Count of subsets with sum equal to X

Last Updated : 28 Feb, 2024

Given an array arr[] of length N and an integer X, the task is to find the number of subsets with a sum equal to X.

Examples:Â

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

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

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.Â

Below is the implementation of the above approach:Â

C++

 `// Naive Approach` `#include ` `using` `namespace` `std;`   `void` `printBool(``int` `n, ``int` `len)` `{`   `  ``while` `(n) {` `    ``if` `(n & 1)` `      ``cout << 1;` `    ``else` `      ``cout << 0;`   `    ``n >>= 1;` `    ``len--;` `  ``}`   `  ``// This is used for padding zeros` `  ``while` `(len) {` `    ``cout << 0;` `    ``len--;` `  ``}` `  ``cout << endl;` `}`   `// Function` `// Prints all the subsets of given set[]` `void` `printSubsetsCount(``int` `set[], ``int` `n, ``int` `val)` `{` `  ``int` `sum; ``// it stores the current sum` `  ``int` `count = 0;` `  ``for` `(``int` `i = 0; i < (1 << n); i++) {` `    ``sum = 0;` `    ``// Print current subset` `    ``for` `(``int` `j = 0; j < n; j++)`   `      ``// (1< 0) {` `        ``sum += set[j]; ``// elements are added one by` `        ``// one of a subset to the sum` `      ``}` `    ``// It checks if the sum is equal to desired sum. If` `    ``// it is true then it prints the elements of the sum` `    ``// to the output` `    ``if` `(sum == val) {` `      ``/*` `             ``* Uncomment printBool(i,n) to get the boolean` `             ``* representation of the selected elements from` `             ``* set. For this example output of this` `             ``* representation will be 0 1 1 1 0 // 2,3,4` `             ``* makes sum 9 1 0 1 0 1 // 1,3,5 also makes sum` `             ``* 9 0 0 0 1 1 // 4,5 also makes sum 9` `             ``*` `             ``* 'i' is used for 'and' operation so the` `             ``* position of set bits in 'i' will be the` `             ``* selected element. and as we have to give` `             ``* padding with zeros to represent the complete` `             ``* set , so length of the set ('n') is passed to` `             ``* the function.` `             ``* */` `      ``//  printBool(i,n);` `      ``count++;` `    ``}` `  ``}` `  ``// it means no subset is found with given sum` `  ``if` `(count == 0) {` `    ``cout << (``"No subset is found"``) << endl;` `  ``}` `  ``else` `{` `    ``cout << count << endl;` `  ``}` `}`   `// Driver code` `int` `main()` `{` `  ``int` `set[] = { 1, 2, 3, 4, 5 };` `  ``printSubsetsCount(set, 5, 9);` `}`   `// This code is contributed by garg28harsh.`

C

 `// Naive Approach`   `#include `   `void` `printBool(``int` `n, ``int` `len)` `{`   `    ``while` `(n) {` `        ``if` `(n & 1)` `            ``printf``(``"1 "``);` `        ``else` `            ``printf``(``"0 "``);`   `        ``n >>= 1;` `        ``len--;` `    ``}`   `    ``// This is used for padding zeros` `    ``while` `(len) {` `        ``printf``(``"0 "``);` `        ``len--;` `    ``}` `    ``printf``(``"\n"``);` `}`   `// Function` `// Prints all the subsets of given set[]` `void` `printSubsetsCount(``int` `set[], ``int` `n, ``int` `val)` `{` `    ``int` `sum; ``// it stores the current sum` `    ``int` `count = 0;` `    ``for` `(``int` `i = 0; i < (1 << n); i++) {` `        ``sum = 0;` `        ``// Print current subset` `        ``for` `(``int` `j = 0; j < n; j++)`   `            ``// (1< 0) {` `                ``sum += set[j]; ``// elements are added one by` `                               ``// one of a subset to the sum` `            ``}` `        ``// It checks if the sum is equal to desired sum. If` `        ``// it is true then it prints the elements of the sum` `        ``// to the output` `        ``if` `(sum == val) {` `            ``/*` `             ``* Uncomment printBool(i,n) to get the boolean` `             ``* representation of the selected elements from` `             ``* set. For this example output of this` `             ``* representation will be 0 1 1 1 0 // 2,3,4` `             ``* makes sum 9 1 0 1 0 1 // 1,3,5 also makes sum` `             ``* 9 0 0 0 1 1 // 4,5 also makes sum 9` `             ``*` `             ``* 'i' is used for 'and' operation so the` `             ``* position of set bits in 'i' will be the` `             ``* selected element. and as we have to give` `             ``* padding with zeros to represent the complete` `             ``* set , so length of the set ('n') is passed to` `             ``* the function.` `             ``* */` `            ``//  printBool(i,n);` `            ``count++;` `        ``}` `    ``}` `    ``// it means no subset is found with given sum` `    ``if` `(count == 0) {` `        ``printf``(``"No subset is found"``);` `    ``}` `    ``else` `{` `        ``printf``(``"%d"``, count);` `    ``}` `}`   `// Driver code` `void` `main()` `{` `    ``int` `set[] = { 1, 2, 3, 4, 5 };` `    ``printSubsetsCount(set, 5, 9);` `}`

