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This approach involves counting the frequency of each character in the string and calculating the length of the longest possible palindrome based on those frequencies. The key insight is that even frequency counts can completely contribute to a palindrome, while only one odd frequency character can be used once in the center of the palindrome.
Time Complexity: O(n) - where n is the length of the string as we iterate through the string to count character frequencies.
Space Complexity: O(1) - We only use a fixed-size array of 128 elements to store character frequencies.
1import java.util.HashMap;
2
3public class LongestPalindrome {
4 public static int longestPalindrome(String s) {
5 HashMap<Character, Integer> frequency = new HashMap<>();
6 for (char c : s.toCharArray()) {
7 frequency.put(c, frequency.getOrDefault(c, 0) + 1);
8 }
9 int length = 0;
10 for (int count : frequency.values()) {
11 length += (count / 2) * 2;
12 if (count % 2 == 1 && length % 2 == 0) {
13 length++;
14 }
15 }
16 return length;
17 }
18
19 public static void main(String[] args) {
20 String s = "abccccdd";
21 System.out.println(longestPalindrome(s)); // Output: 7
22 }
23}In Java, we use a HashMap to keep track of character frequencies. The calculation of the longest possible palindrome is performed similarly, using integer division to add pairs and considering potentially an odd center.
This approach leverages a set data structure to efficiently keep track of characters with odd frequencies. A character encountered an odd number of times will toggle its presence in the set. The result is derived from the total number of characters and the size of this set.
Time Complexity: O(n) - Each character is processed independently.
Space Complexity: O(1) - Fixed memory allocation for set representation.
1
Using Python's set, the solution offers an easy and clear manner to count distinct odd occurrences directly impacting the final outcome for the palindrome computation.