Subset sum problem where Array sum is at most N

Last Updated : 16 Feb, 2023

Given an array arr[] of size N such that the sum of all the array elements does not exceed N, and array queries[] containing Q queries. For each query, the task is to find if there is a subset of the array whose sum is the same as queries[i].

Examples:

Input: arr[] = {1, 0, 0, 0, 0, 2, 3}, queries[] = {3, 7, 6}
Output:
Possible
Not Possible
Possible
Explanation: 3 is spossible. 6 can be obtained by the subset {1, 2, 3}
7 is greater than the sum of all array elements.

Input: arr[] = {0, 1, 2}, queries[] = {1, 2, 3, 0}
Output:
Possible
Possible
Possible
Possible
Explanation: All the sums can be obtained by using the elements.

Approach: The problem can be solved using the approach as in the subset sum problem.
However, the time complexity can be reduced using the fact that the sum can be at most N. As the sum can be at most N, it can be proved that there are at most √2N unique positive elements where all have a frequency of 1.

Say there are √2N unique positive elements starting from 1 to √2N.
Therefore the sum of those numbers is N + √(N/2).
This sum is more than N itself. So there can be at most √2N unique elements.

The above fact can be used and implemented in dynamic programming. Using coordinate compression all those unique elements can be stored in minimum space.

For each element check what is the minimum contribution of that element to achieve a sum j (j in the range [0, N]) or if it is not possible to achieve the sum j. The contribution of each item (say i) depends on the contribution of the other smaller items till the sum (j – i).

Follow the image shown below to understand better the difference of unused states for normal subset and when the sum is N at max:

Comparison:

Say the arr[] = {1, 2, 2, 2, 3, 3}. (Here sum is greater, so does not follow the condition of sum at most N. But here unique elements maintain the threshold. That’s why it is used here just for understanding purpose)

Red cells signify the useless states, these are much more in traditional algorithm than optimized one.

Traditional Subset-Sum vs Frequency Optimized DP, Useless States

Follow the steps mentioned below to implement the approach;

Use coordinate compression on all the unique elements.
Build a 2D dp[][] array where dp[i][j] stores the contribution of ith item to get sum j. (If it is not possible then store –1, and if ith item is not needed then store 0 in dp[i][j]).
Iterate from i = 0 to the maximum element:
- Iterate for j = 0 to N:
  - If the value of dp[i][j-arr[i]] + 1 < dp[i][j] then update it.
  - Otherwise, keep it as it was.
Then iterate from i = 0 to Q:
- Check if that sum (query[i])is possible or not.
- It is not possible if it exceeds the array sum or all the elements together cannot get a certain sum i.e. dp[len][query[i]] = -1. (len is total number of unique elements)

Below is the implementation of the above approach.

C++

   // C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find if the queries
// are possible or not
void findSol(vector<int>& arr,
             vector<int>& queries)
{
    int s = 0;
 
    // Calculating sum of array
    for (auto& item : arr) {
        s += item;
    }
 
    // Coordinate compression,
    // make frequency-value pairs
    map<int, int> mp;
    for (auto& item : arr) {
        mp[item]++;
    }
 
    vector<int> val, freq;
 
    // Frequency mapping
    for (auto& x : mp) {
        val.push_back(x.first);
        freq.push_back(x.second);
    }
 
    int len = val.size();
    vector<vector<int> > dp(len + 1,
                            vector<int>(
                                s + 1, 0));
 
    for (int j = 1; j <= s; ++j) {
        dp[0][j] = -1;
    }
 
    // Loop to build the dp[][]
    for (int i = 1; i <= len; ++i) {
        for (int j = 1; j <= s; ++j) {
            int v = val[i - 1];
            int f = freq[i - 1];
 
            if (dp[i - 1][j] != -1) {
                dp[i][j] = 0;
            }
            else if (j >= v
                     && dp[i][j - v] != -1
                     && dp[i][j - v] + 1 <= f) {
                dp[i][j] = dp[i][j - v] + 1;
            }
            else {
                dp[i][j] = -1;
            }
        }
    }
 
    // Answer queries
    for (auto& q : queries) {
        if (q > s || dp[len][q] == -1) {
            cout << "Not Possible" << endl;
        }
        else {
            cout << "Possible" << endl;
        }
    }
}
 
