Problem Description

You are given a string s that contains only digit characters and an integer k.

The task is to repeatedly transform the string through "rounds" of operations until the string's length becomes less than or equal to k.

Each round consists of these steps:

1. Divide the string into groups: Split s into consecutive groups where each group contains exactly k characters (except possibly the last group, which may have fewer than k characters if the string length isn't divisible by k). For example, if s = "11111222" and k = 3, the groups would be "111", "112", and "22".

2. Replace each group with its digit sum: Calculate the sum of all digits in each group and replace that group with the sum as a string. Using the example above:

   • "111" → 1 + 1 + 1 = 3 → "3"
   • "112" → 1 + 1 + 2 = 4 → "4"
   • "22" → 2 + 2 = 4 → "4"

3. Merge the results: Concatenate all the replacement strings to form a new string. From our example: "3" + "4" + "4" = "344".

4. Check and repeat: If the new string's length is still greater than k, repeat the entire process from step 1. Otherwise, the transformation is complete.

The goal is to return the final string after all rounds have been completed (when the string length is no longer greater than k).

For instance, if s = "11111222" and k = 3:

• Round 1: "11111222" → groups: ["111", "112", "22"] → sums: ["3", "4", "4"] → "344"
• Since length of "344" is 3, which equals k, we stop and return "344".

Intuition

The problem asks us to repeatedly transform a string until it becomes short enough (length ≤ k). This immediately suggests a simulation approach - we need to directly implement what the problem describes rather than finding a mathematical shortcut.

The key insight is that this is a convergent process. Each round potentially reduces the string length because:

• We're replacing groups of digits with their sum
• A sum of multiple digits often results in fewer digits than the original group (e.g., "999" becomes "27", reducing from 3 characters to 2)

Since we don't know exactly how many rounds we'll need (it depends on the specific digits and their distribution), we can't predict the final result without actually performing the operations. This rules out any clever mathematical formula and points us toward a straightforward simulation.

The natural approach is to use a while loop that continues as long as len(s) > k. Inside the loop, we:

1. Process the string in chunks of size k
2. Calculate the sum of digits in each chunk
3. Build a new string from these sums

For implementation, we iterate through the string with a step size of k (using range(0, n, k)), which automatically gives us the starting position of each group. For each group, we sum up the digits from position i to min(i + k, n) - using min ensures we don't go out of bounds for the last group.

The solution is essentially a direct translation of the problem statement into code. There's no optimization needed because each round necessarily makes progress toward the termination condition, and we must perform each round sequentially since each depends on the previous result.

Solution Approach

The solution uses a simulation approach that directly implements the transformation process described in the problem. Here's how the implementation works:

Main Loop Structure: We use a while loop that continues as long as len(s) > k. This ensures we keep transforming the string until it's short enough.

while len(s) > k:
    # transformation logic

Processing Each Round: For each round, we need to:

1. Initialize a temporary list t to store the results of each group's sum. We use a list instead of string concatenation for efficiency.

2. Iterate through groups using range(0, n, k) where n = len(s). This gives us starting positions at indices 0, k, 2k, 3k, etc., automatically dividing the string into groups of size k.

3. Calculate digit sum for each group:

   • For each group starting at position i, we initialize a sum variable x = 0
   • We iterate from i to min(i + k, n) to handle both regular groups and the potentially smaller last group
   • Convert each character to an integer and add to the sum: x += int(s[j])

4. Store the result by converting the sum back to a string and appending to our list: t.append(str(x))

5. Merge all groups by joining the list into a new string: s = "".join(t)

Example Walkthrough: If s = "11111222" and k = 3:

• First iteration: i = 0, process "111" → sum = 3
• Second iteration: i = 3, process "112" → sum = 4
• Third iteration: i = 6, process "22" → sum = 4
• Result: t = ["3", "4", "4"] → s = "344"
• Since len("344") = 3 which is not greater than k = 3, the loop exits

Time Complexity: Each round processes each character once, and the number of rounds depends on how quickly the string length reduces. In the worst case, this is logarithmic in the initial string length.

Space Complexity: O(n) where n is the current string length, for storing the temporary list t in each round.

The beauty of this solution is its simplicity - it directly translates the problem statement into code without any complex optimizations, making it easy to understand and verify for correctness.

