构造 K 个回文字符串

/*
 * Problem: Given a string s and an integer k, determine if you can construct k non-empty palindrome strings using all characters in s.
 * Approach:
 * 1. Count the frequency of each character in the string using Java Streams.
 * 2. Count how many characters have an odd frequency (oddCount).
 * 3. If oddCount is greater than k, return false (since each palindrome can have at most one odd character).
 * 4. If the length of the string is less than k, return false (not enough characters to form k palindromes).
 * 5. Otherwise, return true.
 */

import java.util.Map;
import java.util.function.Function;
import java.util.stream.Collectors;

public class Solution {
    public boolean canConstruct(String s, int k) {
        if (s.length() < k) return false;
        Map<Character, Long> charCount = s.chars()
                                          .mapToObj(c -> (char) c)
                                          .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
        long oddCount = charCount.values().stream()
                                 .filter(count -> count % 2 != 0)
                                 .count();
        return oddCount <= k;
    }
}