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:
- If we include the current element in the resultant subsequence then, decrement K and the above value of current element to the sum S.
- Recursively iterate from next index of the element to the end of the array to find the resultant subsequence.
- 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.
- 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++ 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 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 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# 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
<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)