To solve this problem, you can follow these steps:

1. First, get the length of the string. This will be used to determine the possible values of k.

2. Start a loop from 1 to the length of the string. For each iteration, this value will be a possible k.

3. Check if the length of the string is divisible by the current k. If it is not, continue to the next iteration.

4. If the length of the string is divisible by k, divide the string into substrings of length k.

5. Check if each substring (except the first) is a right-rotation of the previous substring. You can do this by comparing the current substring with the previous substring rotated to the right.

6. If all substrings meet this condition, the current k is a valid k-periodic. Since we are looking for the smallest k, you can stop the loop and return the current k.

7. If no valid k is found after the loop, return the length of the string, as this means the string is not k-periodic for any k less than its length.

Here is a Python code snippet that implements these steps:

```python
def find_k(s):
n = len(s)
for k in range(1, n + 1):
if n % k == 0:
substrings = [s[i:i+k] for i in range(0, n, k)]
if all(substrings[i] == substrings[i-1][-1] + substrings[i-1][:-1] for i in range(1, len(substrings))):
return k
return n
```

You can use this function like this:

```python
print(find_k('aaaaaaaa')) # Output: 1
print(find_k('abbaabbaabba')) # Output: 2
print(find_k('abcdef')) # Output: 6
print(find_k('abccabbcaabc')) # Output: 3
```

Question

To solve this problem, you can follow these steps:

1. First, get the length of the string. This will be used to determine the possible values of k.

2. Start a loop from 1 to the length of the string. For each iteration, this value will be a possible k.

3. Check if the length of the string is divisible by the current k. If it is not, continue to the next iteration.

4. If the length of the string is divisible by k, divide the string into substrings of length k.

5. Check if each substring (except the first) is a right-rotation of the previous substring. You can do this by comparing the current substring with the previous substring rotated to the right.

6. If all substrings meet this condition, the current k is a valid k-periodic. Since we are looking for the smallest k, you can stop the loop and return the current k.

7. If no valid k is found after the loop, return the length of the string, as this means the string is not k-periodic for any k less than its length.

Here is a Python code snippet that implements these steps:

```python
def find_k(s):
    n = len(s)
    for k in range(1, n + 1):
        if n % k == 0:
            substrings = [s[i:i+k] for i in range(0, n, k)]
            if all(substrings[i] == substrings[i-1][-1] + substrings[i-1][:-1] for i in range(1, len(substrings))):
                return k
    return n
```

You can use this function like this:

```python
print(find_k('aaaaaaaa'))  # Output: 1
print(find_k('abbaabbaabba'))  # Output: 2
print(find_k('abcdef'))  # Output: 6
print(find_k('abccabbcaabc'))  # Output: 3
```

Knowee AI · Accepted Answer