Find the largest palindrome made from the product of two 3-digit numbers. (Java)

Question

I am having a bit of trouble solving Project Euler question 4. I am inexperienced in programming and didn\'t really understand other answers. I managed to write a code that

Answer 1

Here is my take on the same question...

It is way more easier to create a list, that we know will be filled from smallest to the largest, because counter numbers only rize...

        List<Integer> palindrome = new ArrayList<>(); // create integer list

    for ( int i = 900; i<999; i++){
        for ( int j = 900; j < 950; j++){
            // both counter numbers are bigger than 900
            // because there will those two int's be, logically
            String newMul = String.valueOf(i * j);  // make it string for palindrome comparation
            String strRev = new StringBuilder(newMul).reverse().toString(); // rotate previous string
                                                                            // for palindrome comparation 
            if(newMul.equals(strRev)){  //if its equal
            palindrome.add(i * j);      //will put multiplication result of i and j into list
            }
        }
    }
    System.out.println(palindrome.get(palindrome.size()-1)); // last number in list is the 
}                                                            // biggest so lets print it out

Basicly, you just need little casting to se if it is a palindrome, but if it is, just multiply i and j to put them into list... It will loop for a long time, but that way you checked all the numbers bigger than 900 for the biggest palindrome...

Hope I helped...

Answer 2

Here is a code to check PALINDROME-ness of a number without converting it to a String.

bool checkPalindrome(int n)
{
    int num = n;
    int s = 0;
    while(num!=0)
    {
      s = s*10 + (num%10);
      num = num/10;
    }
    if(s==n)
      return true;
    return false;
}

As you can see that if n = 120, the reverse is calculated as s = 21 instead of 021, but that is fine, keeping in mind that a palindrome number won't end in 0, because if it does, then it would have to begin with 0 which makes it an invalid number!!!!

Hope this helps!!!

Answer 3

for 3 digits loop can be started from 100 to 1000(exclusive).

You can try this ::

int num1 = 0, num2 = 0;
for(int i=100; i<1000; i++){
    for(int j=100; j<1000; j++){
        String mul = String.valueOf(i*j);
        if(isPalindrome(mul)){
            num1 = i;
            num2 = j;
        }
    }
}
System.out.println(num1*num2);

Implement palindrome method as :

boolean isPalindrome(String str) {
    String strRev = new StringBuilder(str).reverse().toString();
    return str.equals(strRev);
}

This works good but take this as reference and improve as needed Thanks

Answer 4

Here is a method which loops over all numbers from 1-999 and multiplies them with all possible numbers from 1-999. You will need to define a method isPalindrome(int) that will check if a given int is a palindrome.

//save the largest number
int largest = 0;

//loop over every possible product of numbers from 100-999
for (int i = 999; i >= 100; i--) {
 for (int j = i; j >= 100; j--) {
  int curr = i * j;

  //if the current number is a palindrome and is greater than the last found largest
  if (isPalindrome(curr) && curr > largest) {

   //save it as the new largest found number
   largest = curr;
  }
 }
}

The isPalindrome method might look like this:

private static boolean isPalindrome(Integer possible) {
    String toStr = possible.toString();
    int len = toStr.length();
    if (len % 2 == 1) {
        len = len - 1;
    }

    for (int i = 0; i < len / 2; i++) {
        if (toStr.charAt(i) != toStr.charAt(toStr.length() - (1 + i))) {
            return false;
        }
    }

    return true;
}