Beats 100%✅ | Counting | Simple Solution With Added Comments | Time Complexity O(n) 
IntuitionThe idea behind this solution is that if there are more than k characters then it is not possible to make k palindromes, If this is not the case, then
# Intuition
The idea behind this solution is that if there are more than k characters then it is not possible to make k palindromes, If this is not the case, then the count of characters which occured odd times must be less than k.



# Complexity
- Time complexity:
O(n)

- Space complexity:
O(26)

# Code
```cpp []
class Solution {
public:
    bool canConstruct(string s, int k) {
        int n=s.size(),cntodd=0;
        if(k>n)return false;// not possible -> because even if each character is treated individually it would still not be possible to make k palindrome strings.
        vector<int> freq(26,0);// stores the frequency of each character
        for(int i=0;i<n;i++)freq[s[i]-'a']++;
        //count character which has occured odd times if it's less than or equal to k then only it is possible to make k palindromes
        for(int i=0;i<26;i++)cntodd+=(freq[i]&1);
        return cntodd<=k;
    }
};
```
['Counting', 'C++']