Subset Sum Queries in a Range using Bitset

Last Updated : 21 Mar, 2023

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.

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], the answer would be pre[R] – pre[L – 1], where pre[] is the precomputed array.

Below is the implementation of the above approach.

C++

   // CPP Program to answer subset
// sum queries in a given range
#include <bits/stdc++.h>
using namespace std;
 
const int MAX = 1001;
bitset<MAX> 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[0] = 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[0]);
    int M = 4;
    que Queries[M];
    Queries[0].L = 1, Queries[0].R = 2;
    Queries[1].L = 1, Queries[1].R = 5;
    Queries[2].L = 3, Queries[2].R = 6;
    Queries[3].L = 9, Queries[3].R = 30;
    answerQueries(M, Queries, N, arr);
    return 0;
}
 
  

Java

   import java.util.Arrays;
 
// Class to represent query
class Que {
  int L, R;
 
  Que(int L, int R) {
    this.L = L;
    this.R = R;
  }
}
 
public class Main {
  private static final int MAX = 1001;
  private static boolean[] bit = new boolean[MAX];
 
  // Precomputation array
  private static int[] pre = new int[MAX];
 
  public static void answerQueries(int Q, Que[] Queries, int N, int[] arr) {
    // Setting bit at 0th position as 1
    bit[0] = true;
    for (int i = 0; i < N; i++) {
      for (int j = MAX - 1; j >= arr[i]; j--) {
        bit[j] |= bit[j - arr[i]];
      }
    }
 
    // Precompute the array
    for (int i = 1; i < MAX; i++) {
      pre[i] = pre[i - 1] + (bit[i] ? 1 : 0);
    }
 
    // Answer Queries
    for (int i = 0; i < Q; i++) {
      int l = Queries[i].L;
      int r = Queries[i].R;
      System.out.println(pre[r] - pre[l - 1]);
    }
  }
 
  // Driver Code to test above function
  public static void main(String[] args) {
    int[] arr = {1, 2, 2, 3, 5};
    int N = arr.length;
    int M = 4;
    Que[] Queries = {new Que(1, 2), new Que(1, 5), new Que(3, 6), new Que(9, 30)};
    answerQueries(M, Queries, N, arr);
  }
}
 
  

Python3

   from typing import List
 
MAX = 1001
bit = [0] * MAX
 
# precomputation array
pre = [0] * MAX
 
# structure to represent query
class Que:
    def __init__(self, L, R):
        self.L = L
        self.R = R
 
def answerQueries(Q: int, Queries: List[Que], N: int, arr: List[int]) -> None:
    global bit, pre
    # Setting bit at 0th position as 1
    bit[0] = 1
    for i in range(N):
        bit = [b or (bit[j - arr[i]] if j - arr[i] >= 0 else 0) for j, b in enumerate(bit)]
 
    # Precompute the array
    for i in range(1, MAX):
        pre[i] = pre[i - 1] + bit[i]
 
    # Answer Queries
    for i in range(Q):
        l = Queries[i].L
        r = Queries[i].R
        print(pre[r] - pre[l - 1])
 
# Driver Code to test above function
if __name__ == "__main__":
    arr = [1, 2, 2, 3, 5]
    N = len(arr)
    M = 4
    Queries = [Que(1, 2), Que(1, 5), Que(3, 6), Que(9, 30)]
    answerQueries(M, Queries, N, arr)
 
  

C#

   using System;
 
public class GFG
{
    private const int MAX = 1001;
    private static bool[] bit = new bool[MAX];
 
    // Precomputation array
    private static int[] pre = new int[MAX];
 
    // Class to represent query
    public class Que
    {
        public int L, R;
 
        public Que(int L, int R)
        {
            this.L = L;
            this.R = R;
        }
    }
 
    public static void answerQueries(int Q, Que[] Queries, int N, int[] arr)
    {
        // Setting bit at 0th position as 1
        bit[0] = true;
        for (int i = 0; i < N; i++)
        {
            for (int j = MAX - 1; j >= arr[i]; j--)
            {
                bit[j] |= bit[j - arr[i]];
            }
        }
 
        // Precompute the array
        for (int i = 1; i < MAX; i++)
        {
            pre[i] = pre[i - 1] + (bit[i] ? 1 : 0);
        }
 
        // Answer Queries
        for (int i = 0; i < Q; i++)
        {
            int l = Queries[i].L;
            int r = Queries[i].R;
            Console.WriteLine(pre[r] - pre[l - 1]);
        }
    }
 
    // Driver Code to test above function
    public static void Main(String[] args)
    {
        int[] arr = { 1, 2, 2, 3, 5 };
        int N = arr.Length;
        int M = 4;
        Que[] Queries = { new Que(1, 2), new Que(1, 5), new Que(3, 6), new Que(9, 30) };
        answerQueries(M, Queries, N, arr);
    }
}
 
  

Javascript

   // JavaScript Program to answer subset
// sum queries in a given range
 
const MAX = 1001;
let bit = Array(MAX).fill(0);
 
// precomputation array
let pre = Array(MAX).fill(0);
 
// class to represent query
class Que {
constructor(L, R) {
this.L = L;
this.R = R;
}
}
 
function answerQueries(Q, Queries, N, arr) {
// Setting bit at 0th position as 1
bit[0] = 1;
for (let i = 0; i < N; i++) {
for (let j = MAX - 1; j >= arr[i]; j--) {
bit[j] |= bit[j - arr[i]];
}
}// Precompute the array
for (let i = 1; i < MAX; i++) {
    pre[i] = pre[i - 1] + bit[i];
}
 
// Answer Queries
for (let i = 0; i < Q; i++) {
    let l = Queries[i].L;
    let r = Queries[i].R;
    console.log(pre[r] - pre[l - 1]);
}
}
 
// Driver Code to test above function
let arr = [1, 2, 2, 3, 5];
let N = arr.length;
let M = 4;
let Queries = [new Que(1, 2), new Que(1, 5), new Que(3, 6), new Que(9, 30)];
answerQueries(M, Queries, N, arr);
 
  

Output

2 5 4 5

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

Range Sum Queries and Update with Square Root

Nishant Tanwar

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Practice Tags :

Subset Sum Queries in a Range using Bitset

C++

Java

Python3

C#

Javascript

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