easy intuition based solution runs in O(n) with hash table
IntuitionAny even occurence can be easily be used to form a new palindrome.
What about characters with odd frequency ?ApproachCount the total occurences of char
# Intuition
Any even occurence can be easily be used to form a new palindrome.
What about characters with odd frequency ?

# Approach
Count the total occurences of characters with odd freqquencies . if this value say count
count <= k then we can form that many odd length palindromic strings 
ex -: s = "annabelle" k = 3
frequencies of various characters -
a -> 2
n -> 2
b -> 1
e -> 2
l -> 2
an example of pallindromic substrings -> "aba" , "lnnl" , "ee"

Edge case -> when s.length() < k What shall be output?

# Complexity
- Time complexity:
O(N)

- Space complexity:
O(1)

# Code
```cpp []
class Solution {
public:
    bool canConstruct(string s, int k) {
        if(s.length() < k) return false;
        vector<int> freq(26,0);
        for(auto it: s) freq[it-'a']++;
        int count = 0;
        for(int i=0;i<26;i++){
            if(freq[i]&1) count++;
            
        }
        if(count > k) return false;
        return true;
    }
};
```
['Hash Table', 'Counting', 'C++']