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Number of Good Ways to Split a String

Medium

Amazon

2 views

Topics:

ArraysStringsTwo Pointers

You are given a string s.

A split is called good if you can split s into two non-empty strings s_left and s_right where their concatenation is equal to s (i.e., s_left + s_right = s) and the number of distinct letters in s_left and s_right is the same.

Return the number of good splits you can make in s.

For example:

s = "aacaba" should return 2. The good splits are ("aac", "aba") and ("aaca", "ba")
s = "abcd" should return 1 The good split is ("ab", "cd")

Write a function to implement this with the most optimal time and space complexity.

Number of Good Ways to Split a String

Medium

Amazon

2 views

Topics:

ArraysStringsTwo Pointers

You are given a string s.

Return the number of good splits you can make in s.

For example:

s = "aacaba" should return 2. The good splits are ("aac", "aba") and ("aaca", "ba")
s = "abcd" should return 1 The good split is ("ab", "cd")

Write a function to implement this with the most optimal time and space complexity.

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?
Is the input string guaranteed to contain only lowercase English letters?
If there are no good ways to split the string, what should I return?
Could you define more precisely what constitutes a 'good' split if the number of distinct characters on either side is ambiguous?
Does the split have to be non-empty substrings on both sides?

Brute Force Solution

Approach

The brute force strategy for this problem is all about trying out every single way you can split the given string into two parts. For each split, we check if the two parts are 'good' according to the problem's definition, and count the valid ones.

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

Think of cutting the string at every possible position, one at a time.
For each cut, count the number of different characters on the left side of the cut.
Then, count the number of different characters on the right side of the cut.
If the number of different characters on the left side is the same as the number of different characters on the right side, then we found one 'good' split.
Do this for every single possible place you can cut the string.
In the end, count all the 'good' splits you found.

Code Implementation

def num_good_ways_to_split_a_string_brute_force(input_string):
    number_of_good_splits = 0
    string_length = len(input_string)

    for split_index in range(1, string_length):
        # Iterate through all possible split positions
        left_substring = input_string[:split_index]
        right_substring = input_string[split_index:]

        distinct_characters_left = set(left_substring)
        number_distinct_left = len(distinct_characters_left)

        # Count distinct characters on the left

        distinct_characters_right = set(right_substring)
        number_distinct_right = len(distinct_characters_right)

        # Count distinct characters on the right
        if number_distinct_left == number_distinct_right:
            # Only increment if the split is good

            number_of_good_splits += 1

    return number_of_good_splits

Big(O) Analysis

Time Complexity

O(n²) – The described brute force solution iterates through all possible split positions in the string of length n. For each split, it counts the distinct characters on the left and right sides. Counting distinct characters on each side involves iterating through at most n characters. Since we iterate through n possible split positions and perform an O(n) operation to count distinct characters for each split, the overall time complexity is O(n * n). Therefore, the time complexity simplifies to O(n²).

Space Complexity

O(1) – The described brute force solution iterates through all possible splits of the string and counts distinct characters on both sides for each split. The number of distinct characters is counted using a fixed number of variables (potentially frequency counts up to the size of the alphabet, which is constant). No auxiliary data structures like arrays or hash maps that scale with the input string's length (N) are created or used to store intermediate results during the splitting and counting process. Therefore, the space complexity is constant.

Optimal Solution

Approach

To efficiently count the splits, we focus on tracking distinct characters from both ends of the string. By comparing these counts, we can determine valid split points quickly. This avoids repeatedly counting distinct characters for every possible split.

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

First, figure out for each position in the string, how many unique characters are in the string from the beginning up to that position.
Do the same thing, but this time, start from the end of the string and go backwards, figuring out how many unique characters there are from the end up to each position.
Now, go through the string again, looking at each position as a potential split point.
At each position, compare the number of unique characters in the first part (from the beginning up to that position) with the number of unique characters in the second part (from that position to the end).
If those two numbers are the same, that means we have a good split, so increment a counter.
After checking all the positions, the counter will hold the total number of good splits.

Code Implementation

def num_good_splits(string): 
    string_length = len(string)
    prefix_unique_counts = [0] * string_length
    suffix_unique_counts = [0] * string_length

    seen_characters = set()
    for i in range(string_length): 
        seen_characters.add(string[i])
        prefix_unique_counts[i] = len(seen_characters)

    seen_characters = set()
    for i in range(string_length - 1, -1, -1): 
        seen_characters.add(string[i])
        suffix_unique_counts[i] = len(seen_characters)

    good_split_count = 0

    # Iterate through possible split points.
    for i in range(string_length - 1): 
        # Comparing unique char counts.
        if prefix_unique_counts[i] == suffix_unique_counts[i + 1]:
            good_split_count += 1

    return good_split_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the string of length n three times. The first pass calculates the number of distinct characters from the beginning up to each position. The second pass does the same from the end to each position. The third pass iterates through the string again to check the number of unique characters at the split point. Each of these passes takes O(n) time, resulting in a total time complexity of O(n) + O(n) + O(n) which simplifies to O(n).

