3 Line | Best Easy Solution | Beginner Friendly | O(n) Time :)
IntuitionThe problem revolves around the concept of constructing palindromes, which is heavily influenced by the frequency of characters in the string. A palind
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# Intuition
The problem revolves around the concept of constructing palindromes, which is heavily influenced by the frequency of characters in the string. A palindrome can tolerate at most one character with an odd frequency, as this odd character can occupy the center of the palindrome. The rest of the characters must have even frequencies to ensure symmetry. Thus, the number of characters with odd frequencies dictates the minimum number of palindromes required.

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# Approach
1. **Character Frequency Count**:
   - Use a hash map (or `Counter` from Python's `collections`) to count the occurrences of each character in the string.

2. **Count Odd Frequencies**:
   - Iterate through the frequency map and count how many characters have an odd frequency. This count determines the minimum number of palindromes needed.

3. **Validate Conditions**:
   - Check if \( k \) is at least the number of odd frequencies (to satisfy the minimum palindrome requirement).
   - Check if \( k \) is less than or equal to the length of the string (as you cannot form more palindromes than the available characters).

4. **Return the Result**:
   - If both conditions are satisfied, return `True`; otherwise, return `False`.

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# Complexity

- **Time complexity**:
  - Computing the frequency of characters takes \( O(n) \), where \( n \) is the length of the string.
  - Counting odd frequencies involves iterating through the hash map, which has a constant size of 26 (for lowercase English letters), making it \( O(1) \).
  - Overall, the time complexity is **\( O(n) \)**.

- **Space complexity**:
  - Storing the frequency count requires \( O(1) \) space since the hash map has a maximum of 26 entries (one for each lowercase English letter).
  - Overall, the space complexity is **\( O(1) \)**.

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# Code
```python3 []
class Solution:
    def canConstruct(self, s: str, k: int) -> bool:

        freq = Counter(s)
        
        odd_count = sum(1 for count in freq.values() if count % 2 != 0)
       
        return odd_count <= k <= len(s)


        
```
['Python3']