algoadvance

Given a string s consisting of digits and an integer k, you need to repeatedly transform the string. The transformation is defined as follows:

1. Split the string s into groups of size k, so that each group contains k characters. Note that the last group may contain fewer than k characters.
2. Replace each group with its sum of digits (e.g., "123" becomes 6, "4987" becomes 28).
3. Concatenate the sums to form a new string.

Repeat this transformation until the length of the new string becomes less than or equal to k.

Return the resulting string.

Clarifying Questions

1. Will the input string s always consist of digits from ‘0’ to ‘9’?
2. Can k be larger than the length of s?
3. What is the expected output if the input string is already of length k or less?

Assuming:

1. Yes, the string s consists only of digit characters.
2. Yes, although the immediate transformation process will simplify this case.
3. The output should be the string itself if its length is less than or equal to k.

Strategy

1. While the length of the string s is greater than k, perform the following steps:
   • Split the string s into groups of size k.
   • For each group, calculate the sum of its digits.
   • Concatenate these sums to form a new string.
2. Return the final string once its length is less than or equal to k.

Code

def digitSum(s: str, k: int) -> str:
    while len(s) > k:
        new_s = []
        for i in range(0, len(s), k):
            group = s[i:i+k]
            group_sum = sum(int(ch) for ch in group)
            new_s.append(str(group_sum))
        s = ''.join(new_s)
    return s

# Example Usage
print(digitSum("1111122222", 3))  # Output should be a transformed string
print(digitSum("1234", 2))  # Output should be a transformed string

Time Complexity

• Splitting into groups of size k: This takes O(n/k) steps if n is the length of the string.
• Calculating the sum of digits for each group: This takes O(k) per group, and there are n/k groups, resulting in a total of O(n).
• Concatenating the sums to form a new string: This takes O(n/k) since each group sum is at most the number of digits in the original string divided by k.

Overall, the time complexity per transformation step is O(n). Since the length of the string reduces significantly with each transformation, we can denote the maximum number of such transformations as log(n), making the overall time complexity O(n log n).

Edge Cases

• If s is empty, the function should return "".
• If k is greater than or equal to the length of s, the function should return s directly.

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