SubsetSum Printing the list - c++

Question

The problem is to print all subsets that sum up to a value. I wrote code to check if there is a possible subset. Can some one gimme an idea to print the numbers that form the sum. Below is my code. Assume the array contains only +ve nos for simplicity.
void subsetsum(int A[], int target) {
int N = sizeof(A)/sizeof(int), sum = 0;
for(int i = 0; i < N; i++) sum += A[i];
vector<bool> V(sum + 1, 0);
V[0] = 1;
for(int i = 0; i < N; i++)
for(int j = sum; j >= 0; j--) {
if(j + A[i] <= sum && V[j]) V[A[i] + j] = 1;
}
if(V[target]) cout << "Sumbset sum exists" << endl;
else cout << "Sumbset sum doesnt exist" << endl;
}

Answer 1

First you need to generate all the subsets
If [a,b,c,d] is given array, think about generating subsets taking each element from array one at a time.
subsets(X) including y = foreach x in X append y to x
Taking a, we get subsets(a) = { [], [a] }
Take b, we get subsets(a,b) = subsets(a) + (subsets(a) including b)
= { [], [a] } + { [b], [a,b] } = { [], [a], [b], [a,b] }
Take c, subsets(a,b,c) = subsets(a,b) + (subsets(a,b) including c)
= {[], [a],[b],[a,b]} + {[c], [a,c], [b,c], [a,b,c]}
Once you get all subsets, print those whose sums equals target. You can modify the above algo further if you don't need any subsets.

Answer 2

Here's an answer in javascript:
function subsetsum(A, target) {
//int N = sizeof(A)/sizeof(int), sum = 0;
var N = A.length, sum = 0;
//for(int i = 0; i < N; i++) sum += A[i];
for(var i = 0; i < N; i++) sum += A[i];
// vector<bool> V(sum + 1, 0);
var V = [];
V[0] = [];
for(var i = 0; i < N; i++) {
for(var j = sum; j >= 0; j--) {
if(j + A[i] <= sum && V[j]) {
//Join the subset of the memoized result to this result.
V[A[i] + j] = [A[i]].concat(V[j]);
}
}
}
console.log(V);
//evaluates to true if V[target] exists
return !!V[target];
}

Answer 3

function to find power set of a vector<int>
vector<vector<int>> power_set(const vector<int>& nums) {
if (nums.empty()) { return { {} }; }
auto set = power_set(vector<int>(begin(nums) +1, end(nums)));
auto tmp = set;
for (auto& p : tmp) {
p.push_back(nums[0]);
}
set.insert(end(set), begin(tmp), end(tmp));
return set;
}
function that return all sets in the power set that sum to target
vector<vector<int>> test_sum(const vector<vector<int>>& ps, int target) {
vector<vector<int>> v;
for (auto& p : ps) {
int sum = accumulate(begin(p), end(p), 0);
if (sum == target) {
v.push_back(p);
}
}
return v;
}

Answer 4

I modified your code to print the numbers.
void subsetsum(int A[], int target) {
int N = sizeof(A) / sizeof(int), sum = 0;
for (int i = 0; i < N; i++) sum += A[i];
vector<bool> V(sum + 1, 0);
V[0] = 1;
for (int i = 0; i < N; i++)
for (int j = sum; j >= 0; j--) {
if (j + A[i] <= sum && V[j]) V[A[i] + j] = 1;
}
if (V[target]) cout << "Sumbset sum exists" << endl;
else cout << "Sumbset sum doesnt exist" << endl;
if (V[target])
{
for (int i = N - 1; i >= 0; i--)
{
if (V[target - A[i]] == 1) printf("%d, ", A[i]), target -= A[i];
}
printf("\n");
}
}
or Here's my version with vector
bool subsetsum_dp(vector<int>& v, int sum)
{
int n = v.size();
const int MAX_ELEMENT = 100;
const int MAX_ELEMENT_VALUE = 1000;
static int dp[MAX_ELEMENT*MAX_ELEMENT_VALUE + 1]; memset(dp, 0, sizeof(dp));
dp[0] = 1;
for (int i = 0; i < n; i++)
{
for (int j = MAX_ELEMENT*MAX_ELEMENT_VALUE; j >= 0; j--)
{
if (j - v[i] < 0) continue;
if (dp[j - v[i]]) dp[j] = 1;
}
}
if (dp[sum])
{
for (int i = n - 1; i >= 0; i--)
{
if (dp[sum - v[i]] == 1) printf("%d, ", v[i]), sum -= v[i];
}
printf("\n");
}
return dp[sum] ? true : false;
}