Problem Description

You need to process a string s (containing only lowercase English letters) and an integer k through a specific transformation process.

The process consists of two main steps:

1. Convert the string to a number: Replace each letter in the string with its position in the alphabet ('a' becomes 1, 'b' becomes 2, ... 'z' becomes 26). Concatenate all these numbers together to form one large integer.

2. Transform the number k times: For each transformation, calculate the sum of all individual digits in the current number. This sum becomes the new number. Repeat this transformation exactly k times.

Example walkthrough:

• Given s = "zbax" and k = 2
• Step 1 - Convert: 'z' → 26, 'b' → 2, 'a' → 1, 'x' → 24
  • Concatenate: "262124" → integer 262124
• Step 2 - First transformation: Sum the digits: 2 + 6 + 2 + 1 + 2 + 4 = 17
• Step 3 - Second transformation: Sum the digits: 1 + 7 = 8
• Result: 8

The solution approach simulates this process directly:

• First, it converts each character to its alphabet position and concatenates them as a string
• Then it performs k iterations where each iteration sums all digits and converts the result back to a string for the next iteration
• Finally, it returns the integer value after all transformations

Intuition

The problem is asking us to perform a series of transformations, so the natural approach is to simulate exactly what's described. Let's think through why this direct simulation works well:

First, we need to convert letters to numbers. Each letter maps to a specific position ('a' → 1, 'b' → 2, etc.), which we can calculate using ord(c) - ord('a') + 1. The key insight here is that we should convert these numbers to strings immediately and concatenate them, rather than trying to build an actual integer. Why? Because letters like 'z' become 26, which is two digits, so simple multiplication by 10 won't work correctly.

For the digit sum transformations, we need to repeatedly sum the digits of our number. The elegant approach is to keep the number as a string throughout the process. This makes extracting individual digits trivial - we just iterate through each character and convert it to an integer with int(c). After summing, we convert back to a string for the next iteration.

Why maintain everything as strings? Consider that the initial converted number could be very large (imagine a long string with many 'z' characters producing lots of 26s). Working with strings avoids any potential integer overflow issues and makes digit extraction straightforward. We only need the actual integer value at the very end when returning the result.

The simulation approach is optimal here because:

1. We must perform each step exactly as described - there's no mathematical shortcut to skip the conversions
2. The number of operations is bounded by the string length and k, which are given constraints
3. String manipulation provides a clean way to handle both the concatenation and digit extraction operations

Solution Approach

Let's walk through the implementation step by step:

Step 1: Convert string to numeric representation

s = ''.join(str(ord(c) - ord('a') + 1) for c in s)

This line performs the initial conversion from letters to numbers:

• For each character c in the string, we calculate ord(c) - ord('a') + 1 to get its alphabet position
• ord('a') gives us the ASCII value of 'a', so subtracting it from ord(c) gives us the offset from 'a'
• Adding 1 converts the 0-based offset to 1-based position (so 'a' becomes 1, not 0)
• We convert each position to a string using str() and join all these strings together
• For example: "abc" becomes "123", and "zbax" becomes "262124"

Step 2: Perform k transformations

for _ in range(k):
    t = sum(int(c) for c in s)
    s = str(t)

This loop executes exactly k times, performing the digit sum transformation:

• In each iteration, we calculate the sum of all digits in the current string s
• int(c) for c in s converts each character (which represents a digit) back to an integer
• sum() adds up all these individual digits
• We convert the sum back to a string with str(t) and store it in s for the next iteration
• For example: "262124" → sum is 2+6+2+1+2+4 = 17 → "17" for the next iteration

Step 3: Return the final result

return int(s)

After k transformations, we convert the final string back to an integer and return it.

The key pattern here is maintaining the number as a string throughout the process, which simplifies both the initial concatenation and the repeated digit extraction. This approach has a time complexity of O(n + k*m) where n is the length of the input string and m is the maximum number of digits in any intermediate result.

