Unique subsequences of length K with given sum

Last Updated : 02 Nov, 2023

Given an array arr[] of N integers and two numbers K and S, the task is to print all the subsequence of length K with the sum S.
Examples:

Input: N = 5, K = 3, S = 20, arr[] = {4, 6, 8, 2, 12}
Output:
{6, 2, 12}
Explanation:
Only one subsequence of size 3 with a sum 20 is possible i.e., {6, 2, 12} and sum is 6 + 2 + 12 = 20
Input: N = 10, K = 5, S = 25, arr[] = {2, 4, 6, 8, 10, 12, 1, 2, 5, 7}
Output:
{10, 1, 2, 5, 7}
{4, 8, 1, 5, 7}
{4, 8, 10, 1, 2}
{4, 6, 12, 1, 2}
{4, 6, 8, 2, 5}
{2, 10, 1, 5, 7}
{2, 8, 12, 1, 2}
{2, 6, 10, 2, 5}
{2, 6, 8, 2, 7}
{2, 4, 12, 2, 5}
{2, 4, 10, 2, 7}
{2, 4, 8, 10, 1}
{2, 4, 6, 12, 1}
{2, 4, 6, 8, 5}

Approach: The idea is to use Backtracking to print all the subsequence with given sum S. Below are the steps:

Iterate for all the value of the array arr[] and do the following:
1. If we include the current element in the resultant subsequence then, decrement K and the above value of current element to the sum S.
2. Recursively iterate from next index of the element to the end of the array to find the resultant subsequence.
3. If K is 0 and S is 0 then we got our one of the resultant subsequence of length K and sum S, print this subsequence and backtrack for the next resulting subsequence.
4. If we doesn’t include the current element then, find the resultant subsequence by excluding the current element and repeating the above procedure for the rest of the element in the array.
Resultant array in the steps 3 will give all the possible subsequence of length K with given sum S.

Below is the implementation of the above approach:

C++

   // C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find all the subsequences
// of a given length and having sum S
void comb(int* arr, int len, int r,
          int ipos, int* op, int opos,
          int sum)
{
 
    // Termination condition
    if (opos == r) {
 
        int sum2 = 0;
        for (int i = 0; i < opos; i++) {
 
            // Add value to sum
            sum2 = sum2 + op[i];
        }
 
        // Check if the resultant sum
        // equals to target sum
        if (sum == sum2) {
 
            // If true
            for (int i = 0; i < opos; i++)
 
                // Print resultant array
                cout << op[i] << ", ";
 
            cout << endl;
        }
 
        // End this recursion stack
        return;
    }
    if (ipos < len) {
 
        // Check all the combinations
        // using backtracking
        comb(arr, len, r, ipos + 1,
             op, opos, sum);
 
        op[opos] = arr[ipos];
 
        // Check all the combinations
        // using backtracking
        comb(arr, len, r, ipos + 1,
             op, opos + 1, sum);
    }
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 4, 6, 8, 2, 12 };
    int K = 3;
    int S = 20;
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // To store the subsequence
    int op[N] = { 0 };
 
    // Function Call
    comb(arr, N, K, 0, op, 0, S);
    return 0;
}
 
  

Java

   // Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find all the subsequences
// of a given length and having sum S
static void comb(int []arr, int len, int r,
                 int ipos, int[] op, int opos,
                 int sum)
{
 
    // Termination condition
    if (opos == r)
    {
        int sum2 = 0;
        for(int i = 0; i < opos; i++) 
        {
             
           // Add value to sum
           sum2 = sum2 + op[i];
        }
 
        // Check if the resultant sum
        // equals to target sum
        if (sum == sum2)
        {
 
            // If true
            for(int i = 0; i < opos; i++)
                
               // Print resultant array
               System.out.print(op[i] + ", ");
 
            System.out.println();
        }
 
        // End this recursion stack
        return;
    }
    if (ipos < len) 
    {
 
        // Check all the combinations
        // using backtracking
        comb(arr, len, r, ipos + 1,
             op, opos, sum);
              
        op[opos] = arr[ipos];
 
        // Check all the combinations
        // using backtracking
        comb(arr, len, r, ipos + 1,
             op, opos + 1, sum);
    }
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array
    int arr[] = { 4, 6, 8, 2, 12 };
    int K = 3;
    int S = 20;
 
    int N = arr.length;
 
    // To store the subsequence
    int op[] = new int[N];
 
    // Function Call
    comb(arr, N, K, 0, op, 0, S);
}
}
 
// This code is contributed by amal kumar choubey
 
  

