Easiest C++ Solution | Count Method
IntuitionApproachComplexity
Time complexity:

Space complexity:
Code
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->

# Approach
<!-- Describe your approach to solving the problem. -->

# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```cpp []
class Solution {
public:
    bool canConstruct(string s, int k) {
        // If the s.length < k we cannot construct k strings from s, hence answer is false.
        if(s.length()<k) return false;
        if(s.length()==k) return true;
        
        unordered_map<char,int> mp;
        for(auto ch: s) mp[ch]++;

        int oddCharactersCount = 0;
        for(auto ch: mp) if(ch.second%2 == 1) oddCharactersCount++;

        // If the number of characters that have odd counts is > k then, k palindrome 
        // string can't be form, hence return false
        // Otherwise you can construct exactly k palindrome strings and answer is true
        return oddCharactersCount<=k;
    }
};
```
['C++']