Program to find Strong numbers in 1 to 100 in Java

April 1, 2022

Strong Number in 1 to 100 in Java

Here, in this page we will discuss the program to find Strong Number in 1 to 100 in Java. We will discuss the algorithm along its code in java

Strong Number Strong number is a special number whose sum of the factorial of digits is equal to the original number.

Algorithm :

Run a loop from [1, 100] and check if any number between the range is strong number or not.
To check a number is strong or not,
Store the factorial for digits 0 to 9 in an array.
Run a loop till the number is greater than 0, set
Sum = sum + (n%10)!.
At last check if(sum==n)
If yes then print the corresponding number, else check for the next number.

Time and Space Complexity :

Time Complexity : O(n)
Space Complexity : O(1)

Code in Java

Run

class Main{
 
    static int[] factorial = { 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880 };
 
    public static boolean isStrong(int N)
    {
 
        String num = Integer.toString(N);
 
        int sum = 0;
 
        for (int i = 0; i < num.length(); i++) {
            sum += factorial[Integer.parseInt(num.charAt(i) + "")];
        }
 
        return sum == N;
    }
 
    public static void printStrongNumbers(int N)
    {
 
        for (int i = 1; i <= N; i++) {

            if (isStrong(i)) {
                System.out.print(i + " ");
            }
        }
    }
 
    // Driver Code
    public static void main(String[] args) 
    {
        int N=100;
 
        // Function Call
        printStrongNumbers(N);
    }
}

Output

1 2