Python3

   # Python3 program for the above approach
 
# Function to find all the subsequences 
# of a given length and having sum S 
def comb(arr, Len, r, ipos, op, opos, Sum):
 
    # Termination condition
    if (opos == r):
 
        sum2 = 0
        for i in range(opos):
 
            # Add value to sum
            sum2 = sum2 + op[i]
 
        # Check if the resultant sum
        # equals to target sum
        if (Sum == sum2):
 
            # If true
            for i in range(opos):
 
                # Print resultant array
                print(op[i], end = ", ")
 
            print()
 
        # End this recursion stack
        return
 
    if (ipos < Len):
 
        # Check all the combinations
        # using backtracking
        comb(arr, Len, r, ipos + 1, 
             op, opos, Sum)
 
        op[opos] = arr[ipos]
 
        # Check all the combinations
        # using backtracking
        comb(arr, Len, r, ipos + 1, op, 
                          opos + 1, Sum)
 
# Driver code
if __name__ == '__main__':
 
    # Given array
    arr = [ 4, 6, 8, 2, 12 ]
    K = 3
    S = 20
    N = len(arr)
 
    # To store the subsequence
    op = [0] * N
 
    # Function call
    comb(arr, N, K, 0, op, 0, S)
 
# This code is contributed by himanshu77
 
  

C#

   // C# program for the above approach
using System;
 
class GFG{
 
// Function to find all the subsequences
// of a given length and having sum S
static void comb(int []arr, int len, int r,
                 int ipos, int[] op, int opos,
                 int sum)
{
 
    // Termination condition
    if (opos == r)
    {
        int sum2 = 0;
        for(int i = 0; i < opos; i++) 
        {
            
           // Add value to sum
           sum2 = sum2 + op[i];
        }
 
        // Check if the resultant sum
        // equals to target sum
        if (sum == sum2)
        {
 
            // If true
            for(int i = 0; i < opos; i++)
                
               // Print resultant array
               Console.Write(op[i] + ", ");
            Console.WriteLine();
        }
 
        // End this recursion stack
        return;
    }
    if (ipos < len) 
    {
 
        // Check all the combinations
        // using backtracking
        comb(arr, len, r, ipos + 1,
             op, opos, sum);
             
        op[opos] = arr[ipos];
 
        // Check all the combinations
        // using backtracking
        comb(arr, len, r, ipos + 1,
             op, opos + 1, sum);
    }
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given array
    int []arr = { 4, 6, 8, 2, 12 };
    int K = 3;
    int S = 20;
 
    int N = arr.Length;
 
    // To store the subsequence
    int []op = new int[N];
 
    // Function call
    comb(arr, N, K, 0, op, 0, S);
}
}
 
// This code is contributed by amal kumar choubey
 
  

Javascript

   <script>
 
// JavaScript program for the above approach
 
// Function to find all the subsequences
// of a given length and having sum S
function comb(arr, len, r,
                 ipos, op, opos,
                 sum)
{
 
    // Termination condition
    if (opos == r)
    {
        let sum2 = 0;
        for(let i = 0; i < opos; i++) 
        {
             
           // Add value to sum
           sum2 = sum2 + op[i];
        }
 
        // Check if the resultant sum
        // equals to target sum
        if (sum == sum2)
        {
 
            // If true
            for(let i = 0; i < opos; i++)
                
               // Print resultant array
               document.write(op[i] + ", ");
 
            document.write();
        }
 
        // End this recursion stack
        return;
    }
    if (ipos < len) 
    {
 
        // Check all the combinations
        // using backtracking
        comb(arr, len, r, ipos + 1,
             op, opos, sum);
              
        op[opos] = arr[ipos];
 
        // Check all the combinations
        // using backtracking
        comb(arr, len, r, ipos + 1,
             op, opos + 1, sum);
    }
}
  
// Driver Code
 
    // Given array
    let arr = [ 4, 6, 8, 2, 12 ];
    let K = 3;
    let S = 20;
 
    let N = arr.length;
 
    // To store the subsequence
    let op = Array.from({length: N}, (_, i) => 0);
 
    // Function Call
    comb(arr, N, K, 0, op, 0, S);
 
// This code is contributed by sanjoy_62.
</script>
 
  

Output:

6, 2, 12,

Time Complexity: O(2^N * K)
Auxiliary Space: O(N)

Number of K length subsequences with minimum sum

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Unique subsequences of length K with given sum

C++

Java

Python3

C#

Javascript

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