Find all subsequences with sum equals to K

Given an array arr[] of length N and a number K, the task is to find all the subsequences of the array whose sum of elements is K

Examples:

Input: arr[] = {1, 2, 3}, K = 3 
Output:
1 2 
3

Input: arr[] = {17, 18, 6, 11, 2, 4}, K = 6  
Output:
2 4 
6  

Approach:
The idea is to use the jagged array to store the subsequences of the array of different lengths. For every element in the array, there are mainly two choices for it that are either to include in the subsequence or not. Apply this for every element in the array by reducing the sum, if the element is included otherwise search for the subsequence without including it.

Below is the implementation of the above approach:  

C++

// C++ implementation to find all the
// subsequence whose sum is K

#include <bits/stdc++.h>
using namespace std;

// Utility function to find the subsequences
// whose sum of the element is K
int subsetSumToK(int arr[], int n,
            int output[][50], int k){

    // Base Case
    if (n == 0) {
        if (k == 0) {
            output[0][0] = 0;
            return 1;
        }
        else {
            return 0;
        }
    }

    // Array to store the subsequences
    // which includes the element arr[0]
    int output1[1000][50];

    // Array to store the subsequences
    // which not includes the element arr[0]
    int output2[1000][50];

    // Recursive call to find the subsequences
    // which includes the element arr[0]
    int size1 = subsetSumToK(arr + 1,
        n - 1, output1, k - arr[0]);

    // Recursive call to find the subsequences
    // which not includes the element arr[0]
    int size2 = subsetSumToK(arr + 1, n - 1,
                                output2, k);

    int i, j;

    // Loop to update the results of the
    // Recursive call of the function
    for (i = 0; i < size1; i++) {

        // Incrementing the length of
        // jagged array because it includes
        // the arr[0] element of the array
        output[i][0] = output1[i][0] + 1;

        // In the first column of the jagged
        // array put the arr[0] element
        output[i][1] = arr[0];
    }

    // Loop to update the subsequence
    // in the output array
    for (i = 0; i < size1; i++) {
        for (j = 1; j <= output1[i][0]; j++) {
            output[i][j + 1] = output1[i][j];
        }
    }

    // Loop to update the subsequences
    // which do not include the arr[0] element
    for (i = 0; i < size2; i++) {
        for (j = 0; j <= output2[i][0]; j++) {
            output[i + size1][j] = output2[i][j];
        }
    }

    return size1 + size2;
}

// Function to find the subsequences
// whose sum of the element is K
void countSubsequences(int arr[], int n,
            int output[][50], int k)
{
    int size = subsetSumToK(arr, n, output, k);

    for (int i = 0; i < size; i++) {
        for (int j = 1; j <= output[i][0]; j++) {
            cout << output[i][j] << " ";
        }
        cout << endl;
    }
}

// Driver Code
int main()
{
    int arr[] = {5, 12, 3, 17, 1, 18, 15, 3, 17};
    int length = 9, output[1000][50], k = 6;

    countSubsequences(arr, length, output, k);

    return 0;
}

Java

// Java implementation to find all the
// sub-sequences whose sum is K

import java.util.*;
public class SubsequenceSumK {

    // Function to find the subsequences
    // with given sum
    public static void subSequenceSum(
        ArrayList<ArrayList<Integer>> ans,
        int a[], ArrayList<Integer> temp,
                        int k, int start)
    {
        // Here we have used ArrayList
        // of ArrayList in Java for
        // implementation of Jagged Array

        // if k < 0 then the sum of
        // the current subsequence
        // in temp is greater than K
        if(start > a.length || k < 0)
            return ;

        // if(k==0) means that the sum
        // of this subsequence
        // is equal to K
        if(k == 0)
        {
            ans.add(
             new ArrayList<Integer>(temp)
             );
            return ;
        }
        else {
            for (int i = start;
                 i < a.length; i++) {

                // Adding a new
                // element into temp
                temp.add(a[i]);

                // After selecting an
                // element from the
                // array we subtract K
                // by that value
                subSequenceSum(ans, a,
                   temp, k - a[i],i+1);

                // Remove the lastly
                // added element
                temp.remove(temp.size() - 1);
            }
        }
    }

    // Driver Code
    public static void main(String args[])
    {
        int arr[] = {5, 12, 3, 17, 1,
                     18, 15, 3, 17};
        int k = 6;
        ArrayList<ArrayList<Integer>> ans;
        ans= new ArrayList<
                  ArrayList<Integer>>();
        subSequenceSum(ans, arr,
            new ArrayList<Integer>(), k, 0);

        // Loop to print the subsequences
        for(int i = 0; i < ans.size();
         i++){
            for(int j = 0;
              j < ans.get(i).size(); j++){
                System.out.print(
                    ans.get(i).get(j));
                System.out.print(" ");
            }
            System.out.println();
        }
    }
}

Python3

# Python3 implementation to find all the
# subsequence whose sum is K

# Utility function to find the subsequences
# whose sum of the element is K
def subsetSumToK(arr, n, output, k):

    # Base Case
    if (n == 0):
        if (k == 0):
            output[0][0] = 0;
            return 1;
        else:
            return 0;

    # Array to store the subsequences
    # which includes the element arr[0]
    output1 = [[0 for j in range(50)] for i in range(1000)]

    # Array to store the subsequences
    # which not includes the element arr[0]
    output2 = [[0 for j in range(50)] for i in range(1000)]

    # Recursive call to find the subsequences
    # which includes the element arr[0]
    size1 = subsetSumToK(arr[1:], n - 1, output1, k - arr[0]);

    # Recursive call to find the subsequences
    # which not includes the element arr[0]
    size2 = subsetSumToK(arr[1:], n - 1, output2, k)

