CareerCup

import java.util.ArrayList;
import java.util.List;

/**
 * 
 * Find all the subset of an array, having integers from 1 to 9, such that sum of each subset is 10.
 *
 */
public class SubsetSum {
	
	
	public SubsetSum(int[] arr, int sum) {
		super();
		this.arr = arr;
		this.baseSum = sum;
	}
	
	
	
	public List<List<Integer>> find(){
		
		List<List<Integer>> allSubsets = new ArrayList<List<Integer>>();
		
		for( int i =0; i < arr.length; i++ ){
			
			int value = arr[i];
			
			if( value > 0 && value < 10 ){
			
				List<List<Integer>> subsetTails = getSubsets( baseSum - value, i+1 );
					
				for(List<Integer> tail : subsetTails ){
						
					List<Integer> singleSubset = new ArrayList<Integer>();
					singleSubset.add( value );
					singleSubset.addAll( tail );
						
					if( isBaseSumSatisfied( singleSubset ) ){
						allSubsets.add(singleSubset);
					}
				}
			}
		}
		
		return allSubsets;
	}
		
	
	//==== PRIVATE =====
	
	private final int[] arr;
	private final int baseSum;
	
	private List<List<Integer>> getSubsets( int sum, int offset ){
		
		if( sum <= 0 || offset >= arr.length){
			return new ArrayList<List<Integer>>();
		}
				
		List<List<Integer>> subsets = new ArrayList<List<Integer>>();		
		
		
		for( int i = offset; i < arr.length; i++ ){
			
			if( arr[i] > 0 && arr[i] < 10 ){
				
				List<List<Integer>> posibleTails = getSubsets(sum - arr[i], i+1);
				
				if( posibleTails.isEmpty() ){
					List<Integer> singleSubset = new ArrayList<Integer>();
					singleSubset.add( arr[i] );					
					subsets.add(singleSubset);
				}
				else {				
					for( List<Integer> tail : posibleTails ){
						List<Integer> singleSubset = new ArrayList<Integer>();
						singleSubset.add( arr[i] );
						singleSubset.addAll(tail);
						
						subsets.add(singleSubset);
					}
				}
			}
		}
		return subsets;
	}


	private boolean isBaseSumSatisfied(List<Integer> singleSubset ){
		
		int actualSum = 0;
		
		for(int value : singleSubset ){
			actualSum += value;
		}
		return baseSum == actualSum;
	}

}

output:

arr: {11,16,11,9,7,13,5,3,1,14}
[9, 1], sum: 10
[7, 3], sum: 10

dynamic programming problem...same as coin change

1. Sort the given array by ascending or descending mode by using the quick sort algorithm
2. Subtract one by one every element of the given array from the given number (10) and using the binary search algorithm you will find the next pair.
The complexity of this algorithm is O(LogN)

Hey, did u apply for Infosys in India or US? Is there any other way to apply to Infosys, apart from their online application? because I don't think their online system is working that well. There is absolutely no response for any application, not even application receipt or acknowledgement. If there is, please do let me know.
Thanks!

can anybody give solution as a recursive function of c/c++?

This is a dynamic programming problem. Given an array A of numbers:
and the recurrence:

T(i) = Set of sets T(i-1), {A[i]}, {{T(j)} union A[i]} for all j < i s.t. sum in the sets + A[i] <= 10 
       if A(i) <= 10
       
       T(i-1)     if A(i) > 10

Consider the array and the memoization as follows:

A   Set of sets < 10
 
17  {Empty}
6   {6}
9   {6}, {9}
2   {6}, {9}, {2}, {2,6}, {2,9}
7   {6}, {9}, {2}, {2,6}, {2,9}, {7}, {7,2}
8   {6}, {9}, {2}, {2,6}, {2,9}, {7}, {7,2}, {8}, {8,2}
1   {6}, {9}, {2}, {2,6}, {2,9}, {7}, {7,2}, {8}, {8,2}, {1}, {1,6}, {1,9}, {1,2}, {1,2,6}, {1,7}, {1,7,2}, {1,8}
3   {6}, {9}, {2}, {2,6}, {2,9}, {7}, {7,2}, {8}, {8,2}, {1}, {1,6}, {1,9}, {1,2}, {1,2,6}, {1,7}, {1,7,2}, {1,8}
    {3}, {3, 6}, {3, 2}, {3,7}, {3, 1} {3, 1, 6}, {3, 1, 2}
4   {6}, {9}, {2}, {2,6}, {2,9}, {7}, {7,2}, {8}, {8,2}, {1}, {1,6}, {1,9}, {1,2}, {1,2,6}, {1,7}, {1,7,2}, {1,8}
    {3}, {3, 6}, {3, 2}, {3,7}, {3, 1} {3, 1, 6}, {3, 1, 2}, {4}, {4, 6}, {4, 2}, {1, 4}, {1, 2, 4}, {3, 4}, 
    {3, 2, 4}, {3, 1, 4}
7   {6}, {9}, {2}, {2,6}, {2,9}, {7}, {7,2}, {8}, {8,2}, {1}, {1,6}, {1,9}, {1,2}, {1,2,6}, {1,7}, {1,7,2}, {1,8}
    {3}, {3, 6}, {3, 2}, {3,7}, {3, 1} {3, 1, 6}, {3, 1, 2}, {4}, {4, 6}, {4, 2}, {1, 4}, {1, 2, 4}, {3, 4}, 
    {3,2,4}, {3,1,4}, {7}, {7,2}, {1,7}, {1,2,7}, {3,7}

Now look at the sets of sets and select the ones that sum up to 10! But this algo is O(n^2).

public class SubsetSum {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		int num[] = new int[]{11,16,11,9,7,13,5,3,1,14};
		count(num,10);

	}
	
	static void count(int num[],int sum){
		
		for(int i=0;i<Math.pow(2, num.length);i++){
			int set[]= new int[num.length];
			int tsum=0;
			for(int j=0;j<num.length;j++){
				if((i&(1<<j))!= 0){
					set[j] = 1;
					tsum+=num[j];
				}
			}
			if(tsum ==  sum){
			for(int k=0;k<num.length;k++){
				if(set[k] == 1)System.out.print(" "+num[k]);
			}
			System.out.print(" : "+sum+" \n");
			}
		}
		
	}

}

using bit operation we can get all the subsets but its limited by the input size :)
Actually the subset sum problem is a NP complete problem.