DP25 子集和问题 Subset Sum Problem @geeksforgeeks

同分割区间问题：http://blog.youkuaiyun.com/fightforyourdream/article/details/17602699

Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set with sum equal to given sum.

Examples: set[] = {3, 34, 4, 12, 5, 2}, sum = 9
Output:  True  //There is a subset (4, 5) with sum 9.

Let isSubSetSum(int set[], int n, int sum) be the function to find whether there is a subset of set[] with sum equal to sum. n is the number of elements in set[].

The isSubsetSum problem can be divided into two subproblems
…a) Include the last element, recur for n = n-1, sum = sum – set[n-1]
…b) Exclude the last element, recur for n = n-1.
If any of the above the above subproblems return true, then return true.

Following is the recursive formula for isSubsetSum() problem.

isSubsetSum(set, n, sum) = isSubsetSum(set, n-1, sum) || 
                           isSubsetSum(arr, n-1, sum-set[n-1])
Base Cases:
isSubsetSum(set, n, sum) = false, if sum > 0 and n == 0
isSubsetSum(set, n, sum) = true, if sum == 0

The above solution may try all subsets of given set in worst case. Therefore time complexity of the above solution is exponential. The problem is in-fact NP-Complete (There is no known polynomial time solution for this problem).

We can solve the problem in Pseudo-polynomial time using Dynamic programming. We create a boolean 2D table subset[][] and fill it in bottom up manner. The value of subset[i][j] will be true if there is a subset of set[0..j-1] with sum equal to i., otherwise false. Finally, we return subset[sum][n]

package DP;

public class SubsetSum {

	public static void main(String[] args) {
		int[] set = {3, 34, 4, 12, 5, 2};
		int n = set.length;
		int sum = 9;
		System.out.println(isSubsetSumRec(set, n, sum));
		System.out.println(isSubsetSumDP(set, n, sum));
	}
	
	public static boolean isSubsetSumRec(int[] set, int n, int sum){
		if(sum < 0){
			return false;
		}
		if(sum == 0){
			return true;
		}
		if(n==0 && sum>0){
			return false;
		}
		
		// not select last element || select last element
		return isSubsetSumRec(set, n-1, sum) || isSubsetSumRec(set, n-1, sum-set[n-1]);
	}
	
	public static boolean isSubsetSumDP(int[] set, int n, int sum){
		// The value of subset[i][j] will be true if there is a subset of set[0..j-1]
	    //  with sum equal to i
		boolean[][] subset = new boolean[sum+1][n+1];
		
		for(int i=0; i<=n; i++){		// If sum is 0, then answer is true
			subset[0][i] = true;
		}
		
		for(int i=1; i<=n; i++){		// If sum is not 0 and set is empty, then answer is false
			subset[i][0] = false;
		}
		
		for(int i=1; i<=sum; i++){		// sum
			for(int j=1; j<=n; j++){	// n
				subset[i][j] = subset[i][j-1];	// not select last element
				if(i-set[j-1] >= 0){		// select last element
					subset[i][j] = subset[i][j] || subset[i-set[j-1]][j-1];
				}
			}
		}
		
		return subset[sum][n];
	}

}