# Subset Sum Queries in a Range using Bitset

Given an array[] of N positive integers and M queries. Each query consists of two integers L and R represented by a range. For each query, find the count of numbers that lie in the given range which can be expressed as the sum of any subset of given array.

Prerequisite : Subset Sum Queries using Bitset
Examples:

Input : arr[] = { 1, 2, 2, 3, 5 }, M = 4
L = 1, R = 2
L = 1, R = 5
L = 3, R = 6
L = 9, R = 30
Output :
2
5
4
5
Explanation : For the first query, in range [1, 2] all numbers i.e. 1 and 2 can be expressed as a subset sum, 1 as 1, 2 as 2. For the second query, in range [1, 5] all numbers i.e. 1, 2, 3, 4 and 5 can be expressed as subset sum, 1 as 1, 2 as 2, 3 as 3, 4 as 2 + 2 or 1 + 3, 5 as 5. For the third query, in range [3, 6], all numbers i.e. 3, 4, 5 and 6 can be expressed as subset sum. For the last query, only numbers 9, 10, 11, 12, 13 can be expressed as subset sum, 9 as 5 + 2 + 2, 10 as 5 + 2 + 2 + 1, 11 as 5 + 3 + 2 + 1, 12 as 5 + 3 + 2 + 2 and 13 as 5 + 3 + 2 + 2 + 1.

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

Approach : The idea is to use a bitset and iterate over the array to represent all possible subset sums. The current state of bitset is defined by ORing it with the previous state of bitset left shifted X times where X is the current element processed in the array. To answer the queries in O(1) time, we can precompute the count of numbers upto every number and for a range [L, R], answer would be pre[R] – pre[L – 1], where pre[] is the precomputed array.

Below is the implementation of above approach.

 `// CPP Program to answer subset ` `// sum queries in a given range ` `#include ` `using` `namespace` `std; ` ` `  `const` `int` `MAX = 1001; ` `bitset bit; ` ` `  `// precomputation array ` `int` `pre[MAX]; ` ` `  `// structure to represent query ` `struct` `que { ` `    ``int` `L, R; ` `}; ` ` `  `void` `answerQueries(``int` `Q, que Queries[], ``int` `N, ` `                   ``int` `arr[]) ` `{ ` `    ``// Setting bit at 0th position as 1 ` `    ``bit = 1; ` `    ``for` `(``int` `i = 0; i < N; i++) ` `        ``bit |= (bit << arr[i]); ` ` `  `    ``// Precompute the array ` `    ``for` `(``int` `i = 1; i < MAX; i++) ` `        ``pre[i] = pre[i - 1] + bit[i]; ` ` `  `    ``// Answer Queries ` `    ``for` `(``int` `i = 0; i < Q; i++) { ` `        ``int` `l = Queries[i].L; ` `        ``int` `r = Queries[i].R; ` `        ``cout << pre[r] - pre[l - 1] << endl; ` `    ``} ` `} ` ` `  `// Driver Code to test above function ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 2, 3, 5 }; ` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `M = 4; ` `    ``que Queries[M]; ` `    ``Queries.L = 1, Queries.R = 2; ` `    ``Queries.L = 1, Queries.R = 5; ` `    ``Queries.L = 3, Queries.R = 6; ` `    ``Queries.L = 9, Queries.R = 30; ` `    ``answerQueries(M, Queries, N, arr); ` `    ``return` `0; ` `} `

Output:

```2
5
4
5
```

Time Complexity : Each query can be answered in O(1) time and precomputation requires O(MAX) time.

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