Example Walkthrough

Let's walk through the solution with s = "1234567890" and k = 4:

Initial State:

• s = "1234567890" (length = 10)
• Since 10 > 4, we enter the while loop

Round 1:

1. Divide into groups of size 4:

   • Group 1 (i=0): "1234"
   • Group 2 (i=4): "5678"
   • Group 3 (i=8): "90" (last group, only 2 characters)

2. Calculate digit sums:

   • "1234" → 1+2+3+4 = 10 → "10"
   • "5678" → 5+6+7+8 = 26 → "26"
   • "90" → 9+0 = 9 → "9"

3. Merge results: s = "10" + "26" + "9" = "10269"

Round 2:

• s = "10269" (length = 5)
• Since 5 > 4, continue loop

1. Divide into groups:

   • Group 1 (i=0): "1026"
   • Group 2 (i=4): "9" (last group, only 1 character)

2. Calculate digit sums:

   • "1026" → 1+0+2+6 = 9 → "9"
   • "9" → 9 → "9"

3. Merge results: s = "9" + "9" = "99"

Final Check:

• s = "99" (length = 2)
• Since 2 ≤ 4, exit loop
• Return "99"

The algorithm correctly transforms the string from 10 characters down to 2 characters through two rounds of grouping and summing operations.

Solution Implementation

Time and Space Complexity

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

Time Complexity Analysis:

• In each iteration of the while loop, the string is processed in groups of size k, taking O(len(s)) time
• Each iteration reduces the string length by at least a factor of k (since we're replacing k digits with their sum)
• The maximum number of iterations is O(log_k(n)) since the string length decreases geometrically
• The total work done across all iterations forms a geometric series: n + n/k + n/k² + ... ≤ n(1 + 1/k + 1/k² + ...) = n * k/(k-1)
• Since k ≥ 2 (otherwise the loop wouldn't execute), this sum is bounded by 2n
• Therefore, the overall time complexity is O(n)

Space Complexity Analysis:

• The list t stores the intermediate results in each iteration
• In the worst case (first iteration), t can contain up to n/k elements, each being a string representation of a sum
• The total space used for t and the new string s in each iteration is proportional to the current string length
• The maximum space used at any point is O(n) during the first iteration
• Therefore, the space complexity is O(n)

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

Common Pitfalls

1. Off-by-One Error in Group Boundary Calculation

A common mistake is incorrectly handling the last group's boundaries, especially when the string length is not perfectly divisible by k. Developers might write:

Incorrect approach:

for index in range(start_index, start_index + k):  # Wrong! May go out of bounds
    segment_sum += int(s[index])

This will cause an IndexError when processing the last group if it has fewer than k characters.

Correct approach:

for index in range(start_index, min(start_index + k, string_length)):
    segment_sum += int(s[index])

2. Infinite Loop Due to Wrong Loop Condition

Another pitfall is using >= instead of > in the while loop condition:

Incorrect:

while len(s) >= k:  # Wrong! Will loop infinitely when len(s) == k

This creates an infinite loop when the string length equals k exactly, since transforming a string of length k into groups of size k results in a single group, which likely produces a string of length 1 or more (potentially equal to k again).

Correct:

while len(s) > k:  # Correct! Stops when length becomes <= k

3. String Concatenation Performance Issue

Using string concatenation in a loop is inefficient in Python:

Inefficient approach:

result = ""
for start_index in range(0, string_length, k):
    # ... calculate segment_sum ...
    result += str(segment_sum)  # Inefficient! Creates new string each time

String concatenation creates a new string object each time, leading to O(n²) time complexity for this part alone.

Efficient approach:

segment_sums = []
for start_index in range(0, string_length, k):
    # ... calculate segment_sum ...
    segment_sums.append(str(segment_sum))
s = "".join(segment_sums)  # Efficient! Single join operation

4. Forgetting to Handle Single-Digit Sums

Some might assume that digit sums always reduce the string length, but a group of k ones (like "111" when k=3) sums to just 3, producing a single character "3". Not accounting for this in logic or edge case testing can lead to bugs.

5. Not Converting Back to String After Summation

A subtle bug occurs when forgetting to convert the integer sum back to a string:

Incorrect:

segment_sums.append(segment_sum)  # Wrong! Appending integer instead of string

This will cause a TypeError when trying to join the list elements.

Correct:

segment_sums.append(str(segment_sum))  # Convert to string before appending