Hash Divided String

Medium

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Strings

You are given a string s of length n and an integer k, where n is a multiple of k. Your task is to hash the string s into a new string called result, which has a length of n / k.

First, divide s into n / k substrings, each with a length of k. Then, initialize result as an empty string.

For each substring in order from the beginning:

The hash value of a character is the index of that character in the English alphabet (e.g., 'a' → 0, 'b' → 1, ..., 'z' → 25).
Calculate the sum of all the hash values of the characters in the substring.
Find the remainder of this sum when divided by 26, which is called hashedChar.
Identify the character in the English lowercase alphabet that corresponds to hashedChar.
Append that character to the end of result.

Return result.

Example 1:

Input: s = "abcd", k = 2

Output: "bf"

Explanation:

First substring: "ab", 0 + 1 = 1, 1 % 26 = 1, result[0] = 'b'.

Second substring: "cd", 2 + 3 = 5, 5 % 26 = 5, result[1] = 'f'.

Example 2:

Input: s = "mxz", k = 3

Output: "i"

Explanation:

The only substring: "mxz", 12 + 23 + 25 = 60, 60 % 26 = 8, result[0] = 'i'.

Constraints:

1 <= k <= 100
k <= s.length <= 1000
s.length is divisible by k.
s consists only of lowercase English letters.

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:

Can you clarify the expected range of values within the input strings, and what characters might be included (e.g., only lowercase letters, alphanumeric, special characters)?
What should be returned if no valid division of the string based on equal hash values exists? Should I return null, an empty array, or throw an exception?
How is the hash function defined, or can I choose my own? If I can choose, are there any constraints or suggestions for the hash function to avoid collisions or overflow?
If multiple valid divisions of the string exist, should I return only one, or all of them? If only one, is there a preference for which division to return (e.g., the first, the shortest, the longest)?
Can the input string be empty or null? If so, what is the expected behavior in these cases?

Brute Force Solution

Approach

The brute force strategy for this problem involves trying every conceivable way to split the string. It explores all possible divisions, checking each one to see if it meets the problem's requirements. It’s like testing every single possibility, no matter how inefficient, until a valid solution is found.

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

Consider the first possible split: the first character as one part and the rest of the string as another part.
Check if both parts satisfy the hashing requirement (that both parts' hash values can be equal).
If the hashing requirement is satisfied, keep track of this split as a possible solution.
Now, consider the next possible split: the first two characters as one part and the rest of the string as the other part.
Again, check if the hash values of the two resulting parts are equal. If so, record this split.
Continue this process, increasing the size of the first part by one character at a time, until the first part includes all but one character of the original string.
After examining all possible splits, check all of the recorded splits and return 'yes' if there is one or more recorded split, otherwise return 'no'.

Code Implementation

def hash_divided_string_brute_force(input_string):
    possible_splits = []
    string_length = len(input_string)

    for first_part_length in range(1, string_length):
        first_part = input_string[:first_part_length]
        second_part = input_string[first_part_length:]

        # Calculate the hash values for each substring.
        first_part_hash = hash(first_part)
        second_part_hash = hash(second_part)

        # Determine if the hash values are equal, indicating a valid split.
        if first_part_hash == second_part_hash:
            possible_splits.append((first_part, second_part))

    # Check if there is at least one valid split.
    if len(possible_splits) > 0:
        return 'yes'
    else:
        return 'no'

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible splits of the string. For a string of length n, there are n-1 possible split points. For each split, calculating the hash value of both substrings takes O(n) time in the worst case, as it involves iterating through the characters of each substring. Since we have n-1 splits and each split requires O(n) hash calculations, the overall time complexity is O(n * n), which simplifies to O(n²).

Space Complexity

O(N) – The algorithm records possible splits that satisfy the hashing requirement. In the worst-case scenario, every possible split might satisfy the condition, leading to storing approximately N-1 splits. Since the splits are not explicitly stored (the plain English only mentions 'keeping track'), let's assume we implicitly store at most N split positions. Thus, auxiliary space grows linearly with the input size N, where N is the length of the input string.

Optimal Solution

Approach

The core idea is to keep track of the sums of the substrings. We calculate a rolling hash for the original string and compare it to the rolling hashes of potential substrings to efficiently check for matches. This avoids repeatedly calculating the hash of the same substrings.

