Naive Approach

The brute force approach would involve generating all possible combinations of substrings from the input string s, and then checking if we can form k palindromes from these substrings. This approach is highly inefficient due to the combinatorial explosion of possible substring combinations.

• Time Complexity: Exponential, roughly O(2ⁿ) or higher where n is the length of the string s.
• Space Complexity: High, due to the need to store the generated substring combinations. Potentially exponential as well.

This approach is not practical for any reasonably sized input.

Optimal Approach

A more efficient approach revolves around analyzing the character frequencies in the string s. The key insight is that a palindrome can be constructed from characters that appear an even number of times. Characters that appear an odd number of times can be at the center of a palindrome. Therefore, the minimum number of palindromes required is equal to the number of characters that appear an odd number of times.

1. Count Odd Frequencies: Count the number of characters with odd frequencies in the input string s.
2. Check Conditions:
   • If the number of odd frequency characters is greater than k, it's impossible to form k palindromes.
   • If k is greater than the length of s, it's also impossible as each character would need to be its own palindrome (and string can't be split that way).
   • If the number of odd frequency characters is less than or equal to k and k is less than or equal to the string length, it's possible to form k palindromes.

Edge Cases:

• If the input string s is empty or null, return false because we need non-empty palindromes.
• If k is 0, return false as we need to construct non-empty palindromes.
• If k is 1, check if the entire string can be a palindrome (which is always true if you use all characters).
• If k is greater than the length of s, then it's impossible, because each character needs to be its own palindrome.

import java.util.HashMap;
import java.util.Map;

class Solution {
    public boolean canConstruct(String s, int k) {
        if (s == null || s.isEmpty()) {
            return false;
        }

        if (k <= 0) {
            return false;
        }

        if (k > s.length()) {
            return false;
        }

        Map<Character, Integer> charFrequencies = new HashMap<>();
        for (char c : s.toCharArray()) {
            charFrequencies.put(c, charFrequencies.getOrDefault(c, 0) + 1);
        }

        int oddCount = 0;
        for (int frequency : charFrequencies.values()) {
            if (frequency % 2 != 0) {
                oddCount++;
            }
        }

        return oddCount <= k;
    }
}

• Time Complexity: O(n), where n is the length of the string s. This is because we iterate through the string once to count character frequencies and then iterate through the HashMap (which is at most size 26) which is constant time.
• Space Complexity: O(1). The space used by the HashMap is constant because the number of distinct characters is limited (26 lowercase English letters).