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409. Longest Palindrome

Description

Given a string s which consists of lowercase or uppercase letters, return the length of the longest palindrome that can be built with those letters.

Letters are case sensitive, for example, "Aa" is not considered a palindrome here.

 

Example 1:

Input: s = "abccccdd"
Output: 7
Explanation: One longest palindrome that can be built is "dccaccd", whose length is 7.

Example 2:

Input: s = "a"
Output: 1
Explanation: The longest palindrome that can be built is "a", whose length is 1.

 

Constraints:

  • 1 <= s.length <= 2000
  • s consists of lowercase and/or uppercase English letters only.

Solutions

Solution 1: Counting

A valid palindrome string can have at most one character that appears an odd number of times, and the rest of the characters appear an even number of times.

Therefore, we can first traverse the string $s$, count the number of times each character appears, and record it in an array or hash table $cnt$.

Then, we traverse $cnt$, for each character $c$, if $cnt[c]$ is even, then directly add $cnt[c]$ to the answer $ans$; if $cnt[c]$ is odd, then add $cnt[c] - 1$ to $ans$, if $ans$ is even, then increase $ans$ by $1$.

Finally, we return $ans$.

The time complexity is $O(n)$, and the space complexity is $O(C)$. Here, $n$ is the length of the string $s$; and $C$ is the size of the character set, in this problem $C = 128$.

  • class Solution {
    public int longestPalindrome(String s) {
        int[] cnt = new int[128];
        for (int i = 0; i < s.length(); ++i) {
            ++cnt[s.charAt(i)];
        }
        int ans = 0;
        for (int v : cnt) {
            ans += v - (v & 1);
            if (ans % 2 == 0 && v % 2 == 1) {
                ++ans;
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int longestPalindrome(string s) {
        int cnt[128]{};
        for (char& c : s) {
            ++cnt[c];
        }
        int ans = 0;
        for (int v : cnt) {
            ans += v - (v & 1);
            if (ans % 2 == 0 && v % 2 == 1) {
                ++ans;
            }
        }
        return ans;
    }
};

  • class Solution:
    def longestPalindrome(self, s: str) -> int:
        cnt = Counter(s)
        ans = 0
        for v in cnt.values():
            ans += v - (v & 1)
            ans += (ans & 1 ^ 1) and (v & 1)
        return ans


  • func longestPalindrome(s string) (ans int) {
	cnt := [128]int{}
	for _, c := range s {
		cnt[c]++
	}
	for _, v := range cnt {
		ans += v - (v & 1)
		if ans&1 == 0 && v&1 == 1 {
			ans++
		}
	}
	return
}

  • function longestPalindrome(s: string): number {
    let n = s.length;
    let ans = 0;
    let record = new Array(128).fill(0);
    for (let i = 0; i < n; i++) {
        record[s.charCodeAt(i)]++;
    }
    for (let i = 65; i < 128; i++) {
        let count = record[i];
        ans += count % 2 == 0 ? count : count - 1;
    }
    return ans < s.length ? ans + 1 : ans;
}


  • use std::collections::HashMap;

impl Solution {
    pub fn longest_palindrome(s: String) -> i32 {
        let mut map: HashMap<char, i32> = HashMap::new();
        for c in s.chars() {
            map.insert(c, map.get(&c).unwrap_or(&0) + 1);
        }
        let mut has_odd = false;
        let mut res = 0;
        for v in map.values() {
            res += v;
            if v % 2 == 1 {
                has_odd = true;
                res -= 1;
            }
        }
        res + (if has_odd { 1 } else { 0 })
    }
}

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