Space Complexity

O(N) – The algorithm uses two arrays, `left_unique` and `right_unique`, to store the number of unique characters encountered up to each position from the left and right ends of the string, respectively. Each of these arrays has a size equal to the length of the input string, N. Therefore, the auxiliary space required is proportional to N, leading to a space complexity of O(N).

Edge Cases

Case	How to Handle
Null or empty string	Return 0 if the input string is null or empty as there are no ways to split it.
String with length 1	Return 0 because a string of length 1 cannot be split into two non-empty substrings.
String with all identical characters	Return string length - 1 as every split will result in same number of unique characters on both sides.
Very long string approaching maximum allowed string length	Ensure the solution uses efficient data structures (e.g., hash maps) to avoid time limit exceeded errors.
String with only two distinct characters and skewed distribution	The solution should handle skewed distributions of characters without causing significant performance degradation.
Integer overflow when counting unique characters (if applicable)	Use appropriate data types to store counts to prevent integer overflow.
Memory constraints with extremely long strings	Optimize memory usage by avoiding unnecessary data duplication or large intermediate data structures.
String contains unicode characters	The solution should correctly count unicode characters as distinct elements in the string.

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?
Is the input string guaranteed to contain only lowercase English letters?
If there are no good ways to split the string, what should I return?
Could you define more precisely what constitutes a 'good' split if the number of distinct characters on either side is ambiguous?
Does the split have to be non-empty substrings on both sides?

Brute Force Solution

Approach

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

Think of cutting the string at every possible position, one at a time.
For each cut, count the number of different characters on the left side of the cut.
Then, count the number of different characters on the right side of the cut.
If the number of different characters on the left side is the same as the number of different characters on the right side, then we found one 'good' split.
Do this for every single possible place you can cut the string.
In the end, count all the 'good' splits you found.

Code Implementation

def num_good_ways_to_split_a_string_brute_force(input_string):
    number_of_good_splits = 0
    string_length = len(input_string)

    for split_index in range(1, string_length):
        # Iterate through all possible split positions
        left_substring = input_string[:split_index]
        right_substring = input_string[split_index:]

        distinct_characters_left = set(left_substring)
        number_distinct_left = len(distinct_characters_left)

        # Count distinct characters on the left

        distinct_characters_right = set(right_substring)
        number_distinct_right = len(distinct_characters_right)

        # Count distinct characters on the right
        if number_distinct_left == number_distinct_right:
            # Only increment if the split is good

            number_of_good_splits += 1

    return number_of_good_splits

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, figure out for each position in the string, how many unique characters are in the string from the beginning up to that position.
Do the same thing, but this time, start from the end of the string and go backwards, figuring out how many unique characters there are from the end up to each position.
Now, go through the string again, looking at each position as a potential split point.
At each position, compare the number of unique characters in the first part (from the beginning up to that position) with the number of unique characters in the second part (from that position to the end).
If those two numbers are the same, that means we have a good split, so increment a counter.
After checking all the positions, the counter will hold the total number of good splits.

Code Implementation

def num_good_splits(string): 
    string_length = len(string)
    prefix_unique_counts = [0] * string_length
    suffix_unique_counts = [0] * string_length

    seen_characters = set()
    for i in range(string_length): 
        seen_characters.add(string[i])
        prefix_unique_counts[i] = len(seen_characters)

    seen_characters = set()
    for i in range(string_length - 1, -1, -1): 
        seen_characters.add(string[i])
        suffix_unique_counts[i] = len(seen_characters)

    good_split_count = 0

    # Iterate through possible split points.
    for i in range(string_length - 1): 
        # Comparing unique char counts.
        if prefix_unique_counts[i] == suffix_unique_counts[i + 1]:
            good_split_count += 1

    return good_split_count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty string	Return 0 if the input string is null or empty as there are no ways to split it.
String with length 1	Return 0 because a string of length 1 cannot be split into two non-empty substrings.
String with all identical characters	Return string length - 1 as every split will result in same number of unique characters on both sides.
Very long string approaching maximum allowed string length	Ensure the solution uses efficient data structures (e.g., hash maps) to avoid time limit exceeded errors.
String with only two distinct characters and skewed distribution	The solution should handle skewed distributions of characters without causing significant performance degradation.
Integer overflow when counting unique characters (if applicable)	Use appropriate data types to store counts to prevent integer overflow.
Memory constraints with extremely long strings	Optimize memory usage by avoiding unnecessary data duplication or large intermediate data structures.
String contains unicode characters	The solution should correctly count unicode characters as distinct elements in the string.