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Recursive program to print all subsets with given sum

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  • Difficulty Level : Medium
  • Last Updated : 12 Jul, 2021

Given an array and a number, print all subsets with sum equal to given the sum.
Examples: 
 

Input :  arr[] =  {2, 5, 8, 4, 6, 11}, sum = 13
Output : 
5 8
2 11
2 5 6

Input : arr[] =  {1, 5, 8, 4, 6, 11}, sum = 9
Output :
5 4
1 8

 

This problem is an extension of check if there is a subset with given sum. We recursively generate all subsets. We keep track of elements of current subset. If sum of elements in current subset becomes equal to given sum, we print the subset. 
 

C++




// CPP program to print all subsets with given sum
#include <bits/stdc++.h>
using namespace std;
 
// The vector v stores current subset.
void printAllSubsetsRec(int arr[], int n, vector<int> v,
                        int sum)
{
    // If remaining sum is 0, then print all
    // elements of current subset.
    if (sum == 0) {
        for (auto x : v)
            cout << x << " ";
        cout << endl;
        return;
    }
 
    // If no remaining elements,
    if (n == 0)
        return;
 
    // We consider two cases for every element.
    // a) We do not include last element.
    // b) We include last element in current subset.
    printAllSubsetsRec(arr, n - 1, v, sum);
    v.push_back(arr[n - 1]);
    printAllSubsetsRec(arr, n - 1, v, sum - arr[n - 1]);
}
 
// Wrapper over printAllSubsetsRec()
void printAllSubsets(int arr[], int n, int sum)
{
    vector<int> v;
    printAllSubsetsRec(arr, n, v, sum);
}
 
// Driver code
int main()
{
    int arr[] = { 2, 5, 8, 4, 6, 11 };
    int sum = 13;
    int n = sizeof(arr) / sizeof(arr[0]);
    printAllSubsets(arr, n, sum);
    return 0;
}


Java




// Java program to print all subsets with given sum
import java.util.*;
 class Solution
{
 
// The vector v stores current subset.
static void printAllSubsetsRec(int arr[], int n, Vector<Integer> v,
                        int sum)
{
    // If remaining sum is 0, then print all
    // elements of current subset.
    if (sum == 0) {
        for (int i=0;i<v.size();i++)
            System.out.print( v.get(i) + " ");
        System.out.println();
        return;
    }
 
    // If no remaining elements,
    if (n == 0)
        return;
 
    // We consider two cases for every element.
    // a) We do not include last element.
    // b) We include last element in current subset.
    printAllSubsetsRec(arr, n - 1, v, sum);
    Vector<Integer> v1=new Vector<Integer>(v);
    v1.add(arr[n - 1]);
    printAllSubsetsRec(arr, n - 1, v1, sum - arr[n - 1]);
}
 
// Wrapper over printAllSubsetsRec()
static void printAllSubsets(int arr[], int n, int sum)
{
    Vector<Integer> v= new Vector<Integer>();
    printAllSubsetsRec(arr, n, v, sum);
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = { 2, 5, 8, 4, 6, 11 };
    int sum = 13;
    int n = arr.length;
    printAllSubsets(arr, n, sum);
     
}
}
//contributed by Arnab Kundu


Python3




# Python program to print all subsets with given sum
 
# The vector v stores current subset.
def printAllSubsetsRec(arr, n, v, sum) :
 
    # If remaining sum is 0, then print all
    # elements of current subset.
    if (sum == 0) :
        for value in v :
            print(value, end=" ")
        print()
        return
     
 
    # If no remaining elements,
    if (n == 0):
        return
 
    # We consider two cases for every element.
    # a) We do not include last element.
    # b) We include last element in current subset.
    printAllSubsetsRec(arr, n - 1, v, sum)
    v1 = [] + v
    v1.append(arr[n - 1])
    printAllSubsetsRec(arr, n - 1, v1, sum - arr[n - 1])
 
 
# Wrapper over printAllSubsetsRec()
def printAllSubsets(arr, n, sum):
 
    v = []
    printAllSubsetsRec(arr, n, v, sum)
 
