largestPalindrome.java

ass Solution {
    public int largestPalindrome(int n) {
        // if input is 1 then max is 9 
        if(n == 1){
            return 9;
        }
        
        // if n = 3 then upperBound = 999 and lowerBound = 99
        int upperBound = (int) Math.pow(10, n) - 1, lowerBound = upperBound / 10;
        long maxNumber = (long) upperBound * (long) upperBound;
        
        // represents the first half of the maximum assumed palindrom.
        // e.g. if n = 3 then maxNumber = 999 x 999 = 998001 so firstHalf = 998
        int firstHalf = (int)(maxNumber / (long) Math.pow(10, n));
        
        boolean palindromFound = false;
        long palindrom = 0;
        
        while (!palindromFound) {
            // creates maximum assumed palindrom
            // e.g. if n = 3 first time the maximum assumed palindrom will be 998 899
            palindrom = createPalindrom(firstHalf);
            
            // here i and palindrom/i forms the two factor of assumed palindrom
            for (long i = upperBound; upperBound > lowerBound; i--) {
                // if n= 3 none of the factor of palindrom  can be more than 999 or less than square root of assumed palindrom 
                if (palindrom / i > maxNumber || i * i < palindrom) {
                    break;
                }
                
                // if two factors found, where both of them are n-digits,
                if (palindrom % i == 0) {
                    palindromFound = true;
                    break;
                }
            }

            firstHalf--;
        }

        return (int) (palindrom % 1337);
    }

    private long createPalindrom(long num) {
        String str = num + new StringBuilder().append(num).reverse().toString();
        return Long.parseLong(str);
    }
       
}