Easy Approach Solution Optimized Complexity 🚀
IntuitionThe problem is about determining if we can construct k palindromic strings from the given string s. The key property of palindromes is that they can ha
# Intuition
The problem is about determining if we can construct k palindromic strings from the given string s. The key property of palindromes is that they can have at most one character with an odd frequency (for the center of the palindrome), and the remaining characters must appear an even number of times.

# Approach
Check Base Case:

If the length of the string s is less than k, it's impossible to construct k palindromes because each palindrome needs at least one character.
Character Frequency Count:

Use an array of size 26 to count the frequency of each character in the string s. Each index in the array corresponds to a character ('a' to 'z').
Count Odd Frequencies:

Iterate over the frequency array and count how many characters have an odd frequency. These characters can only be used in the center of a palindrome.
If there are more than k characters with odd frequencies, it is impossible to construct k palindromes.
Decision:

If the number of odd frequencies is less than or equal to k, then constructing k palindromes is possible, so return true. Otherwise, return false.

# Complexity
- Time complexity:
𝑂(𝑛), where n is the length of the string s. We iterate through the string once to count frequencies and then iterate through the fixed-size array of size 26.


- Space complexity:
𝑂(1), since the space required is constant and only depends on the fixed-size frequency array for 26 characters.

# Code
```java []
class Solution {
    public boolean canConstruct(String s, int k) {
        int n=s.length();
        if(s.length()<k){
            return false;
        }
        int arr[]=new int[26];
        for(int i=0;i<n;i++){
            arr[s.charAt(i)-'a']+=1;
        }
        int count=0;
        for(int i=0;i<arr.length;i++){
            if(arr[i]%2!=0){
                count ++;
            }
        }
        
        if(count>k){
            return false;
        }else{
            return true;
        }
    }
}
```
['Java']