Calculate Digit Sum of a String

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Strings

You are given a string s consisting of digits and an integer k.

A round can be completed if the length of s is greater than k. In one round, do the following:

Divide s into consecutive groups of size k such that the first k characters are in the first group, the next k characters are in the second group, and so on. Note that the size of the last group can be smaller than k.
Replace each group of s with a string representing the sum of all its digits. For example, "346" is replaced with "13" because 3 + 4 + 6 = 13.
Merge consecutive groups together to form a new string. If the length of the string is greater than k, repeat from step 1.

Return s after all rounds have been completed.

Example 1:

Input: s = "11111222223", k = 3
Output: "135"
Explanation: 
- For the first round, we divide s into groups of size 3: "111", "112", "222", and "23".
  Then we calculate the digit sum of each group: 1 + 1 + 1 = 3, 1 + 1 + 2 = 4, 2 + 2 + 2 = 6, and 2 + 3 = 5. 
  So, s becomes "3" + "4" + "6" + "5" = "3465" after the first round.
- For the second round, we divide s into "346" and "5".
  Then we calculate the digit sum of each group: 3 + 4 + 6 = 13, 5 = 5. 
  So, s becomes "13" + "5" = "135" after second round. 
Now, s.length <= k, so we return "135" as the answer.

Example 2:

Input: s = "00000000", k = 3
Output: "000"
Explanation: 
We divide s into "000", "000", and "00".
Then we calculate the digit sum of each group: 0 + 0 + 0 = 0, 0 + 0 + 0 = 0, and 0 + 0 = 0. 
s becomes "0" + "0" + "0" = "000", whose length is equal to k, so we return "000".

Constraints:

1 <= s.length <= 100
2 <= k <= 100
s consists of digits only.

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:

What is the maximum length of the input string `s` and the maximum value of `k`?
Does the input string `s` only contain digits?
What should I return if the input string `s` is empty?
Is `k` always greater than 0?
Are we expected to handle very large numbers and should consider using a specific data type (e.g., BigInteger)?

Brute Force Solution

Approach

The brute force method for this problem directly simulates the repeated summing and string construction process until the string's length is short enough. We will keep doing the process no matter how long it takes.

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

First, check if the string is already short enough. If it is, we're done.
If not, split the string into groups of digits with sizes defined by the given number. For example, with a given number of 3, you group every three digits together.
Calculate the sum of the digits in each group.
Combine these sums to form a new string.
Repeat the entire process (grouping, summing, and combining) using the newly created string until the string length becomes less than or equal to the given number.

Code Implementation

def calculate_digit_sum_brute_force(input_string, group_size):
    while len(input_string) > group_size:
        new_string = ""
        # Iterate over the string in chunks of size group_size
        for i in range(0, len(input_string), group_size):
            group = input_string[i:i + group_size]
            digit_sum = 0
            for digit_char in group:
                digit_sum += int(digit_char)

            # Append the sum to the new string
            new_string += str(digit_sum)

        # Update input_string for the next iteration
        input_string = new_string

    return input_string

Big(O) Analysis

Time Complexity

O(n²) – The algorithm repeatedly processes the string s until its length is less than or equal to k. In the worst case, if k is small (e.g., 1), the string's length reduces slowly with each iteration. Each iteration involves splitting the string of length n into groups of size k and summing the digits within each group, which takes O(n) time. Since the string length decreases by a factor related to k in each iteration, and in the worst case approaches a linear reduction, the outer loop (repeated string processing) could run up to n/k times. When k is considered a constant (as is typical in these problem constraints), the total time complexity becomes O(n * n), thus O(n²).

Space Complexity

O(N) – The auxiliary space complexity is primarily determined by the temporary string created in each iteration. In the worst-case scenario, the length of this temporary string can be proportional to the length of the input string in that iteration, which can be up to N, where N is the length of the original input string. Specifically, the new string holding sums of digit groups will be stored in memory. Therefore, the auxiliary space used is O(N).

Optimal Solution

Approach

The goal is to repeatedly group digits in the string and sum them until the string's length is at most k. We achieve this by processing the string in chunks and then rebuilding it with the calculated sums. This process repeats until the condition is met.

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

As long as the length of the initial string is bigger than the given number k, keep doing the following steps.
Divide the string into groups of digits, where each group has at most k digits.
For each of these groups, calculate the sum of the digits.
Combine all the sums into a new string. This new string becomes the string we work with in the next repeat.
When the string's length is finally not bigger than k, we have our answer. Stop repeating the steps and provide the result.

Code Implementation

def calculate_digit_sum(input_string, group_size):
    while len(input_string) > group_size:
        new_string = ""
        # Iterate through the string in chunks of size k

        for i in range(0, len(input_string), group_size):
            group = input_string[i:i + group_size]
            digit_sum = 0
            for digit_char in group:
                digit_sum += int(digit_char)
            new_string += str(digit_sum)
        
        # Update the string for the next iteration
        input_string = new_string
    
    return input_string

Big(O) Analysis

Time Complexity

O(n²) – The outer loop iterates as long as the string length n is greater than k. Inside, we process the string by dividing it into groups of size k. Within each group, we calculate the digit sum, taking O(k) time. Since there are approximately n/k groups, processing each string takes O(n/k * k) = O(n) time. In the worst case, the string length reduces slowly (e.g., n becomes n-1 after each operation). Therefore, the outer loop could iterate O(n) times, resulting in a total time complexity of O(n * n) = O(n²).

Space Complexity

O(N) – The primary auxiliary space usage comes from creating new strings in each iteration of the while loop, specifically the new string composed of digit sums. In the worst case, where k is small, we may need to create a new string that is almost the same length as the original string in each iteration, especially if the digit sums don't significantly reduce the string's length. Therefore, the space used to store the new string in each iteration can grow up to O(N), where N is the initial length of the input string. The temporary strings used in string concatenation also contribute to O(N) space.

Edge Cases

Case	How to Handle
Null or Empty String s	Return an empty string immediately as there is no input to process.
String s contains non-numeric characters	Raise an IllegalArgumentException or filter out non-numeric characters before processing, depending on problem constraints.
k equals to 0	Return the original string immediately as no grouping is required.
k is greater than the length of s	Treat k as the length of s effectively, grouping all digits into one sum.
String s has length 1	Return the string itself since no grouping and summing is needed.
Integer overflow during digit summation	Use a larger data type (e.g., long) to store the intermediate sums or check for overflow before each addition to prevent incorrect results.
Maximum string length leading to memory exhaustion	Consider processing the string in chunks or using an iterative approach with constant space to avoid storing excessive intermediate strings.
k is equal to length of s	The output is sum of all digits in the string s.

Null or Empty String s

How to Handle:

Return an empty string immediately as there is no input to process.

String s contains non-numeric characters

How to Handle:

Raise an IllegalArgumentException or filter out non-numeric characters before processing, depending on problem constraints.

k equals to 0

How to Handle:

Return the original string immediately as no grouping is required.

k is greater than the length of s

How to Handle:

Treat k as the length of s effectively, grouping all digits into one sum.

String s has length 1

How to Handle:

Return the string itself since no grouping and summing is needed.

Integer overflow during digit summation

How to Handle:

Use a larger data type (e.g., long) to store the intermediate sums or check for overflow before each addition to prevent incorrect results.

Maximum string length leading to memory exhaustion

How to Handle:

Consider processing the string in chunks or using an iterative approach with constant space to avoid storing excessive intermediate strings.

k is equal to length of s

How to Handle:

The output is sum of all digits in the string s.