CareerCup

Hi
Here's the code

package CareerCup;

/**
 * Created by Deven on 17/08/15.
 */
public class SumProduct {

    public static void main(String[] args) {
        int sum=1037,product=29232;

        int a,b;


        a=(sum +(int)Math.sqrt((sum*sum)-(4*product)))/2;
        b=(sum -(int)Math.sqrt((sum*sum)-(4*product)))/2;
        System.out.println("Answers are:" + a + " "+ b);
    }
}

package com.company;

import java.util.*;

/**
 * Created by ideven on 31/01/15.
 */
public class TopK {
    int[] num={1,2,6,5,4,6,7,8,8,4,2,3,2,2,2,2,2,1,1,1,1};
    public final int K=3;
    Map<Integer,Integer> map= new HashMap<Integer,Integer>();

    public void topK(){

        for(int i=0;i<num.length;i++){
            if(map.containsKey(num[i])){
                int n=map.get(num[i]);
                map.put(num[i],++n);
            }
            else{
                map.put(num[i],1);
            }

        }
        List<Map.Entry> list = new LinkedList<Map.Entry>(map.entrySet());
        // Defined Custom Comparator here
        Collections.sort(list, new Comparator() {
            public int compare(Object o1, Object o2) {
                return ((Comparable) ((Map.Entry) (o2)).getValue())
                        .compareTo(((Map.Entry) (o1)).getValue());
            }
        });


       for(int k=0;k<K;k++){
           System.out.println(list.get(k));
       }
    }

    public static void main(String[] args){
        TopK t= new TopK();
        t.topK();
    }
}

After sorting the array:
Basically there are two cases:
1) If the number of digits are even:
first two digits are multiplied with 10^(mid-1) each and added
second two digits are multiplied with 10^(mid -2) each and added and so on..
2)odd:
first digit is multiplied with 10^mid.. and so on as in 1 above
The theory is digit at highest place value should be least for both numbers..

import java.util.Arrays;


public class Minimum {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub

		int arr[]={1,2,7,8,9};
        
        Arrays.sort(arr);
                
        int mid=arr.length/2;
        int remainder=arr.length%2;
        int power=mid;
        int sum=0;
        int count=0;
        if(remainder==1)
        {
        	sum+=arr[0]*Math.pow(10, power);
        	//System.out.println(sum);
        	power--;
        	for(int i=1;i<arr.length;i++)
        	{
        		count++;
        		
        		sum+=arr[i]*Math.pow(10, power);
        		if(count==2)
        		{
        			power--;
        			count=0;
        		}
        	}
        	
        }
        else if(remainder==0)
        {power--;
        	for(int i=0;i<arr.length;i++)
        	{
        		count++;
        		sum+=arr[i]*Math.pow(10, power);
        		if(count==2)
        		{
        			power--;
        			count=0;
        		}
        	}
        }
        System.out.println(sum);
	}

}