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Perfect Subset Sums

// A Java program to count all subsets with given sum.
import java.util.ArrayList;
public class SubSet_sum_problem
// dp[i][j] is going to store true if sum j is
// possible with array elements from 0 to i.
static boolean[][] dp;

static void display(ArrayList<Integer> v)

// A recursive function to print all subsets with the
// help of dp[][]. Vector p[] stores current subset.
static void printSubsetsRec(int arr[], int i, int sum,
ArrayList<Integer> p)
// If we reached end and sum is non-zero. We print
// p[] only if arr[0] is equal to sun OR dp[0][sum]
// is true.
if (i == 0 && sum != 0 && dp[0][sum])

// If sum becomes 0
if (i == 0 && sum == 0)

// If given sum can be achieved after ignoring
// current element.
if (dp[i-1][sum])
// Create a new vector to store path
ArrayList<Integer> b = new ArrayList<>();
printSubsetsRec(arr, i-1, sum, b);

// If given sum can be achieved after considering
// current element.
if (sum >= arr[i] && dp[i-1][sum-arr[i]])
printSubsetsRec(arr, i-1, sum-arr[i], p);

// Prints all subsets of arr[0..n-1] with sum 0.
static void printAllSubsets(int arr[], int n, int sum)
if (n == 0 || sum < 0)

// Sum 0 can always be achieved with 0 elements
dp = new boolean[n][sum + 1];
for (int i=0; i<n; ++i)
dp[i][0] = true;

// Sum arr[0] can be achieved with single element
if (arr[0] <= sum)
dp[0][arr[0]] = true;

// Fill rest of the entries in dp[][]
for (int i = 1; i < n; ++i)
for (int j = 0; j < sum + 1; ++j)
dp[i][j] = (arr[i] <= j) ? (dp[i-1][j] ||
: dp[i - 1][j];
if (dp[n-1][sum] == false)
System.out.println("There are no subsets of size "+ n +" or smaller with" + " sum "+ sum +".");

// Now recursively traverse dp[][] to find all
// paths from dp[n-1][sum]
ArrayList<Integer> p = new ArrayList<>();
printSubsetsRec(arr, n-1, sum, p);

//Driver Program to test above functions
public static void main(String args[])
int arr[] = {16, 16, 15, 14, 11, 10, 9, 9, 7, 7, 5, 2};
int n = 12;
int sum = 72;
printAllSubsets(arr, n, sum);
//This code is contributed by Sumit Ghosh
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