Easy C++ Beats 100%
IntuitionThe problem boils down to understanding the properties of palindromes and how character frequencies contribute to their formation:
A palindrome can hav
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem boils down to understanding the properties of palindromes and how character frequencies contribute to their formation:

   1. A palindrome can have at most one character with an odd frequency.
  2.  To create multiple palindromes, each palindrome can "use" one character with an odd frequency as its center.
   3. Therefore, the number of characters with odd frequencies `(oddCount) `must not exceed the required number of palindromes `(k).`
# Approach
<!-- Describe your approach to solving the problem. -->
1. Edge Case Checks:

   * If the string length equals` k, `each character can act as its own palindrome.
   * If the string length is less than` k,` it's impossible to create `k` palindromes.
2. Frequency Count:

    * Count the frequency of each character in the string.
3. Odd Frequency Calculation:

  *  Iterate over the frequency array to count characters with odd frequencies.
4. Validation:

   * If the number of odd-frequency characters is greater than `k,` return` false.`
  *  Otherwise, return `true. `
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
Overall -- O(N)

- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(1)

# Code
```cpp []
class Solution {
public:
    bool canConstruct(string s, int k) {
        // If the string length equals the number of palindromes required,
        // it means each character can act as its own palindrome.
        if(s.size() == k) { 
            return true;
        }
        
        // If the string length is less than the number of palindromes required,
        // it's impossible to form `k` palindromes.
        if(s.size() < k) {  
            return false;
        }
        
        // Create a frequency array for all 26 lowercase letters.
        vector<int> freq(26, 0);
        int oddCount = 0; // Counts characters with odd frequencies.
        
        // Calculate the frequency of each character in the string.
        for(char ch : s) {
            freq[ch - 'a']++;
        }

        // Count how many characters have odd frequencies.
        for(int count : freq) {
            if(count % 2 != 0) {
                oddCount++;
            }
        }

        // If the number of odd-frequency characters exceeds `k`,
        // it's impossible to form `k` palindromes.
        if(oddCount > k) {
            return false;
        }

        // Otherwise, it's possible to form `k` palindromes.
        return true;
    }
};

```
['String', 'Greedy', 'Counting', 'C++']