Solution Explanation

The problem asks to repeatedly calculate the digit sum of groups of size k in a string until the string's length is no longer greater than k. The solution involves iteratively applying a digit-summing operation until the termination condition is met.

Algorithm

1. Base Case: If the length of the input string s is less than or equal to k, the string itself is returned.

2. Iteration: If the length of s is greater than k, the following steps are performed repeatedly until the base case is reached:

   • Grouping: The string s is divided into groups of size k. The last group might be shorter than k if the length of s is not a multiple of k.
   • Digit Sum Calculation: For each group, the sum of its digits is calculated.
   • String Construction: The calculated digit sums (converted to strings) are concatenated to form a new string.
   • Update: The new string replaces the original s, and the process repeats from step 2.

Code Explanation (Python)

class Solution:
    def digitSum(self, s: str, k: int) -> str:
        while len(s) > k:
            t = []
            n = len(s)
            for i in range(0, n, k):
                x = 0
                for j in range(i, min(i + k, n)):
                    x += int(s[j])
                t.append(str(x))
            s = "".join(t)
        return s

• The while loop continues as long as the length of s exceeds k.
• Inside the loop:
  • t is an empty list to store the digit sums of each group.
  • The outer for loop iterates through s in steps of k, creating groups. min(i + k, n) handles the last group's potential shorter length.
  • The inner for loop calculates the sum (x) of digits in each group.
  • str(x) converts the sum to a string, which is appended to t.
  • "".join(t) concatenates the strings in t to create the new s.

Time and Space Complexity

• Time Complexity: O(n*logk(n)), where n is the length of the input string. In each iteration, the string's length is reduced by a factor of approximately k. The loop will iterate approximately logk(n) times. The inner loops take O(n) time in total across all iterations.

• Space Complexity: O(n) in the worst case. This is because the intermediate strings generated during the iterations could potentially be of length n in the worst case. In practice, the space usage will likely be less than O(n) because the string length reduces with each iteration.

The solutions in other languages (Java, C++, Go, TypeScript) follow a similar algorithmic approach, differing mainly in syntax and data structure usage. Their time and space complexities remain the same as the Python solution.