctly counts zeros from the left and ones from the right, regardless of their position.

Integer overflow when calculating left and right counts

How to Handle:

Using integers for left and right counts is sufficient as string length is constrained by problem description, preventing integer overflow.