Example Walkthrough

Let's walk through a small example with s = "iiii" and k = 1:

Step 1: Convert string to numeric representation

• 'i' is the 9th letter of the alphabet
• For each 'i': ord('i') - ord('a') + 1 = 105 - 97 + 1 = 9
• Convert each to string and concatenate: "9" + "9" + "9" + "9" = "9999"
• Our string s is now "9999"

Step 2: Perform k transformations (k = 1, so one iteration)

• Iteration 1:
  • Calculate sum of digits: int('9') + int('9') + int('9') + int('9') = 9 + 9 + 9 + 9 = 36
  • Convert sum back to string: s = "36"

Step 3: Return the final result

• Convert "36" to integer: 36
• Return 36

Let's trace through another example with s = "abc" and k = 3:

Step 1: Convert string to numeric representation

• 'a' → position 1, 'b' → position 2, 'c' → position 3
• Concatenate as strings: "1" + "2" + "3" = "123"

Step 2: Perform k transformations (k = 3)

• Iteration 1: Sum digits of "123" → 1 + 2 + 3 = 6 → s = "6"
• Iteration 2: Sum digits of "6" → 6 → s = "6"
• Iteration 3: Sum digits of "6" → 6 → s = "6"

Step 3: Return the final result

• Return 6

Notice how in the second example, once we reach a single digit (6), further transformations don't change the value since the sum of a single digit is itself.

Solution Implementation

Time and Space Complexity

The time complexity is O(n + k * m), where n is the length of the input string s, k is the number of iterations, and m is the maximum number of digits in the intermediate results.

• The initial conversion ''.join(str(ord(c) - ord('a') + 1) for c in s) takes O(n) time as it processes each character once.
• Each iteration of the k loop processes all digits in the current string representation, which can be at most O(m) digits.
• In the worst case, m can be 2n (when all characters map to two-digit numbers like 26 for 'z').
• Since k is typically small and the number of digits decreases rapidly with each iteration, the dominant term is often O(n).

The space complexity is O(n), where n is the length of the input string s.

• The initial string conversion creates a new string that can be at most 2n characters long (each letter converts to at most 2 digits).
• The intermediate string s stores the digit representation, which requires O(n) space in the worst case.
• Additional space for temporary variables during computation is O(1).

The reference answer simplifies to O(n) for both time and space complexity, which is reasonable when k is treated as a small constant and considering that the number of digits decreases exponentially with each transformation.

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

Common Pitfalls

1. Incorrect Character-to-Number Mapping

A frequent mistake is using ord(c) - ord('a') instead of ord(c) - ord('a') + 1, which would map 'a' to 0 instead of 1. This off-by-one error completely changes the resulting number.

Incorrect:

numeric_string = ''.join(str(ord(char) - ord('a')) for char in s)  # 'a' becomes 0

Correct:

numeric_string = ''.join(str(ord(char) - ord('a') + 1) for char in s)  # 'a' becomes 1

2. Attempting Integer Operations Instead of String Concatenation

Some might try to build the initial number mathematically rather than through string concatenation, leading to incorrect results when dealing with two-digit positions (like 'z' = 26).

Incorrect:

# This treats each position as a single digit, losing information
result = 0
for char in s:
    result = result * 10 + (ord(char) - ord('a') + 1)  # Won't work for positions > 9

Correct:

# String concatenation preserves all digits
numeric_string = ''.join(str(ord(char) - ord('a') + 1) for char in s)

3. Not Handling k = 0 Edge Case

While the problem constraints typically ensure k ≥ 1, failing to consider k = 0 could cause issues if the problem specifications change.

Solution:

if k == 0:
    return int(''.join(str(ord(char) - ord('a') + 1) for char in s))

4. Inefficient Repeated String-to-Integer Conversions

Converting the entire string to an integer in each iteration when you only need individual digits is inefficient.

Inefficient:

for _ in range(k):
    num = int(numeric_string)
    digit_sum = 0
    while num > 0:
        digit_sum += num % 10
        num //= 10
    numeric_string = str(digit_sum)

Efficient:

for _ in range(k):
    digit_sum = sum(int(digit) for digit in numeric_string)
    numeric_string = str(digit_sum)

5. Type Confusion Between Final Iterations

Forgetting to maintain consistent types throughout the transformation process or returning the wrong type at the end.

Incorrect:

# Mixing integer and string operations
for _ in range(k):
    result = sum(int(d) for d in str(result))  # result undefined initially
return result  # Might return wrong type

Correct:

# Consistent string handling until final return
numeric_string = ''.join(str(ord(char) - ord('a') + 1) for char in s)
for _ in range(k):
    digit_sum = sum(int(digit) for digit in numeric_string)
    numeric_string = str(digit_sum)
return int(numeric_string)