only consider the parity of freq+bitset|Py3 1-liner|C++ beats 100%
IntuitionFollow the hints.
This problem becomes fairly easy.The key observation: palindrome strings of even lengths have all letters even occurrences, palindrom
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Follow the hints.
This problem becomes fairly easy.

The key observation: palindrome strings of even lengths have all letters even occurrences, palindrome strings of odd lengths have all letters even occurrences but except one letter with odd occurrence.
If n is big enough, say n>=k, it's only to consider the number of odd occurrences for chars.

Python 1-liner is made.
# Approach
<!-- Describe your approach to solving the problem. -->
[Please Turn on English subtitles if necessary]
[https://youtu.be/zCV4ZiBNlKQ?si=4erkSpNzpDiUM4Xs](https://youtu.be/zCV4ZiBNlKQ?si=4erkSpNzpDiUM4Xs)
1. let `n=|s|`
2. If `n<k` return false
3. Initialize the `bitset<26> freq=0`
4. Set `freq.flip(c-'a')` same as `freq[c-'a']^=1` for each c in s
5. if `bitcount(freq)<=k` return `true` else `false`
6. Python code is made
7. 2nd version for C++ is made by using C++20 popcount
8. Python 1-liner is made in use of Counter which is a submodule of dict
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
$O(n+26)$
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
$O(26)$
# Code||C++ bitset 0ms beats 100%
```cpp []
class Solution {
public:
    bool canConstruct(string& s, int k) {
        const int n=s.size();
        if (n<k) return 0;
        bitset<26> freq=0;
        for(char c: s)
            freq.flip(c-'a');
        return freq.count()<=k;
    }
};
```
```cpp [C++ 2nd ver]
class Solution {
public:
    bool canConstruct(string& s, int k) {
        const int n=s.size();
        if (n<k) return 0;
        unsigned freq=0;
        for(char c: s)
            freq^=(1<<(c-'a'));
        return popcount(freq)<=k;
    }
};
```
# Python
```
class Solution:
    def canConstruct(self, s: str, k: int) -> bool:
        n=len(s)
        if n<k: return False
        freq=[False]*26
        for c in s:
            freq[ord(c)-97]^=1
        return freq.count(True)<=k
        
```
# Python 1-liner
```
class Solution:
    def canConstruct(self, s: str, k: int) -> bool:
        return (freq:=Counter(s)) and sum(f&1 for f in freq.values())<=k<=len(s)
```
# Some other Palindrome Problems
If you think this medium question too easy, try the following!
[2108. Find First Palindromic String in the Array](https://leetcode.com/problems/find-first-palindromic-string-in-the-array/solutions/4717919/python-1-liner-vs-several-c-c-sols-19ms-beats-99-86/)<- easy
[1457. Pseudo-Palindromic Paths in a Binary Tree](https://leetcode.com/problems/pseudo-palindromic-paths-in-a-binary-tree/solutions/4616431/dfs-bitset-parity-check-beats-100/)<- in combination with binary tree
[3260. Find the Largest Palindrome Divisible by K](https://leetcode.com/problems/find-the-largest-palindrome-divisible-by-k/solutions/5654462/modular-arithmetic-only-k-7-is-hard-number-theoretical-sol/)<- really hard
['Greedy', 'Bit Manipulation', 'C++', 'Python3']