 
# Driver code
 
arr = [ 2, 5, 8, 4, 6, 11 ]
sum = 13
n = len(arr)
printAllSubsets(arr, n, sum)
 
# This code is contributed by ihritik


C#




// C# program to print all subsets with given sum
using System;
using System.Collections.Generic;
 
class GFG
{
    // The vector v stores current subset.
    static void printAllSubsetsRec(int []arr, int n,
                                    List<int> v, int sum)
    {
        // If remaining sum is 0, then print all
        // elements of current subset.
        if (sum == 0)
        {
            for (int i = 0; i < v.Count; i++)
                Console.Write( v[i]+ " ");
            Console.WriteLine();
            return;
        }
 
        // If no remaining elements,
        if (n == 0)
            return;
 
        // We consider two cases for every element.
        // a) We do not include last element.
        // b) We include last element in current subset.
        printAllSubsetsRec(arr, n - 1, v, sum);
        List<int> v1 = new List<int>(v);
        v1.Add(arr[n - 1]);
        printAllSubsetsRec(arr, n - 1, v1, sum - arr[n - 1]);
    }
 
    // Wrapper over printAllSubsetsRec()
    static void printAllSubsets(int []arr, int n, int sum)
    {
        List<int> v = new List<int>();
        printAllSubsetsRec(arr, n, v, sum);
    }
 
    // Driver code
    public static void Main()
    {
        int []arr = { 2, 5, 8, 4, 6, 11 };
        int sum = 13;
        int n = arr.Length;
        printAllSubsets(arr, n, sum);
    }
}
 
// This code is contributed by Rajput-Ji


PHP




<?php
// PHP program to print all subsets with given sum
 
// The vector v stores current subset.
function printAllSubsetsRec($arr, $n, $v, $sum)
{
    // If remaining sum is 0, then print all
    // elements of current subset.
    if ($sum == 0)
    {
        for ($i = 0; $i < count($v); $i++)
            echo $v[$i] . " ";
        echo "\n";
        return;
    }
 
    // If no remaining elements,
    if ($n == 0)
        return;
 
    // We consider two cases for every element.
    // a) We do not include last element.
    // b) We include last element in current subset.
    printAllSubsetsRec($arr, $n - 1, $v, $sum);
    array_push($v, $arr[$n - 1]);
    printAllSubsetsRec($arr, $n - 1, $v,
                       $sum - $arr[$n - 1]);
}
 
// Wrapper over printAllSubsetsRec()
function printAllSubsets($arr, $n, $sum)
{
    $v = array();
    printAllSubsetsRec($arr, $n, $v, $sum);
}
 
// Driver code
$arr = array( 2, 5, 8, 4, 6, 11 );
$sum = 13;
$n = count($arr);
printAllSubsets($arr, $n, $sum);
 
// This code is contributed by mits
?>


Javascript




<script>
  
        // JavaScript Program for the above approach
 
        // The vector v stores current subset.
        function printAllSubsetsRec(arr, n, v, sum) {
            // If remaining sum is 0, then print all
            // elements of current subset.
            if (sum == 0) {
                for (let x of v)
                    document.write(x + " ");
                document.write("<br>")
                return;
            }
 
            // If no remaining elements,
            if (n == 0)
                return;
 
            // We consider two cases for every element.
            // a) We do not include last element.
            // b) We include last element in current subset.
            printAllSubsetsRec(arr, n - 1, v, sum);
            v.push(arr[n - 1]);
            printAllSubsetsRec(arr, n - 1, v, sum - arr[n - 1]);
            v.pop();
        }
 
        // Wrapper over printAllSubsetsRec()
        function printAllSubsets(arr, n, sum) {
            let v = [];
            printAllSubsetsRec(arr, n, v, sum);
        }
 
        // Driver code
 
        let arr = [2, 5, 8, 4, 6, 11];
        let sum = 13;
        let n = arr.length;
        printAllSubsets(arr, n, sum);
 
    // This code is contributed by Potta Lokesh
 
</script>


Output: 

8 5 
6 5 2 
11 2

 

Time Complexity : O(2n)
Please refer below post for an optimized solution based on Dynamic Programming. 
Print all subsets with given sum using Dynamic Programming
 


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