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Count of subsets with sum equal to X
  • Difficulty Level : Medium
  • Last Updated : 27 Mar, 2021
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Given an array arr[] of length N and an integer X, the task is to find the number of subsets with sum equal to X.
Examples: 

Input: arr[] = {1, 2, 3, 3}, X = 6 
Output:
All the possible subsets are {1, 2, 3}, 
{1, 2, 3} and {3, 3}
 

Input: arr[] = {1, 1, 1, 1}, X = 1 
Output:

Approach: A simple approach is to solve this problem by generating all the possible subsets and then checking whether the subset has the required sum. This approach will have exponential time complexity. However, for smaller values of X and array elements, this problem can be solved using dynamic programming
Let’s look at the recurrence relation first. 

dp[i][C] = dp[i – 1][C – arr[i]] + dp[i – 1][C] 



Let’s understand the states of the DP now. Here, dp[i][C] stores the number of subsets of the sub-array arr[i…N-1] such that their sum is equal to C
Thus, the recurrence is very trivial as there are only two choices i.e. either consider the ith element in the subset or don’t.
Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
#define maxN 20
#define maxSum 50
#define minSum 50
#define base 50
 
// To store the states of DP
int dp[maxN][maxSum + minSum];
bool v[maxN][maxSum + minSum];
 
// Function to return the required count
int findCnt(int* arr, int i, int required_sum, int n)
{
    // Base case
    if (i == n) {
        if (required_sum == 0)
            return 1;
        else
            return 0;
    }
 
    // If the state has been solved before
    // return the value of the state
    if (v[i][required_sum + base])
        return dp[i][required_sum + base];
 
    // Setting the state as solved
    v[i][required_sum + base] = 1;
 
    // Recurrence relation
    dp[i][required_sum + base]
        = findCnt(arr, i + 1, required_sum, n)
          + findCnt(arr, i + 1, required_sum - arr[i], n);
    return dp[i][required_sum + base];
}
 
// Driver code
int main()
{
    int arr[] = { 3, 3, 3, 3 };
    int n = sizeof(arr) / sizeof(int);
    int x = 6;
 
    cout << findCnt(arr, 0, x, n);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
static int maxN = 20;
static int maxSum = 50;
static int minSum = 50;
static int base = 50;
 
// To store the states of DP
static int [][]dp = new int[maxN][maxSum + minSum];
static boolean [][]v = new boolean[maxN][maxSum + minSum];
 
// Function to return the required count
static int findCnt(int []arr, int i,
                   int required_sum, int n)
{
    // Base case
    if (i == n)
    {
        if (required_sum == 0)
            return 1;
        else
            return 0;
    }
 
    // If the state has been solved before
    // return the value of the state
    if (v[i][required_sum + base])
        return dp[i][required_sum + base];
 
    // Setting the state as solved
    v[i][required_sum + base] = true;
 
    // Recurrence relation
    dp[i][required_sum + base] =
          findCnt(arr, i + 1, required_sum, n) +
          findCnt(arr, i + 1, required_sum - arr[i], n);
    return dp[i][required_sum + base];
}
 
// Driver code
public static void main(String []args)
{
    int arr[] = { 3, 3, 3, 3 };
    int n = arr.length;
    int x = 6;
 
    System.out.println(findCnt(arr, 0, x, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 implementation of the approach
import numpy as np
 
maxN = 20
maxSum = 50
minSum = 50
base = 50
 
# To store the states of DP
dp = np.zeros((maxN, maxSum + minSum));
v = np.zeros((maxN, maxSum + minSum));
 
# Function to return the required count
def findCnt(arr, i, required_sum, n) :
 
    # Base case
    if (i == n) :
        if (required_sum == 0) :
            return 1;
        else :
            return 0;
 
    # If the state has been solved before
    # return the value of the state
    if (v[i][required_sum + base]) :
        return dp[i][required_sum + base];
 
    # Setting the state as solved
    v[i][required_sum + base] = 1;
 
    # Recurrence relation
    dp[i][required_sum + base] = findCnt(arr, i + 1,
                                         required_sum, n) + \
                                 findCnt(arr, i + 1,
                                         required_sum - arr[i], n);
     
    return dp[i][required_sum + base];
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 3, 3, 3, 3 ];
    n = len(arr);
    x = 6;
 
    print(findCnt(arr, 0, x, n));
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
     
class GFG
{
 
static int maxN = 20;
static int maxSum = 50;
static int minSum = 50;
static int Base = 50;
 
// To store the states of DP
static int [,]dp = new int[maxN, maxSum + minSum];
static Boolean [,]v = new Boolean[maxN, maxSum + minSum];
 
// Function to return the required count
static int findCnt(int []arr, int i,
                   int required_sum, int n)
{
    // Base case
    if (i == n)
    {
        if (required_sum == 0)
            return 1;
        else
            return 0;
    }
 
    // If the state has been solved before
    // return the value of the state
    if (v[i, required_sum + Base])
        return dp[i, required_sum + Base];
 
    // Setting the state as solved
    v[i, required_sum + Base] = true;
 
    // Recurrence relation
    dp[i, required_sum + Base] =
          findCnt(arr, i + 1, required_sum, n) +
          findCnt(arr, i + 1, required_sum - arr[i], n);
    return dp[i,required_sum + Base];
}
 
// Driver code
public static void Main(String []args)
{
    int []arr = { 3, 3, 3, 3 };
    int n = arr.Length;
    int x = 6;
 
    Console.WriteLine(findCnt(arr, 0, x, n));
}
}
 
// This code is contributed by 29AjayKumar
Output
6

Method 2: Using Tabulation Method:

In this method we use a 2D array of size (arr.size() + 1) * (target + 1) of type integer.
Initialization of Matrix:
mat[0][0] = 1 because If the size of sum is 
if (A[i] > j)
DP[i][j] = DP[i-1][j]
else 
DP[i][j] = DP[i-1][j] + DP[i-1][j-A[i]]

This means that if the current element has a value greater than the ‘current sum value’ we will copy the answer for previous cases

And if the current sum value is greater than the ‘ith’ element we will see if any of the previous states have already experienced the sum=’j’ and any previous states experienced a value ‘j – A[i]’ which will solve our purpose

C++




#include <bits/stdc++.h>
using namespace std;
 
int subsetSum(int a[], int n, int sum)
{
    // Initializing the matrix
    int tab[n + 1][sum + 1];
  // Initializing the first value of matrix
    tab[0][0] = 1;
    for (int i = 1; i <= sum; i++)
        tab[0][i] = 0;
    for (int i = 1; i <= n; i++)
        tab[i][0] = 1;
 
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= sum; j++)
        {
          // if the value is greater than the sum
            if (a[i - 1] > j)
                tab[i][j] = tab[i - 1][j];
            else
            {
                tab[i][j] = tab[i - 1][j] + tab[i - 1][j - a[i - 1]];
            }
        }
    }
 
 
    return tab[n][sum];
}
 
int main()
{
    int n = 4;
    int a[] = {3,3,3,3};
    int sum = 6;
 
    cout << (subsetSum(a, n, sum));
}
Output
6

Time Complexity: O(sum*n), where the sum is the ‘target sum’ and ‘n’ is the size of the array.

Auxiliary Space: O(sum*n), as the size of the 2-D array, is sum*n. 

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