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# Count number of distinct sum subsets within given range

Given a set S of N numbers and a range specified by two numbers L (Lower Bound) and R (Upper Bound). Find the number of distinct values of all possible sums of some subset of S that lie between the given range. Examples :

```Input : S = { 1, 2, 2, 3, 5 }, L = 1 and R = 5
Output : 5
Explanation :  Every number between
1 and 5 can be made out using some subset of S.
{1 as 1, 2 as 2, 3 as 3, 4 as 2 + 2 and 5 as 5}

Input : S = { 2, 3, 5 }, L = 1 and R = 7
Output : 4
Explanation :  Only 4 numbers between
1 and 7 can be made out, i.e. {2, 3, 5, 7}.
3 numbers which are {1, 4, 6} can't be made out in any way.```

Prerequisites : Bitset | Bit Manipulation

Method 1(Simple) : A naive approach is to generate all possible subsets of given set, calculate their sum subset wise and push them into a hashmap. Iterate over the complete given range and count the numbers which exists in the hashmap. Method 2(Efficient) : An efficient way to solve this problem is by using bitset of size 105. Update the bitset for every element X by left shifting the bitset and doing bitwise OR with previous bitset so that the bitset at the new possible sums become 1. Then by using the concept of Prefix Sums, precompute the required count of numbers between 1 and i for prefix[1..i] to answer each query in O(1) if there are more than query being asked simultaneously. For a query L and R, answer would be simply prefix[R] – prefix[L – 1]

For e.g. S = { 2, 3, 5 }, L = 1 and R = 7 Considering a bitset of size 32 for simplicity. Initially 1 is at 0th position of bitset 00000000000000000000000000000001 For incoming 2, left shifting the bitset by 2 and doing OR with previous bitset 00000000000000000000000000000101 Similarly for 3, 00000000000000000000000000101101 for 5, 00000000000000000000010110101101 This final bitset contains 1 at those positions(possible sums) which can be made out using some subset of S. Hence between position 1 and 7, there are 4 set bits, thus the required answer.

Steps to solve the problem:

•  Initialize a bitset BS of size SZ, where SZ is a large enough constant. Initialize the 0th position of the bitset to 1.
•  Iterate over the elements S[i] of S from 0 to N-1.
• Left shift the bitset BS by S[i] positions and take the bitwise OR with the previous bitset BS.
• Initialize an array prefix of size SZ, and initialize all elements to 0.
• Iterate over the elements of prefix from 1 to SZ-1.
• Set prefix[i] = prefix[i-1] + BS[i].
• Calculate the answer as prefix[R] – prefix[L-1].

Below is the implementation of above approach in C++ :

## CPP

 `// CPP Program to count the number``// distinct values of sum of some``// subset in a range``#include ` `using` `namespace` `std;` `// Constant size for bitset``#define SZ 100001` `int` `countOfpossibleNumbers(``int` `S[], ``int` `N,``                           ``int` `L, ``int` `R)``{``    ``// Creating a bitset of size SZ``    ``bitset BS;``    ` `    ``// Set 0th position to 1``    ``BS[0] = 1;``    ` `    ``// Build the bitset``    ``for` `(``int` `i = 0; i < N; i++) {``        ` `        ``// Left shift the bitset for each``        ``// element and taking bitwise OR``        ``// with previous bitset``        ``BS = BS | (BS << S[i]);``    ``}``    ` `    ``int` `prefix[SZ];``    ` `    ``// Initializing the prefix array to zero``    ``memset``(prefix, 0, ``sizeof``(prefix));``    ` `    ``// Build the prefix array``    ``for` `(``int` `i = 1; i < SZ; i++) {``        ``prefix[i] = prefix[i - 1] + BS[i];``    ``}``    ` `    ``// Answer the given query``    ``int` `ans = prefix[R] - prefix[L - 1];``    ` `    ``return` `ans;``}` `// Driver Code to test above functions``int` `main()``{``    ``int` `S[] = { 1, 2, 3, 5, 7 };``    ``int` `N = ``sizeof``(S) / ``sizeof``(S[0]);``    ` `    ``int` `L = 1, R = 18;``    ` `    ``cout << countOfpossibleNumbers(S, N, L, R);``        ` `    ``return` `0;``}`

## Java

 `// Java Program to count the number``// distinct values of sum of some``// subset in a range` `import` `java.util.*;``import` `java.lang.*;``import` `java.io.*;`  `class` `GFG``{``  ` `    ``// Constant size for bitset``    ``static` `int` `SZ = ``100001``;` `    ``static` `int` `countOfpossibleNumbers(``int``[] S, ``int` `N, ``int` `L,``                                      ``int` `R)``    ``{` `        ``// Initially BS is 1``        ``int` `BS = ``1``;` `        ``for` `(``int` `i = ``0``; i < N; i++) {` `            ``// Left shift the bitset for each``            ``// element and taking bitwise OR``            ``// with previous bitset``            ``BS = BS | (BS << S[i]);``        ``}` `        ``// Convert BS to bitset array``        ``String BS_ = Integer.toBinaryString(BS);``        ``BS_ = String.format(``"%"` `+ (-SZ) + ``"s"``, BS_).replace(``' '``, ``'0'``);`  `        ``// Initializing the prefix array to zero``        ``int``[] prefix = ``new` `int``[SZ];``        ``for` `(``int` `i = ``0``; i < SZ; i++)``            ``prefix[i] = ``0``;` `        ``// Build the prefix array``        ``for` `(var i = ``1``; i < SZ; i++) {``            ``prefix[i]``                ``= prefix[i - ``1``] + ((BS_.charAt(i) == ``'1'``) ? ``1` `: ``0``);``        ``}` `        ``// Answer the given query``        ``int` `ans = prefix[R] - prefix[L - ``1``];` `        ``return` `ans;``    ``}` `    ``// Driver Code to test above functions``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] S = { ``1``, ``2``, ``3``, ``5``, ``7` `};``        ``int` `N = S.length;` `        ``int` `L = ``1``;``        ``int` `R = ``18``;` `        ``// Function call``        ``System.out.println(countOfpossibleNumbers(S, N, L, R));``    ``}``}` `// This code is contributed by phasing17`

## Python3

 `# Python3 Program to count the number``# distinct values of sum of some``# subset in a range` `# Constant size for bitset``SZ ``=` `100001` `def` `countOfpossibleNumbers(S, N, L, R):` `    ``# Initially BS is 1``    ``BS ``=` `1` `    ``for` `i ``in` `range``(N):` `        ``# Left shift the bitset for each``        ``# element and taking bitwise OR``        ``# with previous bitset``        ``BS ``=` `BS | (BS << S[i])` `    ``# Convert BS to bitset array``    ``BS ``=` `bin``(BS)[``2``::]``    ``BS ``=` `[``int``(i) ``for` `i ``in` `BS.zfill(SZ)][::``-``1``]` `    ``# Initializing the prefix array to zero``    ``prefix ``=` `[``0` `for` `i ``in` `range``(SZ)]` `    ``# Build the prefix array``    ``for` `i ``in` `range``(``1``, SZ):``        ``prefix[i] ``=` `prefix[i ``-` `1``] ``+` `BS[i]` `    ``# Answer the given query``    ``ans ``=` `prefix[R] ``-` `prefix[L ``-` `1``]` `    ``return` `ans` `# Driver Code to test above functions``S ``=` `[``1``, ``2``, ``3``, ``5``, ``7``]``N ``=` `len``(S)``L, R ``=` `1``, ``18` `print``(countOfpossibleNumbers(S, N, L, R))` `# This code is contributed by phasing17`

## C#

 `// C# Program to count the number``// distinct values of sum of some``// subset in a range``using` `System;``class` `GFG``{``  ` `    ``// Constant size for bitset``    ``static` `int` `SZ = 100001;` `    ``static` `int` `countOfpossibleNumbers(``int``[] S, ``int` `N, ``int` `L,``                                      ``int` `R)``    ``{` `        ``// Initially BS is 1``        ``int` `BS = 1;` `        ``for` `(``int` `i = 0; i < N; i++) {` `            ``// Left shift the bitset for each``            ``// element and taking bitwise OR``            ``// with previous bitset``            ``BS = BS | (BS << S[i]);``        ``}` `        ``// Convert BS to bitset array``        ``string` `BS_ = Convert.ToString(BS, 2);``        ``BS_ = BS_.PadRight(SZ, ``'0'``);` `        ``// Initializing the prefix array to zero``        ``int``[] prefix = ``new` `int``[SZ];``        ``for` `(``int` `i = 0; i < SZ; i++)``            ``prefix[i] = 0;` `        ``// Build the prefix array``        ``for` `(``var` `i = 1; i < SZ; i++) {``            ``prefix[i]``                ``= prefix[i - 1] + ((BS_[i] == ``'1'``) ? 1 : 0);``        ``}` `        ``// Answer the given query``        ``int` `ans = prefix[R] - prefix[L - 1];` `        ``return` `ans;``    ``}` `    ``// Driver Code to test above functions``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``int``[] S = { 1, 2, 3, 5, 7 };``        ``int` `N = S.Length;` `        ``int` `L = 1;``        ``int` `R = 18;` `        ``// Function call``        ``Console.WriteLine(``            ``countOfpossibleNumbers(S, N, L, R));``    ``}``}` `// This code is contributed by phasing17`

## Javascript

 `// JS Program to count the number``// distinct values of sum of some``// subset in a range` `// Constant size for bitset``const SZ = 100001;` `function` `countOfpossibleNumbers(S, N, L, R)``{` `    ``//Initially BS is 1``    ``var` `BS  = 1;``    ` `    ` `    ``for` `(``var` `i = 0; i < N; i++) {``        ` `        ``// Left shift the bitset for each``        ``// element and taking bitwise OR``        ``// with previous bitset``        ``BS = BS | (BS << S[i]);``    ``}``    ` `    ``//Convert BS to bitset array``    ``BS = BS.toString(2);``    ``BS = BS.padEnd(SZ, ``"0"``);``    ``BS = BS.split(``""``);``    ``BS = BS.map(str => {``  ``return` `Number(str);``});``    ` `    ``// Initializing the prefix array to zero``    ``var` `prefix = ``new` `Array(SZ).fill(0);` `    ` `    ``// Build the prefix array``    ``for` `(``var` `i = 1; i < SZ; i++) {``        ``prefix[i] = prefix[i - 1] + BS[i];``    ``}` `    ``// Answer the given query``    ``var` `ans = prefix[R] - prefix[L - 1];` `    ``return` `ans;``}` `// Driver Code to test above functions``var` `S = [ 1, 2, 3, 5, 7 ];``var` `N = S.length;``    ` `var` `L = 1;``var` `R = 18;``    ` `console.log(countOfpossibleNumbers(S, N, L, R));` `// This code is contributed by phasing17`

Output:

`18`

Time Complexity: O(S*Z) where S*Z is the maximum sum for given constraints, i.e. 105

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