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Sum of Digits of String After Convert

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Strings

You are given a string s consisting of lowercase English letters, and an integer k. Your task is to convert the string into an integer by a special process, and then transform it by summing its digits repeatedly k times. More specifically, perform the following steps:

  1. Convert s into an integer by replacing each letter with its position in the alphabet (i.e. replace 'a' with 1, 'b' with 2, ..., 'z' with 26).
  2. Transform the integer by replacing it with the sum of its digits.
  3. Repeat the transform operation (step 2) k times in total.

For example, if s = "zbax" and k = 2, then the resulting integer would be 8 by the following operations:

  1. Convert: "zbax" ➝ "(26)(2)(1)(24)" ➝ "262124" ➝ 262124
  2. Transform #1: 262124 ➝ 2 + 6 + 2 + 1 + 2 + 4 ➝ 17
  3. Transform #2: 17 ➝ 1 + 7 ➝ 8

Return the resulting integer after performing the operations described above.

Example:

s = "leetcode", k = 2

Output: 6

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:

  1. What is the maximum length of the input string `s`?
  2. The problem statement asks for converting letters to digits (a=1, b=2 etc). What are the constraints on the input string, does it only contain lowercase English letters?
  3. What is the maximum value of `k`?
  4. After converting to digits, can the sum of digits in any intermediate step exceed the maximum integer size?
  5. Should I handle the case where the input string is empty?

Brute Force Solution

Approach

This problem involves converting a string into a number, then repeatedly summing its digits until a single-digit number remains. The brute force approach simulates this process exactly as described by the problem.

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

  1. First, turn each letter of the input string into its corresponding number (a becomes 1, b becomes 2, and so on).
  2. Then, add up all the numbers you just got to create one big number.
  3. If this number is already a single digit, you're done!
  4. Otherwise, take that number and add up its digits to create a new number.
  5. Repeat the process of adding up the digits until you finally get a number that is just a single digit.

Code Implementation

def sum_of_digits_brute_force(input_string, k_iterations):
    string_sum = 0
    # Convert each letter to number and sum
    for character in input_string:
        string_sum += ord(character) - ord('a') + 1

    # Perform k iterations
    for _ in range(k_iterations):
        new_sum = 0

        # Calculate the digit sum
        while string_sum > 0:
            new_sum += string_sum % 10
            string_sum //= 10

        string_sum = new_sum

    return string_sum

Big(O) Analysis

Time Complexity
O(n * k)The initial conversion of the string 's' to a number involves iterating through each character, which takes O(n) time, where n is the length of the string. Then, we iterate up to 'k' times to sum the digits of the resulting number. In each iteration, we iterate through the digits of the number, which, in the worst case, can be proportional to the initial string length 'n'. Therefore, the repeated summing operation takes O(k * n) time. The total time complexity becomes O(n) + O(k * n), which simplifies to O(n * k) as the n term is dominated.
Space Complexity
O(1)The provided solution primarily involves iterative processes of converting characters to numbers and summing digits. It uses a few integer variables to store intermediate sums during conversion and subsequent digit summing. The amount of memory used by these variables does not scale with the input string's length (N). Thus, the auxiliary space required remains constant, resulting in O(1) space complexity.

Optimal Solution

Approach

The goal is to convert letters to numbers, add them up, and then repeat the adding process a few times. We efficiently calculate the sums in each round without storing large intermediary strings or doing unnecessary conversions.

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

  1. First, change each letter in the input string into its corresponding number. 'a' becomes 1, 'b' becomes 2, and so on. Then, add up all those numbers to get a single number.
  2. After the first sum, check how many times we need to repeat the summing process.
  3. For each repetition, take the previous sum and split it into its individual digits. Add those digits together to get a new sum.
  4. Keep doing this until we've repeated the summing process the specified number of times.
  5. The final sum is the answer.

Code Implementation

def sum_of_digits_of_string_after_convert(input_string, k_iterations):
    initial_sum = 0
    for char in input_string:
        # Convert each letter to its number and add to initial sum
        initial_sum += ord(char) - ord('a') + 1

    current_sum = initial_sum
    for _ in range(k_iterations):
        next_sum = 0

        # Iterate through the digits of the current sum
        while current_sum > 0:
            next_sum += current_sum % 10
            current_sum //= 10

        # Update the current sum for the next iteration
        current_sum = next_sum

    return current_sum

Big(O) Analysis

Time Complexity
O(n + k*log(n))Converting the input string of length n to its numerical representation involves iterating through the string once, resulting in O(n) time complexity. The subsequent summing process repeats k times. In each repetition, we calculate the sum of the digits of the previous sum. The number of digits in a number is logarithmic to its value (base 10). The maximum initial sum is bounded by 26*n, hence the number of digits is at most log(26n), which is approximately log(n). Therefore, each of the k summing steps takes O(log(n)) time. The total time complexity is thus O(n + k*log(n)).
Space Complexity
O(1)The algorithm's space complexity is O(1) because it primarily uses a few integer variables to store intermediate sums and the number of repetitions remaining. The conversion process from characters to numbers happens directly within the summation without storing the converted string. The space used by these variables remains constant irrespective of the input string's length or the number of repetitions (k), so the auxiliary space is constant.

Edge Cases

CaseHow to Handle
Null or empty string inputReturn 0 immediately as there are no digits to sum.
Input string contains non-alphabetic charactersFilter out non-alphabetic characters before processing the string to avoid incorrect conversions.
Large k value causing multiple iterationsEnsure the solution has reasonable time complexity even with large k, preventing potential timeouts.
String 'a' * very large numberThe sum will overflow if not using long or arbitrary precision integers, so use appropriate data types to store the sum.
k is zeroReturn the initial conversion result without performing any further sum of digits operations.
String resulting in a single-digit sum after one conversionThe while loop should terminate correctly if the sum becomes a single digit before k iterations are completed.
Maximum string length that may cause integer overflowUtilize long integers to handle potential overflow during the digit sum operations.
The intermediate sum is very large but the initial string is shortUse long datatype to accommodate for potentially large sums arising from conversion and repetitive summation.