409. Longest Palindrome

LeetCode easy original: C# #csharp #easy #hash-table #leetcode #string

Task text is translated from Russian for the selected interface language. Code is left unchanged.

LeetCode Static

Если задана string s, состоящая из строчных или прописных букв, return длину самого длинного палиндрома, который можно построить из этих букв. Буквы чувствительны к регистру, наExample, "Aa" не считается палиндромом.

Example:

Input: s = "abccccdd"

Output: 7

C# solution

matched/original

using System.Collections.Generic;
public class Solution {
    public int LongestPalindrome(string s) {
        Dictionary<char, int> charCount = new Dictionary<char, int>();
        foreach (char c in s) {
            if (charCount.ContainsKey(c)) {
                charCount[c]++;
            } else {
                charCount[c] = 1;
            }
        }
        int length = 0;
        bool oddFound = false;
        foreach (int count in charCount.Values) {
            if (count % 2 == 0) {
                length += count;
            } else {
                length += count - 1;
                oddFound = true;
            }
        }
        return oddFound ? length + 1 : length;
    }
}

C++ solution

auto-draft, review before submit

#include <bits/stdc++.h>
using namespace std;

// Auto-generated C++ draft from the C# solution. Review containers, LINQ and helper types before submit.
class Solution {
public:
    public int LongestPalindrome(string s) {
        unordered_map<char, int> charCount = new unordered_map<char, int>();
        foreach (char c in s) {
            if (charCount.count(c)) {
                charCount[c]++;
            } else {
                charCount[c] = 1;
            }
        }
        int length = 0;
        bool oddFound = false;
        foreach (int count in charCount.Values) {
            if (count % 2 == 0) {
                length += count;
            } else {
                length += count - 1;
                oddFound = true;
            }
        }
        return oddFound ? length + 1 : length;
    }
}

Java solution

matched/original

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

public class Solution {
    public int longestPalindrome(String s) {
        Map<Character, Integer> charCount = new HashMap<>();
        for (char c : s.toCharArray()) {
            charCount.put(c, charCount.getOrDefault(c, 0) + 1);
        }
        int length = 0;
        boolean oddFound = false;
        for (int count : charCount.values()) {
            if (count % 2 == 0) {
                length += count;
            } else {
                length += count - 1;
                oddFound = true;
            }
        }
        return oddFound ? length + 1 : length;
    }
}

JavaScript solution

matched/original

function longestPalindrome(s) {
    const charCount = {};
    for (const char of s) {
        charCount[char] = (charCount[char] || 0) + 1;
    }
    let length = 0;
    let oddFound = false;
    for (const count of Object.values(charCount)) {
        if (count % 2 === 0) {
            length += count;
        } else {
            length += count - 1;
            oddFound = true;
        }
    }
    return oddFound ? length + 1 : length;
}

Python solution

matched/original

def longestPalindrome(s):
    charCount = {}
    for char in s:
        charCount[char] = charCount.get(char, 0) + 1
    length = 0
    oddFound = False
    for count in charCount.values:
        if count % 2 == 0:
            length += count
        else:
            length += count - 1
            oddFound = True
    return length + 1 if oddFound else length

Go solution

matched/original

package main

func longestPalindrome(s string) int {
    charCount := make(map[rune]int)
    for _, char := range s {
        charCount[char]++
    }
    length := 0
    oddFound := false
    for _, count := range charCount {
        if count % 2 == 0 {
            length += count
        } else {
            length += count - 1
            oddFound = true
        }
    }
    if oddFound {
        return length + 1
    }
    return length
}

Algorithm

Создайте словарь для подсчета количества каждого символа в строке.

Пройдитесь по словарю и добавьте четное количество каждого символа к длине палиндрома. Если встречается нечетное количество символа, добавьте (count - 1) к длине палиндрома.

Если есть хотя бы один символ с нечетным количеством, добавьте 1 к длине палиндрома для центрального символа.

😎

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