# Subset sum queries using bitset

Given an array arr[] and a number of queries, where in each query we have to check whether a subset whose sum is equal to given number exists in the array or not.

Examples:

```Input : arr[]   = {1, 2, 3};
query[] = {5, 3, 8}
Output : Yes, Yes, No
There is a subset with sum 5, subset is {2, 3}
There is a subset with sum 3, subset is {1, 2}
There is no subset with sum 8.

Input : arr[] = {4, 1, 5};
query[] = {7, 9}
Output : No, Yes
There is no subset with sum 7.
There is a subset with sum 9, subset is {4, 5}
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to use bitset container in C++. Using bitset, we can precalculate the existence all the subset sums in an array in O(n) and answer subsequent queries in just O(1).

We basically use an array of bits bit[] to represent the subset sum of elements in the array. Size of bit[] should be at least sum of all array elements plus 1 to answer all queries. We keep of bit[x] as 1 if x is a subset sum of given array, else false. Note that indexing is assumed to begin with 0.

```For every element arr[i] of input array,
we do following

// bit[x] will be 1 if x is a subset
// sum of arr[], else 0
bit = bit | (bit << arr[i])```

How does this work?

```Let us consider arr[] = {3, 1, 5}, we need
to whether a subset sum of x exists or not,
where 0 ≤ x ≤ Σarri.

We create a bitset bit[10] and reset all the
bits to 0, i.e., we make it 0000000000.

Set the 0th bit, because a subset sum of 0
exists in every array.
Now, the bit array is 0000000001

Apply the above technique for all the elements
of the array :
Current bitset = 0000000001

After doing "bit = bit | (bit << 3)",
bitset becomes	0000001001

After doing "bit | (bit << 1)",
bitset becomes	0000011011

After doing "bit | (bit << 5)",
bitset becomes	1101111011
```

Finally, we have the bit array as 1101111011, so, if bit[x] is 1 then a subset sum of x exists otherwise not. We can clearly observe that a subset sum of all the numbers from 0 to 9 except 2 and 7 exists in the array.

Here is a C++ implementation :

```// C++ program to answer subset sum queries using bitset
#include <bits/stdc++.h>
using namespace std;

// Maximum allowed query value
# define MAXSUM 10000

// function to check whether a subset sum equal to n
// exists in the array or not.
void processQueries(int query[], int nq, bitset<MAXSUM> bit)
{
// One by one process subset sum queries
for (int i=0; i<nq; i++)
{
int x = query[i];

// If x is beyond size of bit[]
if (x >= MAXSUM)
{
cout << "NA, ";
continue;
}

// Else if x is a subset sum, then x'th bit
// must be set
bit[x]? cout << "Yes, " : cout << "No, ";
}
}

// function to store all the subset sums in bit vector
void preprocess(bitset<MAXSUM> &bit, int arr[], int n)
{
// set all the bits to 0
bit.reset();

// set the 0th bit because subset sum of 0 exists
bit[0] = 1;

// Process all array elements one by one
for (int i = 0; i < n; ++i)

// Do OR of following two
// 1) All previous sums. We keep previous value
//    of bit.
// 2) arr[i] added to every previous sum. We
//    move all previous indexes arr[i] ahead.
bit |= (bit << arr[i]);
}

// Driver program
int main()
{
int arr[] = {3, 1, 5};
int query[] = {8, 7};

int n  = sizeof(arr) / sizeof(arr[0]);
int nq = sizeof(query) / sizeof(query[0]);

// a vector of MAXSUM number of bits
bitset<MAXSUM> bit;

preprocess(bit, arr, n);
processQueries(query, nq,  bit);

return 0;
}
```

Output:

```Yes, No,
```

Time complexity : O(n) for pre-calculating and O(1) for subsequent queries, where n is the number of elements in the array.

Refer http://stackoverflow.com/questions/12459563/what-is-the-size-of-bitset-in-c for space requirements of this approach.

This article is contributed by Avinash Kumar Saw. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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