Find smallest number K , if exists, such that product of its digits is N. Eg:when N = 6, smallest number is k=16(1*6=6) and not k=23(2*3=6)

Question

I have made this program using array concept in java. I am getting Exception as ArrayIndexOutOfBound while trying to generate product.

I made the function generateFNos(int max) to generate factors of the given number. For example a number 6 will have factors 1,2,3,6. Now,i tried to combine the first and the last digit so that the product becomes equal to 6. I have not used the logic of finding the smallest number in that array right now. I will do it later.

Question is Why i am getting Exception as ArrayIndexOutOfBound? [i couldn't figure out]

Below is my code

public class SmallestNoProduct {

    public static void generateFNos(int max) {
        int ar[] = new int[max];
        int k = 0;
        for (int i = 1; i <= max; i++) {
            if (max % i == 0) {
                ar[k] = i;
                k++;
            }
        }
        smallestNoProduct(ar);
    }

    public static void smallestNoProduct(int x[]) {
        int j[] = new int[x.length];
        int p = x.length;
        for (int d = 0; d < p / 2;) {
            String t = x[d++] + "" + x[p--];
            int i = Integer.parseInt(t);
            j[d] = i;
        }
        for (int u = 0; u < j.length; u++) {
            System.out.println(j[u]);
        }
    }

    public static void main(String s[]) {
        generateFNos(6);
    }
}
****OutputShown****

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 6
    at SmallestNoProduct.smallestNoProduct(SmallestNoProduct.java:36)
    at SmallestNoProduct.generateFNos(SmallestNoProduct.java:27)
    at SmallestNoProduct.main(SmallestNoProduct.java:52)

@Edit

The improved Code using array only.

public class SmallestNoProduct {
    public static void generateFNos(int max) {
        int s = 0;
        int ar[] = new int[max];
        int k = 0;
        for (int i = 1; i <= max; i++) {
            if (max % i == 0) {
                ar[k] = i;
                k++;
                s++;
            }
        }
        for (int g = 0; g < s; g++) {
            System.out.println(ar[g]);
        }
        smallestNoProduct(ar, s);
    }

    public static void smallestNoProduct(int x[], int s) {
        int j[] = new int[x.length];

        int p = s - 1;
        for (int d = 0; d < p;) {
            String t = x[d++] + "" + x[p--];
            System.out.println(t);

            int i = Integer.parseInt(t);
            j[d] = i;
        }
        /*for (int u = 0; u < j.length; u++) {
            System.out.println(j[u]);
        }*/
    }

    public static void main(String s[]) {
        generateFNos(6);
    }
}

Answer 1

Maybe it better:

public class SmallestNoProduct {

    public static int smallest(int n) {
        int small = n*n;
        for(int i = 1; i < Math.sqrt(n); i++) {
            if(n%i == 0) {
                int temp = Integer.parseInt(""+i+""+n/i);
                int temp2 = Integer.parseInt(""+n/i+""+i);
                temp = temp2 < temp? temp2: temp;
                if(temp < small) {
                    small = temp;
                }
            }
        }
        return small;
    }

    public static void main(String[] args) {
        System.out.println(smallest(6));  //6
        System.out.println(smallest(10)); //25
        System.out.println(smallest(100)); //205
    }

}

Answer 2

Problem lies in this line

String t=x[d++]+""+x[p--];

x[p--] will try to fetch 7th position value, as p is length of array x i.e. 6 which results in ArrayIndexOutOfBound exception. Array index starts from 0, so max position is 5 and not 6.

You can refer this question regarding postfix expression.

Note: I haven't checked your logic, this answer is only to point out the cause of exception.

Answer 3

We are unnecessarily using array here...

below method should work....

public int getSmallerMultiplier(int n)
{
  if(n >0 && n <10) // if n is 6
    return (1*10+n); // it will be always (1*10+6) - we cannot find smallest number than this
  else
  {
     int number =10;
     while(true)
     {
         //loop throuogh the digits of n and check for their multiplication
         number++;
     }
  }
}

Answer 4

int num = n;
for(i=9;i>1;i--)
{
   while(n%d==0)
   {
      n=n/d;
      arr[i++] = d;
   }
}

if(num<=9)
  arr[i++] = 1;

//printing array in reverse order;
for(j=i-1;j>=0;j--)
 system.out.println(arr[j]);