✅🔥BEATS 100% PROOF🔥||🌟JAVA | C++ | C# | PYTHON💡|| CONCISE CODE✅||
Proof 💯Problem Analysis 🧠We are given a string s and an integer k.
The goal is to determine if we can use all the characters in s to construct exactly k palindr
# Proof 💯
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# Problem Analysis 🧠
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We are given a string `s` and an integer `k`.  
The goal is to determine if we can use all the characters in `s` to construct exactly `k` palindrome strings.  

**Key Observations:**
1. A palindrome allows at most one character to have an odd frequency.  
2. Therefore, the minimum number of palindromes required is equal to the number of characters with odd frequencies.  
3. If the number of odd frequencies is greater than `k`, it is **impossible** to construct `k` palindromes.
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# Intuition 💡
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1. Count the frequency of each character in the string `s`.  
2. Determine the number of characters with **odd frequencies** (let's call this `oddCount`).  
3. Check the conditions:
   - If `k > s.length()`, return `false` (not enough characters to form `k` palindromes).  
   - If `oddCount > k`, return `false` (cannot form enough palindromes due to odd characters).  
   - Otherwise, return `true`.
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# Approach 🚀
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### 1. Count Odd Frequencies  
- Use a **bitmask** to track the parity (odd or even) of character frequencies efficiently.  
- Toggle the bit corresponding to each character in the bitmask.  
- Use `Integer.bitCount()` to count the number of odd frequencies.

### 2. Verify Conditions  
- If the number of odd frequencies is less than or equal to `k`, return `true`.  
- Otherwise, return `false`.
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# Complexity 🔍
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### 1. Time Complexity 🕒  
- **O(n):** Iterate through the string to update the bitmask.  

### 2. Space Complexity 🗄️  
- **O(1):** Constant space for the bitmask.

# Concept of Tags Used 📌
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- **Bit Manipulation:** Efficiently tracks the odd/even frequency of characters.  
- **Counting and Validation:** Verifies constraints to decide the output.
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# Code 📒
```java []
class Solution {
    public boolean canConstruct(String s, int k) {
        if (s.length() < k)
            return false;
        if (s.length() == k)
            return true;

        int odd = 0;

        for (char chr : s.toCharArray())
            odd ^= 1 << (chr - 'a');

        return Integer.bitCount(odd) <= k;
    }
}
```
```C++ []
class Solution {
public:
    bool canConstruct(string s, int k) {
        if (s.length() < k)
            return false;
        if (s.length() == k)
            return true;

        int odd = 0;

        for (char chr : s)
            odd ^= (1 << (chr - 'a'));

        return __builtin_popcount(odd) <= k;
    }
};
```
```python []
class Solution:
    def canConstruct(self, s: str, k: int) -> bool:
        if len(s) < k:
            return False
        if len(s) == k:
            return True

        odd = 0

        for chr in s:
            odd ^= (1 << (ord(chr) - ord('a')))

        return bin(odd).count('1') <= k

```
```Csharp []
public class Solution {
    public bool CanConstruct(string s, int k) {
        if (s.Length < k)
            return false;
        if (s.Length == k)
            return true;

        int odd = 0;

        foreach (char chr in s)
            odd ^= (1 << (chr - 'a'));

        return CountSetBits(odd) <= k;
    }

    private int CountSetBits(int n) {
        int count = 0;
        while (n > 0) {
            count += (n & 1);
            n >>= 1;
        }
        return count;
    }
}
```

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['String', 'Bit Manipulation', 'Counting', 'Python', 'C++', 'Java', 'C#']