 ⚡⚡⚡One-Liner Solution with Efficient Checks ✅⚡⚡⚡
IntuitionThe problem involves determining if a string s can be rearranged into k palindromes. Each palindrome can have at most one character with an odd frequen


# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem involves determining if a string `s` can be rearranged into `k` palindromes. Each palindrome can have at most one character with an odd frequency (in the middle), and the rest must appear in pairs. This means the number of characters with odd frequencies must not exceed `k`.

# Approach
<!-- Describe your approach to solving the problem. -->
1. First, check if the length of the string `s` is at least `k`. If not, it's impossible to form `k` non-empty palindromes.
2. Count the frequency of each character in `s`.
3. Calculate how many characters have an odd frequency.
4. If the number of odd frequencies is greater than `k`, return `False`. Otherwise, return `True`.

# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
$$O(n)$$, where $$n$$ is the length of the string. Calculating character frequencies involves iterating through the string, and the `set` operation adds a negligible overhead for unique character extraction.

- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
$$O(1)$$ if we consider the set of characters fixed (e.g., ASCII or Unicode range). Otherwise, $$O(u)$$, where $$u$$ is the number of unique characters in `s`.

# Code
```python
class Solution:
    def canConstruct(self, s: str, k: int) -> bool:
        return len(s) >= k and sum(s.count(i) % 2 for i in set(s)) <= k
['Python3']