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3 weeks ago+ 0 comments

Python 3

import sys
t = int(input().strip())
for a0 in range(t):
    n = int(input().strip())
    count =0
    for i in range(n-1,10000,-1):
        if str(i)==str(i)[::-1]:
            for j in range(100,1000):
                if i%j==0 and i/j <=999:
                    count +=1
                    print(i)
                    break;
        if count ==1:
            break;

3 weeks ago+ 0 comments

Java has a NavigableSet interface (implemented by TreeSet) that is helpful for this situation. So first we construct the set, and then query for the result using the lower method.

public class Solution {
  
  static boolean isPalindrome(int test) {
    String c = String.valueOf(test);
    return c.equals(new StringBuilder(c).reverse().toString());
  }
  
  static NavigableSet<Integer> createPalindromesSet() {
    NavigableSet<Integer> numbers = new TreeSet<>();
    for(int i = 100;i<1_000;i++){
      for(int j = Math.max(i,100_000/i); j<1_000;j++) {
        int product = i * j;
        if(product%11==0 && isPalindrome(product)) {
          numbers.add(product);;
        }
      }
    }
    return numbers;
  }

  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int t = in.nextInt();
    NavigableSet<Integer> numbers = createPalindromesSet();
    for(int a0 = 0; a0 < t; a0++){
        int n = in.nextInt();
        System.out.println(numbers.lower(n));
    }
    in.close();
  }
}

3 weeks ago+ 0 comments

Java Solution;

import java.util.; import java.text.; import java.math.; import java.util.regex.;

public class Solution {

public static void main(String[] args){
    Scanner in = new Scanner(System.in);
    int t = in.nextInt();
    for(int a0 = 0; a0 < t; a0++){
        int n = in.nextInt();

        System.out.println(largestPalindrome(n));
    }
}

public static long palindrome(long a){
    long rev = 0;
    while(a!=0){
        rev = rev*10 + a%10;
        a = a/10;
    }
    return rev;
}

public static boolean check(long a){
    return a == palindrome(a);
}

public static long largestPalindrome(long n){
        long largest = 0;
        long res=0;
        for(int i=999; i>=100; i--){
            for(int j=999; j>=100; j--){
                res = j*i;
                if(check(res) && res<n){
                    largest = Math.max(largest,res);
                }
            }
        }
    return largest;
}

}

4 weeks ago+ 0 comments

Python 3

#!/bin/python3

import sys

def checkPalindrome(n):
    number_str = str(n)
    length = len(number_str)
    isPalindrome = True

    if (length % 2 != 0):
        length -= 1

    i = 0
    while i < length / 2:
        if number_str[i] != number_str[len(number_str) - (i + 1)]:
            isPalindrome = False
            break
        i += 1

    return isPalindrome

def findProduct(n):
    palindrome = 0

    for i in range(101, 1000):
        for j in range(i, 1000):
            product = i * j
            if product >= n or product <= palindrome:
                continue
            if checkPalindrome(product):
                palindrome = product
            
    return palindrome

t = int(input().strip())
for a0 in range(t):
    n = int(input().strip())
    print(findProduct(n))

1 month ago+ 0 comments

Python3 Solution

pl = []
for i in range(100, 1000):
    for j in range(100, 1000):
        pro = i * j
        if str(pro)[::-1] == str(pro):
            pl.append(pro)
pl.sort(reverse=True)
for i in range(int(input())):
    n = int(input())
    for j in pl:
        if j < n:
            print(j)
            break

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