Problem Description

You are given a string s of length n and an integer k, where n is a multiple of k. Your goal is to transform the string s into a new compressed string called result that has a length of n / k.

The transformation process works as follows:

1. Divide the string: Split s into n / k consecutive substrings, where each substring has exactly k characters.

2. Process each substring: For each substring (processing them in order from left to right):

   • Calculate the hash value of each character, where the hash value is the character's position in the alphabet ('a' → 0, 'b' → 1, 'c' → 2, ..., 'z' → 25)
   • Sum up all the hash values of the characters in the current substring
   • Take the remainder when this sum is divided by 26 (this gives you hashedChar)
   • Convert hashedChar back to a lowercase letter (0 → 'a', 1 → 'b', ..., 25 → 'z')
   • Append this character to the result string

3. Return the result: After processing all substrings, return the final result string.

For example, if s = "abcd" and k = 2:

• First substring: "ab" → hash values: 0 + 1 = 1 → 1 % 26 = 1 → 'b'
• Second substring: "cd" → hash values: 2 + 3 = 5 → 5 % 26 = 5 → 'f'
• Result: "bf"

The solution iterates through the string in chunks of size k, computes the sum of character positions for each chunk, takes modulo 26, and converts back to a character to build the final result.

Intuition

The problem is essentially asking us to create a hash function that compresses a string by mapping every k characters to a single character. This is a straightforward simulation problem where we need to follow the given instructions step by step.

The key insight is recognizing that this is a chunking and hashing operation:

• We're dividing the input into equal-sized chunks
• Each chunk gets reduced to a single character through a hashing process
• The hash function is simply the sum of character positions modulo 26

Why does this approach make sense?

1. Fixed chunk size: Since n is guaranteed to be a multiple of k, we know we can cleanly divide the string into chunks of size k without any remainder. This makes the iteration pattern simple - we can jump by k positions each time.

2. Character mapping: The problem defines a clear mapping from characters to numbers ('a' → 0, 'b' → 1, etc.). This is just ord(char) - ord('a'), which gives us the alphabetical position.

3. Modulo operation: Taking the sum modulo 26 ensures our result always maps back to a valid lowercase letter (since there are 26 letters in the alphabet). No matter how large the sum gets, sum % 26 will always be in the range [0, 25].

4. Reverse mapping: Converting back from a number to a character is the inverse operation: chr(ord('a') + hashedChar).

The solution naturally flows from these observations - we iterate through the string in steps of k, calculate the hash for each substring, and build our result character by character. There's no need for complex data structures or algorithms; it's purely a matter of following the hashing recipe provided in the problem statement.

Solution Approach

The solution implements a direct simulation of the hashing process described in the problem. Let's walk through the implementation step by step:

1. Initialize the result container:

ans = []

We use a list to collect the hashed characters, which will be joined into a string at the end. This is more efficient than string concatenation in Python.

2. Iterate through the string in chunks of size k:

for i in range(0, len(s), k):

The loop starts at index 0 and jumps by k positions each iteration. This ensures we process each substring of length k exactly once. For example, if len(s) = 6 and k = 2, we'll have i values of 0, 2, and 4.

3. Calculate the hash sum for each substring:

t = 0
for j in range(i, i + k):
    t += ord(s[j]) - ord("a")

For each substring starting at position i:

• We iterate through the next k characters (from index i to i + k - 1)
• Convert each character to its hash value using ord(s[j]) - ord("a")
  • ord("a") gives 97, ord("b") gives 98, etc.
  • Subtracting ord("a") normalizes these to 0, 1, 2, ... 25
• Accumulate the sum in variable t

4. Apply modulo and convert to character:

hashedChar = t % 26
ans.append(chr(ord("a") + hashedChar))

• Take the sum modulo 26 to get a value in range [0, 25]
• Convert this number back to a character:
  • ord("a") gives the ASCII value of 'a' (97)
  • Adding hashedChar gives us the ASCII value of our target character
  • chr() converts the ASCII value back to a character
• Append the resulting character to our answer list

5. Build and return the final string:

return "".join(ans)

Join all the hashed characters into a single string and return it.

Time Complexity: O(n) where n is the length of the input string, as we visit each character exactly once.

Space Complexity: O(n/k) for storing the result string, which has length n/k.

The algorithm is straightforward - no complex data structures or advanced techniques are needed. It's a pure simulation that follows the problem's hashing recipe exactly as specified.

Example Walkthrough

Let's walk through the solution with a concrete example: s = "mauigqpe" and k = 4.