// Driver Code
int main()
{
    vector<int> arr = { 1, 0, 0, 0, 0, 2, 3 };
    vector<int> queries = { 3, 7, 6 };
 
    // Function call
    findSol(arr, queries);
    return 0;
}
 
  

Java

   // Java code to implement the approach
import java.util.*;
class GFG
{
  
// Function to find if the queries
// are possible or not
static void findSol(ArrayList<Integer> arr, ArrayList<Integer> queries)
{
    int s = 0;
  
    // Calculating sum of array
    for (Integer item : arr) {
        s += item;
    }
  
    // Coordinate compression,
    // make frequency-value pairs
    HashMap<Integer, Integer> mp = new HashMap<>();
    for (Integer item : arr) {
        if(mp.containsKey(item))
        mp.put(item,mp.get(item)+1);
        else
        mp.put(item,1);
    }
  
    ArrayList<Integer> val = new ArrayList<Integer>(); 
    ArrayList<Integer> freq = new ArrayList<Integer>();
     
    // Frequency mapping
    for (Map.Entry<Integer,Integer> x : mp.entrySet()) 
    {
        val.add(x.getKey());
        freq.add(x.getValue());
    }
  
    int len = val.size();
    int dp[][] = new int[len+1][s+1];
  
    for (int j = 1; j <= s; ++j) {
        dp[0][j] = -1;
    }
  
    // Loop to build the dp[][]
    for (int i = 1; i <= len; ++i) {
        for (int j = 1; j <= s; ++j) {
            int v = val.get(i - 1);
            int f = freq.get(i - 1);
  
            if (dp[i - 1][j] != -1) {
                dp[i][j] = 0;
            }
            else if (j >= v
                     && dp[i][j - v] != -1
                     && dp[i][j - v] + 1 <= f) {
                dp[i][j] = dp[i][j - v] + 1;
            }
            else {
                dp[i][j] = -1;
            }
        }
    }
  
    // Answer queries
    for(Integer q:queries)
    {
        if (q > s || dp[len][q] == -1) {
            System.out.println("Not Possible");
        }
        else {
            System.out.println("Possible");
        }
    }
}
  
// Driver Code
public static void main(String[] args)
{
    ArrayList<Integer> arr = new ArrayList<Integer>(
            Arrays.asList(1, 0, 0, 0, 0, 2, 3 ));
    ArrayList<Integer> queries = new ArrayList<Integer>(
            Arrays.asList(3, 7, 6 ));
  
    // Function call
    findSol(arr, queries);
}
}
 
// This code is contributed by Pushpesh Raj.
 
  

Python3

   # Python3 code to implement the approach
 
# Function to find if the queries
# are possible or not
def findSol(arr, queries):
 
    # calculating sum of array
    s = sum(arr)
 
    # Coordinate compression,
    # make frequency-value pairs
    mp = dict()
    for item in arr:
        if item not in mp:
            mp[item] = 1
        else:
            mp[item] += 1
 
    val = []
    freq = []
 
    # Frequency mapping
    for x in mp:
        val.append(x)
        freq.append(mp[x])
 
    len_ = len(val)
    dp = [[0 for i in range(s + 1)] for j in range(len_ + 1)]
 
    for j in range(1, s + 1):
        dp[0][j] = -1
 
    # Loop to build dp[][]
    for i in range(1, len_ + 1):
        for j in range(1, s + 1):
            v = val[i - 1]
            f = freq[i - 1]
 
            if dp[i - 1][j] != -1:
                dp[i][j] = 0
 
            elif j >= 0 and dp[i][j - v] != -1 and dp[i][j - v] + 1 <= f:
                dp[i][j] = dp[i][j - v] + 1
            else:
                dp[i][j] = -1
 
    # Answer queries
    for q in queries:
        if q > s or dp[len_][q] == -1:
            print("Not Possible")
        else:
            print("Possible")
 
# Driver Code
arr = [1, 0, 0, 0, 0, 2, 3]
queries = [3, 7, 6]
 
# Function call
findSol(arr, queries)
 
# This code is contributed by phasing17
 
  

C#

   // C# program to implement above approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to find if the queries
    // are possible or not
    public static void findSol(List<int> arr, List<int> queries)
    {
        int s = 0;
 