Java

 `import` `java.io.*;` `// Naive Approach` `class` `GFG {`   `    ``static` `void` `printBool(``int` `n, ``int` `len)` `    ``{`   `        ``while` `(n>``0``) {` `            ``if` `((n & ``1``) == ``1``)` `                ``System.out.print(``1``);` `            ``else` `                ``System.out.print(``0``);`   `            ``n >>= ``1``;` `            ``len--;` `        ``}`   `        ``// This is used for padding zeros` `        ``while` `(len>``0``) {` `            ``System.out.print(``0``);` `            ``len--;` `        ``}` `        ``System.out.println();` `    ``}`   `    ``// Function` `    ``// Prints all the subsets of given set[]` `    ``static` `void` `printSubsetsCount(``int` `set[], ``int` `n, ``int` `val)` `    ``{` `        ``int` `sum; ``// it stores the current sum` `        ``int` `count = ``0``;` `        ``for` `(``int` `i = ``0``; i < (``1` `<< n); i++) {` `            ``sum = ``0``;` `            ``// Print current subset` `            ``for` `(``int` `j = ``0``; j < n; j++)`   `                ``// (1< ``0``) {` `                    ``sum += set[j]; ``// elements are added one` `                                   ``// by` `                    ``// one of a subset to the sum` `                ``}` `            ``// It checks if the sum is equal to desired sum.` `            ``// If it is true then it prints the elements of` `            ``// the sum to the output` `            ``if` `(sum == val) {` `                ``/*` `                 ``* Uncomment printBool(i,n) to get the` `                 ``* boolean representation of the selected` `                 ``* elements from set. For this example` `                 ``* output of this representation will be 0 1` `                 ``* 1 1 0 // 2,3,4 makes sum 9 1 0 1 0 1 //` `                 ``* 1,3,5 also makes sum 9 0 0 0 1 1 // 4,5` `                 ``* also makes sum 9` `                 ``*` `                 ``* 'i' is used for 'and' operation so the` `                 ``* position of set bits in 'i' will be the` `                 ``* selected element. and as we have to give` `                 ``* padding with zeros to represent the` `                 ``* complete set , so length of the set ('n')` `                 ``* is passed to the function.` `                 ``* */` `                ``//  printBool(i,n);` `                ``count++;` `            ``}` `        ``}` `        ``// it means no subset is found with given sum` `        ``if` `(count == ``0``) {` `            ``System.out.println(``"No subset is found"``);` `        ``}` `        ``else` `{` `            ``System.out.println(count);` `        ``}` `    ``}` `  `  `// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `set[] = { ``1``, ``2``, ``3``, ``4``, ``5` `};` `        ``printSubsetsCount(set, ``5``, ``9``);` `    ``}` `}` `// This code is contributed by Abhijeet Kumar(abhijeet19403)`

