# Subsets having Sum between A and B

Given a set of N integers. Find how many subsets of given array have sum between A and B(inclusive).

Constraints:

1 ≤ N ≤ 34,

-2 * 10^{7}≤ arr_{i}≤ 2 * 10^{7}

-5 * 10^{8}≤ A, B ≤ 5 * 10^{8}

Examples:

Input : S[] = { 1, -2, 3 }, A = -1, B = 2 Output : 5Explanation:1) 0 = 0 (the empty subset) 2) {1} = 1 3) {1, -2} = 1 + (-2) = -1 4) {-2, 3} = (-2) + 3 = 1 5) {1, -2, 3} = 1 + (-2) + 3 = 2

**Method 1 (Brute Force)**: We can generate all subsets of the given numbers i.e. Power Set and find the number of subsets that would give a sum between A and B. But this will have 2^{34} operations atmost, which is not very efficient. Hence, below is an efficient approach to solve this problem.

**Method 2 (Meet In The Middle)**: This basically reduces time complexity from O(2^{N}) to O(2^{N/2})

We divide the set into two sets [0…N/2] and [(N/2 + 1)…(N-1)] and generate all subsets sums individually for the two sets which will be 2 * 2^{17} operations. Now, what we can do is to find the combinations of these sets that would give the desired sum. This again can be done in an efficient way, sort one of the summed up set and binary search the values that will yield the sum for the particular value of the other set. Sort the second set and for each element in the first set, search for the lower bound of A – S2[i] (let say ‘low’) and upper bound of B – S2[i]

(let say ‘high’). Subtract (high – low) to get the desired answer.

For e.gS = { 1, 2, -1, 0 }, A = 1, B = -1.

After dividing S into two sets, S1 = { 1, 2 } and S2 = { -1, 0 }.

Power set of S1 = { {0}, {1}, {2}, {1, 2} } and Power set of S2 = { {0}, {-1}, {0}, {-1, 0} }

Subset Sum of S1 = { 0, 1, 2, 3 } and Subset Sum of S2 = { 0, -1, 0, -1 }

Now sort the S2 { -1, -1, 0, 0 } and for every value in S1, we binary search values that would yield the desired sum. For 0 we search for (-1) – 0 = -1 for lower bound and 1 – 0 = 1 for upper bound in S2, for 1 we search for (-1) – 1 = -2 and 1 – 1 = 0 in S2 and so on.

Below is the implementation in C++.

`// CPP Program to find the Number of Subsets that ` `// have sum between A and B ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `/* Function to Generate all subsets of a set ` ` ` `start --> Starting Index of the Set for the ` ` ` `first/second half Set ` ` ` `setSize --> Number of element in half Set ` ` ` `S --> Original Complete Set ` ` ` `res --> Store the subsets sums */` `void` `generateSubsets(` `int` `start, ` `int` `setSize, ` `int` `S[], ` ` ` `vector<` `int` `>& res) ` `{ ` ` ` `// setSize of power set of a set with setSize ` ` ` `// N is (2^n - 1) ` ` ` `unsigned ` `int` `pow_setSize = ` `pow` `(2, setSize); ` ` ` ` ` `// Store the sum of particular subset of set ` ` ` `int` `sum; ` ` ` ` ` `// Run from counter 000..0 to 111..1 ` ` ` `for` `(` `int` `counter = 0; counter < pow_setSize; counter++) { ` ` ` ` ` `// set the sum initially to zero ` ` ` `sum = 0; ` ` ` ` ` `for` `(` `int` `j = 0; j < setSize; j++) { ` ` ` ` ` `// Check if jth bit in the counter is set ` ` ` `// If set then pront jth element from set ` ` ` `if` `(counter & (1 << j)) ` ` ` `sum += S[j + start]; ` ` ` `} ` ` ` ` ` `// Store the sum in a vector ` ` ` `res.push_back(sum); ` ` ` `} ` `} ` ` ` `int` `numberOfSubsets(` `int` `S[], ` `int` `N, ` `int` `A, ` `int` `B) ` `{ ` ` ` `// Vectors to store the subsets sums ` ` ` `// of two half sets individually ` ` ` `vector<` `int` `> S1, S2; ` ` ` ` ` `// Generate subset sums for the first half set ` ` ` `generateSubsets(0, N / 2, S, S1); ` ` ` ` ` `// Generate subset sums for the second half set ` ` ` `if` `(N % 2 != 0) ` ` ` `generateSubsets(N / 2, N / 2 + 1, S, S2); ` ` ` `else` ` ` `generateSubsets(N / 2, N / 2, S, S2); ` ` ` ` ` `// Sort the second half set ` ` ` `sort(S2.begin(), S2.end()); ` ` ` ` ` `// Vector Iterator for S1 and S2; ` ` ` `vector<` `int` `>::iterator low, high; ` ` ` ` ` `// number of required subsets with desired Sum ` ` ` `int` `ans = 0; ` ` ` ` ` `for` `(` `int` `i = 0; i < S1.size(); i++) { ` ` ` ` ` `// search for lower bound ` ` ` `low = lower_bound(S2.begin(), S2.end(), A - S1[i]); ` ` ` ` ` `// search for upper bound ` ` ` `high = upper_bound(S2.begin(), S2.end(), B - S1[i]); ` ` ` ` ` `// Add up to get the desired answer ` ` ` `ans += (high - low); ` ` ` `} ` ` ` `return` `ans; ` `} ` ` ` `// Driver Program to test above functions ` `int` `main() ` `{ ` ` ` `int` `S[] = { 1, -2, 3 }; ` ` ` `int` `N = ` `sizeof` `(S) / ` `sizeof` `(S[0]); ` ` ` ` ` `int` `A = -1, B = 2; ` ` ` ` ` `// Find the number of subsets with desired Sum ` ` ` `cout << numberOfSubsets(S, N, A, B) << endl; ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

5

**Time Complexity:** O(2 * 2^{N/2}), where N is the size of set.

## Recommended Posts:

- Partition an array of non-negative integers into two subsets such that average of both the subsets is equal
- Sum of subsets of all the subsets of an array | O(3^N)
- Sum of subsets of all the subsets of an array | O(2^N)
- Sum of subsets of all the subsets of an array | O(N)
- Sum of XOR of all possible subsets
- Number of subsets with zero sum
- Sum of bitwise AND of all possible subsets of given set
- Sum of the sums of all possible subsets
- Sum of bitwise OR of all possible subsets of given set
- Number of subsets with a given AND value
- Sum of the products of all possible Subsets
- Number of subsets with a given OR value
- Number of subsets with sum divisible by M | Set 2
- Sum of squares of all Subsets of given Array
- Count of subsets with sum equal to X

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.