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Perfect Sum Problem (Print all subsets with given sum)

  • Difficulty Level : Hard
  • Last Updated : 02 Mar, 2021

Given an array of integers and a sum, the task is to print all subsets of the given array with a sum equal to a given sum.
Examples: 

Input : arr[] = {2, 3, 5, 6, 8, 10}
        sum = 10
Output : 5 2 3
         2 8
         10

Input : arr[] = {1, 2, 3, 4, 5}
        sum = 10
Output : 4 3 2 1 
         5 3 2 
         5 4 1 

Asked in Tesco

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This problem is mainly an extension of Subset Sum Problem. Here we not only need to find if there is a subset with the given sum but also need to print all subsets with a given sum.
Like the previous post, we build a 2D array dp[][] such that dp[i][j] stores true if sum j is possible with array elements from 0 to i. 
After filling dp[][], we recursively traverse it from dp[n-1][sum]. For the cell being traversed, we store the path before reaching it and consider two possibilities for the element. 
1) Element is included in the current path. 
2) Element is not included in the current path.
Whenever the sum becomes 0, we stop the recursive calls and print the current path.
Below is an implementation of the above idea.

C++




// C++ program to count all subsets with
// given sum.
#include <bits/stdc++.h>
using namespace std;
 
// dp[i][j] is going to store true if sum j is
// possible with array elements from 0 to i.
bool** dp;
 
void display(const vector<int>& v)
{
    for (int i = 0; i < v.size(); ++i)
        printf("%d ", v[i]);
    printf("\n");
}
 
// A recursive function to print all subsets with the
// help of dp[][]. Vector p[] stores current subset.
void printSubsetsRec(int arr[], int i, int sum, vector<int>& p)
{
    // If we reached end and sum is non-zero. We print
    // p[] only if arr[0] is equal to sun OR dp[0][sum]
    // is true.
    if (i == 0 && sum != 0 && dp[0][sum])
    {
        p.push_back(arr[i]);
        // Display Only when Sum of elements of p is equal to sum
          if (arr[i] == sum)
              display(p);
        return;
    }
 
    // If sum becomes 0
    if (i == 0 && sum == 0)
    {
        display(p);
        return;
    }
 
    // If given sum can be achieved after ignoring
    // current element.
    if (dp[i-1][sum])
    {
        // Create a new vector to store path
        vector<int> b = p;
        printSubsetsRec(arr, i-1, sum, b);
    }
 
    // If given sum can be achieved after considering
    // current element.
    if (sum >= arr[i] && dp[i-1][sum-arr[i]])
    {
        p.push_back(arr[i]);
        printSubsetsRec(arr, i-1, sum-arr[i], p);
    }
}
 
// Prints all subsets of arr[0..n-1] with sum 0.
void printAllSubsets(int arr[], int n, int sum)
{
    if (n == 0 || sum < 0)
       return;
 
    // Sum 0 can always be achieved with 0 elements
    dp = new bool*[n];
    for (int i=0; i<n; ++i)
    {
        dp[i] = new bool[sum + 1];
        dp[i][0] = true;
    }
 
    // Sum arr[0] can be achieved with single element
    if (arr[0] <= sum)
       dp[0][arr[0]] = true;
 
    // Fill rest of the entries in dp[][]
    for (int i = 1; i < n; ++i)
        for (int j = 0; j < sum + 1; ++j)
            dp[i][j] = (arr[i] <= j) ? dp[i-1][j] ||
                                       dp[i-1][j-arr[i]]
                                     : dp[i - 1][j];
    if (dp[n-1][sum] == false)
    {
        printf("There are no subsets with sum %d\n", sum);
        return;
    }
 
    // Now recursively traverse dp[][] to find all
    // paths from dp[n-1][sum]
    vector<int> p;
    printSubsetsRec(arr, n-1, sum, p);
}
 
// Driver code
int main()
{
    int arr[] = {1, 2, 3, 4, 5};
    int n = sizeof(arr)/sizeof(arr[0]);
    int sum = 10;
    printAllSubsets(arr, n, sum);
    return 0;
}

Java




// A Java program to count all subsets with given sum.
import java.util.ArrayList;
public class SubSet_sum_problem
{
    // dp[i][j] is going to store true if sum j is
    // possible with array elements from 0 to i.
    static boolean[][] dp;
      
    static void display(ArrayList<Integer> v)
    {
       System.out.println(v);
    }
      
