✅✅ easiest approach ever 🔥🔥
Intuition
The problem revolves around the properties of palindromes. A string can only form a palindrome if it has at most one character with an odd frequency.
Intuition
The problem revolves around the properties of palindromes. A string can only form a palindrome if it has at most one character with an odd frequency. Hence, the number of palindromes we can form from a string depends on the frequency of its characters.

The key insight is:

If the length of the string is less than k, it is impossible to form k substrings.
The number of characters with an odd frequency determines the minimum number of palindromes required.
Approach
Base Cases:

If the length of s (n) is less than k, return false since we cannot split the string into more substrings than its total characters.
If n == k, return true because each character can be a palindrome substring.
Character Frequency Count:

Use an unordered map to count the frequency of each character in s.
Count Odd Frequencies:

Traverse the frequency map and count the characters with odd frequencies (count).
Check Feasibility:

If the number of odd frequencies (count) exceeds k, return false since we cannot reduce the odd frequencies below this number.
Otherwise, return true.
Complexity
Time Complexity:

Building the frequency map takes O(n), where 𝑛n is the length of the string.Traversing the map to count odd frequencies takes 
O(26), as the map contains at most 26 entries (for lowercase English letters).
Overall: O(n).

Space Complexity:

The map stores at most 26 entries for the lowercase alphabet.
Overall: O(1) (constant space, independent of n).


# Code
```cpp []
class Solution {
public:
    bool canConstruct(string s, int k) {
        int n=s.length();

        if(n<k)
        {
            return false;
        }

        if(n==k)
        {
            return true;
        }


        unordered_map<char,int>map;

        int count=0;

        for(auto it:s)
        {
            map[it]++;
        }



        for(auto it:map)
        {
            if(it.second%2==1)
            {
                count++;
            }
        }

        if(count>k)
        {
            return false;
        }


        return true;
    }
};
```

![Screenshot 2024-11-06 113013.png](https://assets.leetcode.com/users/images/62ed3c9e-296a-4f70-a4c8-24c3c1c72303_1736569213.0044127.png)

['C++']