New_PG

sumarray.java

package StringProgram;

/*
Count of numbers between range having only non-zero digits whose sum of digits is N and number is divisible by M
Difficulty Level : Hard
Last Updated : 23 Apr, 2020
Given a range [L, R] and two positive integers N and M. The task is to count the numbers in the range containing only non-zero digits whose sum of digits is equal to N and the number is divisible by M.

Examples:

Input: L = 1, R = 100, N = 8, M = 2
Output: 4
Only 8, 26, 44 and 62 are valid numbers

Input: L = 1, R = 200, N = 4, M = 11
Output: 2
Only 22 and 121 are valid numbers
 */

import java.util.ArrayList;

public class sumarray {

    public static int solution(int l,int r, int m , int n)
    {
        ArrayList<Integer> all_number = new ArrayList<>();
        for (int i = l; i < r; i++) {
            all_number.add(i);
        }

        ArrayList<Integer> Sec_List = new ArrayList<>();
        for (int i = 0; i <all_number.size() ; i++) {

            int temp = all_number.get(i);
            int temp2 = all_number.get(i);
            int sum = 0;
            while(temp > 0)
            {
                int a = temp%10;
                sum = sum +a;
                temp/=10;
            }
            if (m==sum && temp2%n==0){
                Sec_List.add(temp2);
            }
        }

        ArrayList<Integer> Final_List=new ArrayList<>();
        for (int i = 0; i < Sec_List.size(); i++) {
            int temp =Sec_List.get(i);
            int temp2=Sec_List.get(i);

            int sum = 0;
            while (temp>0)
            {
                int a=temp%10;
                if (a!=0)
                {
                   sum = sum*10+a;

                }
                else
                {
                    break;
                }
                temp/=10;
            }
            int temp3 = sum;
            int final_sum = 0;
            while(temp3>0)
            {
                int a = temp3%10;
                final_sum=final_sum*10+a;
                temp3/=10;


            }
            if (final_sum==0) {
                break;
            }
            Final_List.add(final_sum);
        }


//        System.out.println(all_number);
//        System.out.println(Sec_List);
//        System.out.println();
////        System.out.println(count);
//        System.out.println(Final_List);
//        System.out.println(Final_List.size());
//        int count = Final_List.size();
        return Final_List.size();

    }

    public static void main(String[] args) {

        int l=1;
        int r=100;
        int m=8;
        int n=2;
        System.out.println(solution(l,r,m,n));

    }
}