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

First, compute a hash value for the original string using a good hashing function.
Then, go through all the possible substring lengths.
For each length, calculate the hash value for the substring from the beginning of the original string.
Now, move through the original string, calculating the hash value for each substring of the current length using a 'rolling hash' technique (updating the hash from the previous substring rather than recomputing it from scratch).
Compare the hash value of each substring to the hash value of the original string.
If the hash values match, check if the substrings are actually equal to avoid collisions (rare but possible).
Count how many substrings are equal to the original string. The substring is equal to the original string if its length matches the length of the original string and its value is the same.
Return the count.

Code Implementation

def hash_divided_string(original_string):
    string_length = len(original_string)
    if string_length == 0:
        return 0

    prime_number = 31
    modulo_value = 10**9 + 9

    # Calculate hash value for the original string.
    original_string_hash = 0
    for char in original_string:
        original_string_hash = (original_string_hash * prime_number + ord(char)) % modulo_value

    substring_count = 0

    for substring_length in range(1, string_length + 1):
        # Check substring length matches original string length.
        if substring_length == string_length:

            substring_hash = 0
            for i in range(substring_length):
                substring_hash = (substring_hash * prime_number + ord(original_string[i])) % modulo_value

            if substring_hash == original_string_hash:
                if original_string[:substring_length] == original_string:
                    substring_count += 1

    return substring_count

Big(O) Analysis

Time Complexity

O(n²) – The solution iterates through all possible substring lengths, which can range from 1 to n (the length of the original string). For each substring length, it iterates through the original string to calculate the rolling hash of each substring of that length. The outer loop iterates up to n times, and the inner loop iterates up to n times to generate and compare substring hashes. Comparing the substring (in case of a hash collision) takes O(n) time, but this is inside the nested loops. Thus the dominant cost comes from the nested loops resulting in O(n*n) or O(n²) time complexity.

Space Complexity

O(1) – The algorithm primarily uses a few variables to store hash values, substring lengths, and counters. These variables occupy a constant amount of space, irrespective of the input string's length, denoted as N. No auxiliary data structures, like arrays or hash maps, are created that scale with the input size. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input string	Return an empty list or throw an IllegalArgumentException, depending on requirements, as there's nothing to divide.
String length is 1 or less	Return an empty list because the string cannot be divided into two non-empty substrings.
String contains only one distinct character repeated multiple times	Handle this scenario where hash values of any substrings would likely collide frequently potentially impacting performance, or create an infinite loop if the comparison is based on hash alone; compare the substrings directly to confirm equality.
Very long input string leading to potential integer overflow in hash calculation	Use a sufficiently large data type for hash values (e.g., long) or consider using a rolling hash function with modular arithmetic to prevent overflow.
Extreme skew in character distribution in the input string	If some characters appear much more frequently than others, the substring hashes might collide more often, so direct string comparison should be used after hash comparison to confirm a match.
No valid division exists where the hashes of the substrings are equal	Return an empty list indicating no such division is possible.
Multiple valid divisions exist	Return the first valid division found or all valid divisions, depending on the requirements.
Input string contains Unicode characters	Ensure the hash function is compatible with Unicode characters or convert the Unicode string to a compatible encoding like UTF-8 before hashing.

Null or empty input string

How to Handle:

Return an empty list or throw an IllegalArgumentException, depending on requirements, as there's nothing to divide.

String length is 1 or less

How to Handle:

Return an empty list because the string cannot be divided into two non-empty substrings.

String contains only one distinct character repeated multiple times

How to Handle:

Handle this scenario where hash values of any substrings would likely collide frequently potentially impacting performance, or create an infinite loop if the comparison is based on hash alone; compare the substrings directly to confirm equality.

Very long input string leading to potential integer overflow in hash calculation

How to Handle:

Use a sufficiently large data type for hash values (e.g., long) or consider using a rolling hash function with modular arithmetic to prevent overflow.

Extreme skew in character distribution in the input string

How to Handle:

If some characters appear much more frequently than others, the substring hashes might collide more often, so direct string comparison should be used after hash comparison to confirm a match.

No valid division exists where the hashes of the substrings are equal

How to Handle:

Return an empty list indicating no such division is possible.

Multiple valid divisions exist

How to Handle:

Return the first valid division found or all valid divisions, depending on the requirements.

Input string contains Unicode characters

How to Handle:

Ensure the hash function is compatible with Unicode characters or convert the Unicode string to a compatible encoding like UTF-8 before hashing.