Step 1: Divide into chunks

• String length: 8 characters
• Number of chunks: 8 / 4 = 2 chunks
• Chunks: "maui" and "gqpe"

Step 2: Process first chunk "maui"

• Calculate hash values:
  • 'm' → 12 (m is the 13th letter, so position 12)
  • 'a' → 0
  • 'u' → 20
  • 'i' → 8
• Sum: 12 + 0 + 20 + 8 = 40
• Apply modulo: 40 % 26 = 14
• Convert to character: 14 → 'o' (since 'a' + 14 = 'o')

Step 3: Process second chunk "gqpe"

• Calculate hash values:
  • 'g' → 6
  • 'q' → 16
  • 'p' → 15
  • 'e' → 4
• Sum: 6 + 16 + 15 + 4 = 41
• Apply modulo: 41 % 26 = 15
• Convert to character: 15 → 'p' (since 'a' + 15 = 'p')

Step 4: Build result

• First chunk produced: 'o'
• Second chunk produced: 'p'
• Final result: "op"

The algorithm compressed the 8-character string "mauigqpe" into a 2-character string "op" by hashing every 4 consecutive characters into a single character.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string s. The algorithm iterates through the string in chunks of size k, processing n/k chunks total. For each chunk, it performs k character operations (converting to ASCII values and summing). Therefore, the total operations are (n/k) * k = n, resulting in linear time complexity.

The space complexity is O(n/k) for the answer list that stores the hashed characters. Since each chunk of k characters produces one hashed character, the answer list contains n/k elements. If we ignore the space consumption of the answer string (as mentioned in the reference), the space complexity would be O(1) as we only use a constant amount of extra space for variables t, i, and j.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Off-by-One Errors in Character Position Calculation

A common mistake is incorrectly calculating the character's position in the alphabet. Developers might forget to subtract ord('a') or might use 1-based indexing instead of 0-based.

Incorrect Implementation:

# Wrong: Using 1-based indexing
substring_sum += ord(s[j]) - ord('a') + 1  # 'a' would be 1, not 0

# Wrong: Forgetting to normalize
substring_sum += ord(s[j])  # Would give ASCII values like 97, 98, etc.

Solution: Always remember that the problem specifies 0-based positions ('a' → 0), so use ord(s[j]) - ord('a') consistently.

2. Integer Overflow in Languages with Fixed Integer Sizes

While Python handles arbitrary precision integers automatically, in languages like Java or C++, accumulating large sums might cause integer overflow if the substring is very long or contains many 'z' characters.

Potential Issue in Other Languages:

// In Java, this could overflow for large k values
int sum = 0;
for (int j = i; j < i + k; j++) {
    sum += s.charAt(j) - 'a';  // Could overflow if k is very large
}

Solution: Apply modulo 26 during accumulation to prevent overflow:

substring_sum = 0
for j in range(i, i + k):
    substring_sum = (substring_sum + ord(s[j]) - ord('a')) % 26

3. Incorrect Loop Boundaries

Developers might incorrectly set up the inner loop boundaries, especially when dealing with the end index.

Incorrect Implementation:

# Wrong: Off by one - would include k+1 characters
for j in range(i, i + k + 1):
    substring_sum += ord(s[j]) - ord('a')

# Wrong: Would miss the last character
for j in range(i, i + k - 1):
    substring_sum += ord(s[j]) - ord('a')

Solution: Use range(i, i + k) which correctly iterates from index i to i + k - 1, giving exactly k characters.

4. String Concatenation Performance

Using string concatenation in a loop can be inefficient in Python due to string immutability.

Inefficient Implementation:

result = ""
for i in range(0, len(s), k):
    # ... calculate hashed_character
    result += hashed_character  # Creates a new string object each time

Solution: Use a list to collect characters and join at the end:

result = []
for i in range(0, len(s), k):
    # ... calculate hashed_character
    result.append(hashed_character)
return ''.join(result)

5. Assuming Input Validity

The problem states that n is a multiple of k, but in production code, you might want to validate this assumption.

Defensive Programming:

def stringHash(self, s: str, k: int) -> str:
    if len(s) % k != 0:
        raise ValueError(f"String length {len(s)} is not divisible by k={k}")
  
    # Rest of the implementation...

6. Misunderstanding the Modulo Operation

Some developers might apply modulo 26 at the wrong step or forget it entirely.

Incorrect Implementation:

# Wrong: Applying modulo to each character individually
for j in range(i, i + k):
    substring_sum += (ord(s[j]) - ord('a')) % 26  # Unnecessary here

# Wrong: Forgetting modulo entirely
hashed_position = substring_sum  # Could be > 25, causing chr() to produce non-lowercase letters

Solution: Apply modulo 26 only once after summing all characters in the substring:

hashed_position = substring_sum % 26