Subset sum queries using bitset
Last Updated :
23 Jul, 2025
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 ? ?arri.
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.
Implementation:
CPP
// 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;
}
import java.util.BitSet;
public class SubsetSumQueries {
// Maximum allowed query value
static final int MAXSUM = 10000;
// function to check whether a subset sum equal to n
// exists in the array or not.
static void processQueries(int[] query, int nq, BitSet 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) {
System.out.print("NA, ");
continue;
}
// Else if x is a subset sum, then x'th bit
// must be set
System.out.print(bit.get(x) ? "Yes, " : "No, ");
}
}
static void preprocess(BitSet bit, int[] arr, int n) {
// Set the 0th bit because subset sum of 0 exists
bit.set(0);
// 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.
for (int j = MAXSUM - arr[i] - 1; j >= 0; j--) {
if (bit.get(j)) {
bit.set(j + arr[i]);
}
}
bit.set(arr[i]);
}
}
// Driver program
public static void main(String[] args) {
int[] arr = {3, 1, 5};
int[] query = {8, 7};
int n = arr.length;
int nq = query.length;
// a bit vector
BitSet bit = new BitSet(MAXSUM);
preprocess(bit, arr, n);
processQueries(query, nq, bit);
}
}
using System;
using System.Collections;
public class SubsetSumQueries
{
// Maximum allowed query value
const int MAXSUM = 10000;
// function to check whether a subset sum equal to n
// exists in the array or not.
static void processQueries(int[] query, int nq, BitArray 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)
{
Console.Write("NA, ");
continue;
}
// Else if x is a subset sum, then x'th bit
// must be set
Console.Write(bit[x] ? "Yes, " : "No, ");
}
}
static void preprocess(BitArray bit, int[] arr, int n)
{
// Set the 0th bit because subset sum of 0 exists
bit.Set(0, true);
// 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.
for (int j = MAXSUM - arr[i] - 1; j >= 0; j--)
{
if (bit.Get(j))
{
bit.Set(j + arr[i], true);
}
}
bit.Set(arr[i], true);
}
}
// Driver program
public static void Main(string[] args)
{
int[] arr = { 3, 1, 5 };
int[] query = { 8, 7 };
int n = arr.Length;
int nq = query.Length;
// a bit vector
BitArray bit = new BitArray(MAXSUM);
preprocess(bit, arr, n);
processQueries(query, nq, bit);
}
}
// JavaScript program to answer subset sum queries using bitset
// Maximum allowed query value
const MAXSUM = 10000;
// function to check whether a subset sum equal to n
// exists in the array or not.
function processQueries(query, nq, bit) {
let output = "";
for (let i = 0; i < nq; i++) {
const x = query[i];
if (x >= MAXSUM) {
output += "NA, ";
continue;
}
bit[x] ? output += "Yes, " : output += "No, ";
}
console.log(output.slice(0, -2));
}
// function to store all the subset sums in bit vector
function preprocess(bit, arr, n) {
// set all the bits to 0
for (let i = 0; i < MAXSUM; i++) {
bit[i] = false;
}
// set the 0th bit because subset sum of 0 exists
bit[0] = true;
// Process all array elements one by one
for (let 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.
for (let j = MAXSUM - 1; j >= arr[i]; j--) {
bit[j] = bit[j] || bit[j - arr[i]];
}
}
}
// Driver program
function main() {
const arr = [3, 1, 5];
const query = [8, 7];
const n = arr.length;
const nq = query.length;
// a vector of MAXSUM number of bits
const bit = new Array(MAXSUM);
preprocess(bit, arr, n);
processQueries(query, nq, bit);
}
main();
# Maximum allowed query value
MAXSUM = 10000
# function to check whether a subset sum equal to n
# exists in the array or not.
def processQueries(query, nq, bit):
# One by one process subset sum queries
for i in range(nq):
x = query[i]
# If x is beyond size of bit[]
if x >= MAXSUM:
print("NA, ", end="")
continue
# Else if x is a subset sum, then x'th bit
# must be set
print("Yes, ", end="")
print("No, ", end="")
print()
# function to store all the subset sums in bit vector
def preprocess(bit, arr, n):
# Process all array elements one by one
for i in range(n):
# 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
if __name__ == '__main__':
import array
arr = array.array('i', [3, 1, 5])
query = array.array('i', [8, 7])
n = len(arr)
nq = len(query)
# a bit vector
bit = 0
preprocess(bit, arr, n)
processQueries(query, nq, bit)
Time complexity : O(n * MAX_ELEMENT ) for pre-calculating since left shift operator takes O(q) for p<<q . It takes O(1) for subsequent queries, where n is the number of elements in the array.
Auxiliary Space: O(n)
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