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 = 6Output:3

All the possible subsets are {1, 2, 3},

{1, 2, 3} and {3, 3}

Input:arr[] = {1, 1, 1, 1}, X = 1Output:4

**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 **i ^{th}** 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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