Python3

 `#  Naive Approach` `def` `printBool(n, ``len``):` `    ``while` `n:` `        ``if` `n & ``1``:` `            ``print``(``"1 "``)` `        ``else``:` `            ``print``(``"0 "``)` `        ``n ``=` `n >> ``1` `        ``len` `-``=` `1`   `    ``#  This is used for padding zeros` `    ``while` `len``:` `        ``print``(``"0 "``)` `        ``len` `-``=` `1` `    ``print``()`   `#  Function` `#  Prints all the subsets of given set[]`   `def` `printSubsetsCount(``set``, n, val):` `    ``sum` `=` `0`  `# it stores the current sum` `    ``count ``=` `0` `    ``for` `i ``in` `range``(``0``, ``1` `<< n):` `        ``sum` `=` `0`   `        ``# Print current subset` `        ``for` `j ``in` `range``(``0``, n):`   `            ``#  (1< ``0``:` `                ``sum` `+``=` `set``[j]  ``# elements are added one by` `                ``# one of a subset to the sum`   `        ``#  It checks if the sum is equal to desired sum. If` `        ``#  it is true then it prints the elements of the sum` `        ``#  to the output`   `        ``if` `(``sum` `=``=` `val):` `            ``#` `            ``#   Uncomment printBool(i,n) to get the boolean` `            ``#   representation of the selected elements from` `            ``#   set. For this example output of this` `            ``#   representation will be 0 1 1 1 0 // 2,3,4` `            ``#   makes sum 9 1 0 1 0 1 // 1,3,5 also makes sum` `            ``#   9 0 0 0 1 1 // 4,5 also makes sum 9` `            ``#` `            ``#   'i' is used for 'and' operation so the` `            ``#   position of set bits in 'i' will be the` `            ``#   selected element. and as we have to give` `            ``#   padding with zeros to represent the complete` `            ``#   set , so length of the set ('n') is passed to` `            ``#   the function.` `            ``#` `            ``#   printBool(i,n);` `            ``count ``+``=` `1`   `    ``#  it means no subset is found with given sum` `    ``if` `(count ``=``=` `0``):`   `        ``print``(``"No subset is found"``)`   `    ``else``:` `        ``print``(count)`   `#  Driver code` `set` `=` `[``1``, ``2``, ``3``, ``4``, ``5``]` `printSubsetsCount(``set``, ``5``, ``9``)`

C#

 `// Naive Approach` `using` `System;` `class` `GFG {`   `  ``static` `void` `printBool(``int` `n, ``int` `len)` `  ``{`   `    ``while` `(n > 0) {` `      ``if` `((n & 1) == 1)` `        ``Console.Write(1);` `      ``else` `        ``Console.Write(0);`   `      ``n >>= 1;` `      ``len--;` `    ``}`   `    ``// This is used for padding zeros` `    ``while` `(len > 0) {` `      ``Console.Write(0);` `      ``len--;` `    ``}` `    ``Console.WriteLine();` `  ``}`   `  ``// Function` `  ``// Prints all the subsets of given set[]` `  ``static` `void` `printSubsetsCount(``int``[] ``set``, ``int` `n, ``int` `val)` `  ``{` `    ``int` `sum; ``// it stores the current sum` `    ``int` `count = 0;` `    ``for` `(``int` `i = 0; i < (1 << n); i++) {` `      ``sum = 0;` `      ``// Print current subset` `      ``for` `(``int` `j = 0; j < n; j++)`   `        ``// (1< 0) {` `          ``sum += ``set``[j]; ``// elements are added one` `          ``// by` `          ``// one of a subset to the sum` `        ``}` `      ``// It checks if the sum is equal to desired sum.` `      ``// If it is true then it prints the elements of` `      ``// the sum to the output` `      ``if` `(sum == val) {` `        ``/*` `                 ``* Uncomment printBool(i,n) to get the` `                 ``* boolean representation of the selected` `                 ``* elements from set. For this example` `                 ``* output of this representation will be 0 1` `                 ``* 1 1 0 // 2,3,4 makes sum 9 1 0 1 0 1 //` `                 ``* 1,3,5 also makes sum 9 0 0 0 1 1 // 4,5` `                 ``* also makes sum 9` `                 ``*` `                 ``* 'i' is used for 'and' operation so the` `                 ``* position of set bits in 'i' will be the` `                 ``* selected element. and as we have to give` `                 ``* padding with zeros to represent the` `                 ``* complete set , so length of the set ('n')` `                 ``* is passed to the function.` `                 ``* */` `        ``//  printBool(i,n);` `        ``count++;` `      ``}` `    ``}` `    ``// it means no subset is found with given sum` `    ``if` `(count == 0) {` `      ``Console.WriteLine(``"No subset is found"``);` `    ``}` `    ``else` `{` `      ``Console.WriteLine(count);` `    ``}` `  ``}`   `  ``// Driver code` `  ``public` `static` `void` `Main(``string``[] args)` `  ``{` `    ``int``[] ``set` `= { 1, 2, 3, 4, 5 };` `    ``printSubsetsCount(``set``, 5, 9);` `  ``}` `}`   `// This code is contributed by Karandeep1234`

