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}

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 ≤ Σarr_{i}. 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; ` `} ` |

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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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