906. Super Palindromes

LeetCode hard original: C# #array #csharp #hard #leetcode #string #two-pointers

Le texte du problème est traduit du russe pour la langue sélectionnée. Le code reste inchangé.

LeetCode Static

Если задан entier tableau nums, переместите все четные числа в начало tableauа, а затем все нечетные. return любой tableau, удовлетворяющий этому условию.

Exemple:

Input: left = "4", right = "1000"

Output: 4

C# solution

correspondant/original

public class Solution {
    private bool IsPalindrome(long x) {
        string s = x.ToString();
        char[] arr = s.ToCharArray();
        Array.Reverse(arr);
        return s == new string(arr);
    }
    public int SuperpalindromesInRange(string left, string right) {
        long leftNum = long.Parse(left);
        long rightNum = long.Parse(right);
        int count = 0;
        for (int i = 1; i < 100000; i++) {
            string s = i.ToString();
            long palin1 = long.Parse(s + new string(s.Reverse().ToArray()));
            long palin2 = long.Parse(s + new string(s.Reverse().Skip(1).ToArray()));
            if (palin1 * palin1 > rightNum) {
                break;
            }
            if (palin1 * palin1 >= leftNum && IsPalindrome(palin1 * palin1)) {
                count++;
            }
            if (palin2 * palin2 >= leftNum && palin2 * palin2 <= rightNum && IsPalindrome(palin2 * palin2)) {
                count++;
            }
        }
        return count;
    }
}

C++ solution

brouillon automatique, à relire avant soumission

#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:
    private bool IsPalindrome(long x) {
        string s = x.ToString();
        char[] arr = s.ToCharArray();
        Array.Reverse(arr);
        return s == new string(arr);
    }
    public int SuperpalindromesInRange(string left, string right) {
        long leftNum = long.Parse(left);
        long rightNum = long.Parse(right);
        int count = 0;
        for (int i = 1; i < 100000; i++) {
            string s = i.ToString();
            long palin1 = long.Parse(s + new string(s.Reverse().ToArray()));
            long palin2 = long.Parse(s + new string(s.Reverse().Skip(1).ToArray()));
            if (palin1 * palin1 > rightNum) {
                break;
            }
            if (palin1 * palin1 >= leftNum && IsPalindrome(palin1 * palin1)) {
                count++;
            }
            if (palin2 * palin2 >= leftNum && palin2 * palin2 <= rightNum && IsPalindrome(palin2 * palin2)) {
                count++;
            }
        }
        return count;
    }
}

Java solution

correspondant/original

class Solution {
    private boolean isPalindrome(long x) {
        String s = Long.toString(x);
        return s.equals(new StringBuilder(s).reverse().toString());
    }

    public int superpalindromesInRange(String left, String right) {
        long leftNum = Long.parseLong(left);
        long rightNum = Long.parseLong(right);
        int count = 0;

        for (int i = 1; i < 100000; i++) {
            String s = Integer.toString(i);
            long palin1 = Long.parseLong(s + new StringBuilder(s).reverse().toString());
            long palin2 = Long.parseLong(s + new StringBuilder(s.substring(0, s.length() - 1)).reverse().toString());

            if (palin1 * palin1 > rightNum) {
                break;
            }
            if (palin1 * palin1 >= leftNum && isPalindrome(palin1 * palin1)) {
                count++;
            }
            if (palin2 * palin2 >= leftNum && palin2 * palin2 <= rightNum && isPalindrome(palin2 * palin2)) {
                count++;
            }
        }

        return count;
    }
}

Python solution

correspondant/original

def is_palindrome(x):
    return str(x) == str(x)[::-1]

def superpalindromesInRange(left, right):
    left, right = int(left), int(right)
    count = 0
    max_root = int(right**0.5) + 1

    for i in range(1, 100000):
        s = str(i)
        palin1 = int(s + s[::-1])
        palin2 = int(s + s[-2::-1])

        if palin1**2 > right:
            break
        if palin1**2 >= left and is_palindrome(palin1**2):
            count += 1
        if palin2**2 >= left and palin2**2 <= right and is_palindrome(palin2**2):
            count += 1

    return count

Go solution

correspondant/original

public class Solution {
    private bool IsPalindrome(long x) {
        string s = x.ToString();
        char[] arr = s.ToCharArray();
        Array.Reverse(arr);
        return s == new string(arr);
    }

    public int SuperpalindromesInRange(string left, string right) {
        long leftNum = long.Parse(left);
        long rightNum = long.Parse(right);
        int count = 0;

        for (int i = 1; i < 100000; i++) {
            string s = i.ToString();
            long palin1 = long.Parse(s + new string(s.Reverse().ToArray()));
            long palin2 = long.Parse(s + new string(s.Reverse().Skip(1).ToArray()));

            if (palin1 * palin1 > rightNum) {
                break;
            }
            if (palin1 * palin1 >= leftNum && IsPalindrome(palin1 * palin1)) {
                count++;
            }
            if (palin2 * palin2 >= leftNum && palin2 * palin2 <= rightNum && IsPalindrome(palin2 * palin2)) {
                count++;
            }
        }

        return count;
    }
}

Algorithm

find все палиндромы до корня из right.

Проверить, являются ли квадраты этих палиндромов палиндромами и лежат ли в диапазоне [left, right].

Подсчитать количество таких суперпалиндромов.

😎

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