Javascript

 `// Naive Approach`   `function` `printBool( n,  len)` `{`   `  ``while` `(n!=0) {` `    ``if` `((n & 1)!=0)` `     ``console.log(1);` `    ``else` `      ``console.log(0);`   `    ``n/=2 ;` `    ``len--;` `  ``}`   `  ``// This is used for padding zeros` `  ``while` `(len!=0) {` `    ``console.log(0);` `    ``len--;` `  ``}` `  ``cout << endl;` `}`   `// Function` `// Prints all the subsets of given set[]` `function` `printSubsetsCount(set, n, val)` `{` `  ``let sum; ``// it stores the current sum` `  ``let count = 0;` `  ``for` `(let i = 0; i < Math.pow(2,n); i++) {` `    ``sum = 0;` `    ``// Print current subset` `    ``for` `(let j = 0; j < n; j++)`   `      ``// (1< 0) {` `        ``sum += set[j]; ``// elements are added one by` `        ``// one of a subset to the sum` `      ``}` `    ``// It checks if the sum is equal to desired sum. If` `    ``// it is true then it prints the elements of the sum` `    ``// to the output` `    ``if` `(sum == val) {` `      ``/*` `             ``* Uncomment printBool(i,n) to get the boolean` `             ``* representation of the selected elements from` `             ``* set. For this example output of this` `             ``* representation will be 0 1 1 1 0 // 2,3,4` `             ``* makes sum 9 1 0 1 0 1 // 1,3,5 also makes sum` `             ``* 9 0 0 0 1 1 // 4,5 also makes sum 9` `             ``*` `             ``* 'i' is used for 'and' operation so the` `             ``* position of set bits in 'i' will be the` `             ``* selected element. and as we have to give` `             ``* padding with zeros to represent the complete` `             ``* set , so length of the set ('n') is passed to` `             ``* the function.` `             ``* */` `      ``//  printBool(i,n);` `      ``count++;` `    ``}` `  ``}` `  ``// it means no subset is found with given sum` `  ``if` `(count == 0) {` `    ``console.log(``"No subset is found"``);` `  ``}` `  ``else` `{` `    ``console.log(count);` `  ``}` `}`   `// Driver code`   `  ``let set = [ 1, 2, 3, 4, 5 ];` `  ``printSubsetsCount(set, 5, 9);`   `// This code is contributed by garg28harsh.`

Output

```3
```

Time Complexity: O(2n), as we are generating all the subsets of the given set. Since there are 2^n subsets, therefore it requires O(2^n) time to generate all the subsets.
Auxiliary Space: O(1), No extra space is required.

However, for smaller values of X and array elements, this problem can be solved using dynamic programming.Â
Let’s look at the recurrence relation first.Â

This method is valid for all the integers.

`dp[i][C] = dp[i - 1][C - arr[i]] + dp[i - 1][C] `

Let’s understand the state 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.Â
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:Â

C++

 `// C++ implementation of the approach` `#include ` `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`

Javascript

 ``

Output

```6
```

Time Complexity: O(n * (maxSum + minSum))

The time complexity of the above approach is O(n*(maxSum + minSum)). Here, n is the size of the given array and maxSum + minSum is the total range of values that the required sum can take.

Space Complexity: O(n * (maxSum + minSum))

The space complexity of the approach is also O(n*(maxSum + minSum)). Here, n is the size of the given array and maxSum + minSum is the total range of values that the required sum can take. We need an extra 2-D array of size n*(maxSum + minSum) to store the states of the DP.