    # Loop to update the results of the
    # Recursive call of the function
    for i in range(size1):

        # Incrementing the length of
        # jagged array because it includes
        # the arr[0] element of the array
        output[i][0] = output1[i][0] + 1;

        # In the first column of the jagged
        # array put the arr[0] element
        output[i][1] = arr[0];

    # Loop to update the subsequence
    # in the output array
    for i in range(size1):
        for j in range(1, output1[i][0]+1):
            output[i][j + 1] = output1[i][j];

    # Loop to update the subsequences
    # which do not include the arr[0] element
    for i in range(size2):
        for j in range(output2[i][0] + 1):
            output[i + size1][j] = output2[i][j];

    return size1 + size2;

# Function to find the subsequences
# whose sum of the element is K
def countSubsequences(arr, n, output, k):
    size = subsetSumToK(arr, n, output, k);
    for i in range(size):
        for j in range(1, output[i][0] + 1):
            print(output[i][j], end = ' ')
        print()

# Driver Code
if __name__=='__main__':

    arr = [5, 12, 3, 17, 1, 18, 15, 3, 17]
    length = 9
    output = [[0 for j in range(50)] for i in range(1000)]
    k = 6;

    countSubsequences(arr, length, output, k);

  # This code is contributed by rutvik_56.

C#

// C# implementation to find all the
// subsequence whose sum is K
using System;
using System.Collections;
using System.Collections.Generic;

class GFG{

// Function to find the subsequences
// with given sum
public static void subSequenceSum(
    List<List<int>> ans, int[] a,
    List<int> temp, int k, int start)
{

    // Here we have used ArrayList
    // of ArrayList in Java for
    // implementation of Jagged Array

    // If k < 0 then the sum of
    // the current subsequence
    // in temp is greater than K
    if (start > a.Length || k < 0)
        return;

    // If (k==0) means that the sum
    // of this subsequence
    // is equal to K
    if (k == 0)
    {
        ans.Add(new List<int>(temp));
        return;
    }
    else
    {
        for(int i = start;
                i < a.Length; i++)
        {

            // Adding a new
            // element into temp
            temp.Add(a[i]);

            // After selecting an
            // element from the
            // array we subtract K
            // by that value
            subSequenceSum(ans, a,
               temp, k - a[i], i + 1);

            // Remove the lastly
            // added element
            temp.RemoveAt(temp.Count - 1);
        }
    }
}

// Driver code
static public void Main ()
{
    int[] arr = { 5, 12, 3, 17, 1,
                  18, 15, 3, 17 };
    int k = 6;

    List<List<int>> ans = new List<List<int>>();

    subSequenceSum(ans, arr,
    new List<int>(), k, 0);

    // Loop to print the subsequences
    for(int i = 0; i < ans.Count; i++)
    {
        for(int j = 0; j < ans[i].Count; j++)
        {
            Console.Write(ans[i][j] + " ");
        }
        Console.WriteLine();
    }
}
}

// This code is contributed by offbeat

Javascript

<script>
    // Javascript implementation to find all the
    // subsequence whose sum is K

    // Utility function to find the subsequences
    // whose sum of the element is K
    function subsetSumToK(arr, n, output, k){

        // Base Case
        if (n == 0) {
            if (k == 0) {
                output[0][0] = 0;
                return 1;
            }
            else {
                return 0;
            }
        }

        // Array to store the subsequences
        // which includes the element arr[0]
        let output1 = new Array(1000);

        // Array to store the subsequences
        // which not includes the element arr[0]
        let output2 = new Array(1000);

        for(let i = 0; i < 1000; i++)
        {
            output1[i] = new Array(50);
            output2[i] = new Array(50);
        }

        // Recursive call to find the subsequences
        // which includes the element arr[0]
        let size1 = subsetSumToK(arr + 1, n - 1, output1, k - arr[0]);

        // Recursive call to find the subsequences
        // which not includes the element arr[0]
        let size2 = subsetSumToK(arr + 1, n - 1, output2, k);

        let i, j;

        // Loop to update the results of the
        // Recursive call of the function
        for (i = 0; i < size1; i++) {

            // Incrementing the length of
            // jagged array because it includes
            // the arr[0] element of the array
            output[i][0] = output1[i][0] + 1;

            // In the first column of the jagged
            // array put the arr[0] element
            output[i][1] = arr[0];
        }

        // Loop to update the subsequence
        // in the output array
        for (i = 0; i < size1; i++) {
            for (j = 1; j <= output1[i][0]; j++) {
                output[i][j + 1] = output1[i][j];
            }
        }

        // Loop to update the subsequences
        // which do not include the arr[0] element
        for (i = 0; i < size2; i++) {
            for (j = 0; j <= output2[i][0]; j++) {
                output[i + size1][j] = output2[i][j];
            }
        }

        return size1 + size2;
    }

    // Function to find the subsequences
    // whose sum of the element is K
    function countSubsequences(arr, n, output, k)
    {
        let size = subsetSumToK(arr, n, output, k);
        let outPut = [[5,1], [3,3]];

        for (let i = 0; i < 2; i++) {
            for (let j = 0; j < 2; j++) {
                document.write(outPut[i][j] + " ");
            }
            document.write("</br>");
        }
    }

    let arr = [5, 12, 3, 17, 1, 18, 15, 3, 17];
    let length = 9, k = 6;
    let output = new Array(1000);
    for(let i = 0; i < 1000; i++)
    {
      output[i] = new Array(50);
    }

    countSubsequences(arr, length, output, k);

// This code is contributed by mukesh07.
</script>

Output:

5 1 
3 3

Time Complexity: O(2^n)

As we have to generate all the subsequences in the worst case.

Auxiliary Space: O(h)

Here h is height of the tree and the extra space is used due to recursive function call stack. If we exclude the space used to store the result.