Understand how palindromes work! Easiest possible way!
IntuitionAt first, this problem might seem a bit tricky if you haven’t worked with palindromes before. But the key idea here is to understand how the frequencie
# Intuition
At first, this problem might seem a bit tricky if you haven’t worked with palindromes before. But the key idea here is to understand how the frequencies of characters in a string determine whether it can form a palindrome.

So, what’s special about a palindrome? It’s a string that reads the same forwards and backward, like "aba" or "abba". For that to happen, the characters need to follow a specific pattern:

    Characters with even frequencies are easy to work with—they can always form symmetric pairs on both sides of the palindrome.
    Characters with odd frequencies are trickier. Each one needs to sit in the middle of its own palindrome (because it can’t pair with anything else).

Let’s look at some examples:

 The string abc isn’t a palindrome because all characters have odd frequencies.
    The string aba is a palindrome—a appears twice (even), and b appears once (odd, so it sits in the middle).
    The string abca isn’t a palindrome, but if we rearrange it to abba, it becomes one because all characters can pair symmetrically.

For this problem, we’re trying to figure out if we can split the string into exactly k palindromes. The trick is to think about how the odd and even frequencies will contribute to the palindromes.


# Approach

Count character frequencies:

First, I counted how many times each character appears in the string. This gives us a clear picture of how many odd and even frequencies we’re dealing with.

---
Categorize frequencies:

Characters with even frequencies can form pairs and are easy to fit into palindromes.
        Characters with odd frequencies are the ones that define the minimum number of palindromes we need, because each odd frequency requires its own palindrome (or at least a central spot in one).

---
Find the range of possible palindromes:

    The minimum number of palindromes is equal to the count of characters with odd frequencies.

    The maximum number of palindromes is the total number of characters since, in the worst case, we could just make single-character palindromes.

---

Check if k is valid:
If k is at least the minimum and at most the maximum number of palindromes, we return true. Otherwise, it’s impossible to split the string into k palindromes.



# Complexity
- Time complexity:
O(n) because we only loop through the string once to count the frequencies and then process a fixed-size array of 26 letters.


- Space complexity:
 O(1) since the character frequency array is always of size 26, no matter how long the string is.

# Code
```java []
class Solution {
    public boolean canConstruct(String s, int k) {
        int []arr=new int[26];
        int ind=0;
       for(char ch:s.toCharArray()){
         arr[ch-'a']++;
       }
        int total_x=0;
        int total_y=0;
        int only_x=0;
        int only_y=0;
       for(int i=0;i<26;i++){
        if(arr[i]%2==0){
            total_x+=arr[i];
            only_x++;
        }else{
            total_y+=arr[i];
            only_y++;
        }

       }
       int min=only_y;
       int max=total_x+total_y;
       if(k>=min && k<=max ){
        return true;
       }
       return false;
        
    }
}
```

```python []
class Solution:
    def canConstruct(self, s: str, k: int) -> bool:
        from collections import Counter

        # Count character frequencies
        freq = Counter(s)

        # Count characters with odd frequencies
        odd_count = sum(1 for count in freq.values() if count % 2 != 0)

        # Minimum and maximum possible palindromes
        min_palindromes = odd_count
        max_palindromes = len(s)

        # Check if k is within range
        return min_palindromes <= k <= max_palindromes

```
```cpp []
class Solution {
public:
    bool canConstruct(string s, int k) {
        vector<int> freq(26, 0);

        
        for (char ch : s) {
            freq[ch - 'a']++;
        }

      
        int oddCount = 0;
        for (int count : freq) {
            if (count % 2 != 0) {
                oddCount++;
            }
        }

        
        int minPalindromes = oddCount;
        int maxPalindromes = s.length();

       
        return k >= minPalindromes && k <= maxPalindromes;
    }
};
```

['Hash Table', 'String', 'Greedy', 'Counting', 'C++', 'Java', 'Python3']