        // Calculating sum of array
        foreach (int item in arr){
            s += item;
        }
 
        // Coordinate compression,
        // make frequency-value pairs
        SortedDictionary<int, int> mp = new SortedDictionary<int, int>();
        foreach (int item in arr) {
            if(mp.ContainsKey(item)){
                mp[item] = mp[item] + 1;
            }else{
                mp.Add(item, 1);
            }
        }
 
        List<int> val = new List<int>(); 
        List<int> freq = new List<int>();
         
        // Frequency mapping
        foreach (KeyValuePair<int,int> x in mp) 
        {
            val.Add(x.Key);
            freq.Add(x.Value);
        }
 
        int len = val.Count;
        int[,] dp = new int[len+1, s+1];
 
        for (int j = 1; j <= s; ++j) {
            dp[0, j] = -1;
        }
 
        // Loop to build the dp[][]
        for (int i = 1; i <= len; ++i) {
            for (int j = 1; j <= s; ++j) {
                int v = val[i - 1];
                int f = freq[i - 1];
 
                if (dp[i - 1, j] != -1) {
                    dp[i, j] = 0;
                }
                else if (j >= v && dp[i, j - v] != -1 && dp[i, j - v] + 1 <= f) {
                    dp[i, j] = dp[i, j - v] + 1;
                }
                else{
                    dp[i, j] = -1;
                }
            }
        }
 
        // Answer queries
        foreach(int q in queries)
        {
            if (q > s || dp[len, q] == -1) {
                Console.Write("Not Possible\n");
            }
            else {
                Console.Write("Possible\n");
            }
        }
    }
 
    // Driver Code
    public static void Main(string[] args){
         
        List<int> arr = new List<int>{
                                1, 0, 0, 0, 0, 2, 3
                            };
        List<int> queries = new List<int>{
                                3, 7, 6
                            };
 
        // Function call
        findSol(arr, queries);
 
    }
}
 
// This code is contributed by subhamgoyal2014.
 
  

Javascript

   <script>
 
// JavaScript code to implement the approach
 
// Function to find if the queries
// are possible or not
function findSol(arr,queries)
{
    let s = 0;
 
    // Calculating sum of array
    for (let item of arr) {
        s += item;
    }
 
    // Coordinate compression,
    // make frequency-value pairs
    let mp = new Map();
    for (let item of arr) {
        if(mp.has(item))
            mp.set(item,mp.get(item)+1);
        else mp.set(item,1);
    }
 
    let val = [], freq = [];
 
    // Frequency mapping
    for (let [x,y] of mp) {
        val.push(x);
        freq.push(y);
    }
 
    let len = val.length;
    let dp = new Array(len + 1).fill(0).map(()=>new Array(s+1).fill(0));
 
    for (let j = 1; j <= s; ++j) {
        dp[0][j] = -1;
    }
 
    // Loop to build the dp[][]
    for (let i = 1; i <= len; ++i) {
        for (let j = 1; j <= s; ++j) {
            let v = val[i - 1];
            let f = freq[i - 1];
 
            if (dp[i - 1][j] != -1) {
                dp[i][j] = 0;
            }
            else if (j >= v
                     && dp[i][j - v] != -1
                     && dp[i][j - v] + 1 <= f) {
                dp[i][j] = dp[i][j - v] + 1;
            }
            else {
                dp[i][j] = -1;
            }
        }
    }
 
    // Answer queries
    for (let q of queries) {
        if (q > s || dp[len][q] == -1) {
            console.log("Not Possible");
        }
        else {
            console.log("Possible");
        }
    }
}
 
// Driver Code
 
let arr = [ 1, 0, 0, 0, 0, 2, 3 ];
let queries = [ 3, 7, 6 ];
 
// Function call
findSol(arr, queries);
 
// This code is contributed by shinjanpatra
 
</script>
 
  

Output

Possible Not Possible Possible

Time Complexity: O(N * √N)
Auxiliary Space: O(N * √N)

Smallest Value that cannot be represented as sum of any subset of a given array

sarvjot

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Practice Tags :

Subset sum problem where Array sum is at most N

C++

Java

Python3

C#

Javascript

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