    // A recursive function to print all subsets with the
    // help of dp[][]. Vector p[] stores current subset.
    static void printSubsetsRec(int arr[], int i, int sum,
                                         ArrayList<Integer> p)
    {
        // If we reached end and sum is non-zero. We print
        // p[] only if arr[0] is equal to sun OR dp[0][sum]
        // is true.
        if (i == 0 && sum != 0 && dp[0][sum])
        {
            p.add(arr[i]);
            display(p);
            p.clear();
            return;
        }
      
        // If sum becomes 0
        if (i == 0 && sum == 0)
        {
            display(p);
            p.clear();
            return;
        }
      
        // If given sum can be achieved after ignoring
        // current element.
        if (dp[i-1][sum])
        {
            // Create a new vector to store path
            ArrayList<Integer> b = new ArrayList<>();
            b.addAll(p);
            printSubsetsRec(arr, i-1, sum, b);
        }
      
        // If given sum can be achieved after considering
        // current element.
        if (sum >= arr[i] && dp[i-1][sum-arr[i]])
        {
            p.add(arr[i]);
            printSubsetsRec(arr, i-1, sum-arr[i], p);
        }
    }
      
    // Prints all subsets of arr[0..n-1] with sum 0.
    static void printAllSubsets(int arr[], int n, int sum)
    {
        if (n == 0 || sum < 0)
           return;
      
        // Sum 0 can always be achieved with 0 elements
        dp = new boolean[n][sum + 1];
        for (int i=0; i<n; ++i)
        {
            dp[i][0] = true
        }
      
        // Sum arr[0] can be achieved with single element
        if (arr[0] <= sum)
           dp[0][arr[0]] = true;
      
        // Fill rest of the entries in dp[][]
        for (int i = 1; i < n; ++i)
            for (int j = 0; j < sum + 1; ++j)
                dp[i][j] = (arr[i] <= j) ? (dp[i-1][j] ||
                                           dp[i-1][j-arr[i]])
                                         : dp[i - 1][j];
        if (dp[n-1][sum] == false)
        {
            System.out.println("There are no subsets with" +
                                                  " sum "+ sum);
            return;
        }
      
        // Now recursively traverse dp[][] to find all
        // paths from dp[n-1][sum]
        ArrayList<Integer> p = new ArrayList<>();
        printSubsetsRec(arr, n-1, sum, p);
    }
     
    //Driver Program to test above functions
    public static void main(String args[])
    {
        int arr[] = {1, 2, 3, 4, 5};
        int n = arr.length;
        int sum = 10;
        printAllSubsets(arr, n, sum);
    }
}
//This code is contributed by Sumit Ghosh

Output:



4 3 2 1
5 3 2
5 4 1

Another Approach : 
For each element in the array, first decide to take it or not in the subset. Define a function which will take care of all this. The function is called once in the main function. The static class fields are declared which will be operated by our function. At each call, the function checks for the condition of the fields. In our case, it checks if the current sum is equal to the given sum and accordingly increment the respective class field. If not, it makes function calls by taking all the case. So the number of function calls will be equal to the number of cases. So here, two calls are made – one by taking the element in the subset and increment the current sum and another by not taking the element.
Below is the implementation :

Java




// Java code to find the number of
// possible subset with given sum
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG {
     
    static int count;
    static int sum;
    static int n;
     
    // Driver code
    public static void main (String[] args) {
        count = 0;
        n = 5;
        sum = 10;
 
        int[] pat = {2, 3, 5, 6, 8, 10};
         
        f(pat, 0, 0);
         
        System.out.println(count);
    }
     
    // Function to select or not the array element
    // to form a subset with given sum
    static void f(int[] pat, int i, int currSum) {
        if (currSum == sum) {
            count++;
            return;
        }
        if (currSum < sum && i < n) {
            f(pat, i+1, currSum + pat[i]);
            f(pat, i+1, currSum);
        }
    }
}

Output : 

4 3 2 1 
5 3 2 
5 4 1 

This article is contributed by Jas Kaur. 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 write comments if you find anything incorrect, or you want to share more information about the topic discussed above.




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