Integers - Maximum and Count

Integers - Maximum and Count 

Problem:

N integers are passed as the input. The program must repeat the steps given below. 

Step 1: Find the maximum M of the available integers and check if M is a single digit integer. If M is a single digit integer exit the program after printing M and the count of integers C present finally. 

Step 2: If M is not a single digit integer then divide M into M/2 and M-M/2. Remove M from the list of integers and add M/2 and M-M/2 to the list of integers. Now repeat steps 1 and 2 until M is a single digit integer. 

Boundary Condition(s): 2 <= N <= 10^5 

Example Input/Output 1: 

Input: 

4 

12 6 20 8 

Output: 

8 8 

Explanation: 

20 is the maximum. So the integers become 12 6 10 10 8. Now 12 is the maximum. So the integers become 6 6 6 10 10 8. Now 10 in the maximum. So the integers become 6 6 6 5 5 10 8. Again 10 in the maximum. So the integers become 6 6 6 5 5 5 5 8. Now 8 is the maximum and is a single digit integer. So 8 and 8 are printed. The second 8 denotes the count of integers present. 

Example Input/Output 2: 

Input: 6 

4 5 9 7 8 2 

Output: 

9 6 

Example Input/Output 3: 

Input: 

9 

2 3 3 2 3 2 2 100 55 

Output: 

7 31

Program:

import java.util.*;

public class Hello {

    public static void fun(int k,int [] a)

    {

        if(k < 10)

        {

            a[k]++;

        }

        else

        {

            fun(k/2,a);

            fun(k-(k/2),a);

        }

    }

    public static void main(String[] args) 

    {

Scanner sc = new Scanner(System.in); 

int n = sc.nextInt(); 

int a [] = new int [10]; 

while(n-- > 0)

{

    int k = sc.nextInt(); 

    fun(k,a);

}

for(int i = 9;i>=0;i--)

{

    if(a[i]!=0)

    {

        System.out.print(i+" ");

        break;

    }

}

int s = 0;

for(int i=0;i<=9;i++)

{

    s+= a[i];

}

System.out.print(s);

}

}

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