Method 2: Using Tabulation Method:

`This method is valid only for those arrays which contains positive elements.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 sum is 0 then there exists null subset {} whose sum is 0`
`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 â€˜current sum valueâ€™ we will copy the answer for previous cases

And if the current sum value is greater than the â€˜ithâ€™ element we will see if any of the previous states have already experienced the sum=â€™jâ€™ and any previous states experienced a value â€˜j â€“ A[i]â€™ which will solve our purpose

C++

 `#include ` `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++)` `    ``{` `        ``for` `(``int` `j = 0; 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));` `}`

Java

 `import` `java.io.*;` `import` `java.lang.*;` `import` `java.util.*;`   `class` `GFG{`   `static` `int` `subsetSum(``int` `a[], ``int` `n, ``int` `sum)` `{` `    `  `    ``// Initializing the matrix` `    ``int` `tab[][] = ``new` `int``[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++) ` `    ``{` `        ``for``(``int` `j = ``0``; 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];` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `n = ``4``;` `    ``int` `a[] = { ``3``, ``3``, ``3``, ``3` `};` `    ``int` `sum = ``6``;`   `    ``System.out.print(subsetSum(a, n, sum));` `}` `}`   `// This code is contributed by Kingash`

Python3

 `def` `subset_sum(a: ``list``, n: ``int``, ``sum``: ``int``):` `  `  `    ``# Initializing the matrix` `    ``tab ``=` `[[``0``] ``*` `(``sum` `+` `1``) ``for` `i ``in` `range``(n ``+` `1``)]` `    ``tab[``0``][``0``] ``=` `1` `    ``for` `i ``in` `range``(``1``, ``sum` `+` `1``):` `        ``tab[``0``][i] ``=` `0` `    `  `    ``for` `i ``in` `range``(``1``, n``+``1``):` `        ``for` `j ``in` `range``(``sum` `+` `1``):` `            ``if` `a[i``-``1``] <``=` `j:` `                ``tab[i][j] ``=` `tab[i``-``1``][j] ``+` `tab[i``-``1``][j``-``a[i``-``1``]]` `            ``else``:` `                ``tab[i][j] ``=` `tab[i``-``1``][j]` `    ``return` `tab[n][``sum``]`   `if` `__name__ ``=``=` `'__main__'``:` `    ``a ``=` `[``3``, ``3``, ``3``, ``3``]` `    ``n ``=` `4` `    ``sum` `=` `6` `    ``print``(subset_sum(a, n, ``sum``))`   `    ``# This code is contributed by Premansh2001.`

C#

 `using` `System;`   `class` `GFG{`   `static` `int` `subsetSum(``int` `[]a, ``int` `n, ``int` `sum)` `{` `    `  `    ``// Initializing the matrix` `    ``int` `[,]tab = ``new` `int``[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++) ` `    ``{` `        ``for``(``int` `j = 0; 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];` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``int` `n = 4;` `    ``int` `[]a = { 3, 3, 3, 3 };` `    ``int` `sum = 6;`   `    ``Console.Write(subsetSum(a, n, sum));` `}` `}`   `// This code is contributed by shivanisinghss2110`

Javascript

 ``

Output

```6
```

Time Complexity: O(sum*n), where the sum is the â€˜target sumâ€™ and â€˜nâ€™ is the size of the array.
Auxiliary Space: O(sum*n), as the size of the 2-D array, is sum*n.Â

What if the value of elements starts from 0?Â

In the case of elements with value 0, your answer can be incorrect with the above solution. Here’s the reason why:-

Consider the below example:

arr[] = {0,1,1,1}

target sum = 3

correct output : 2

As you can see from the dp table, if there is a 0 in the array then it will not take part in the count if it is in the starting position (dp[1][1]).

Â

but if the zero is at the end of the array, then the table would be:

Â

So just sort the array in descending order to achieve the correct output.

Method 3: Space Optimization:

We can solve this problem by just taking care of last state and current state so we can wrap up this problem in O(target+1) space complexity.

Example:-

vector<int> arr = { 3, 3, 3, 3 }; with targetSum of 6;

dp[0][arr[0]] — tells about what if at index 0 we need arr[0] to achieve the targetSum and fortunately we have that solve so mark them 1;

=====dp[0][3]=1

target

Index

0 1 2 3 4 5 6
0 1 0 0 1 0 0 0
1 1 0 0 2 0 0 1
2 1 0 0 3 0 0 3
3 1 0 0 4 0 0 6
`at dp[2][6] --- tells tell me is at index 2 can count some subsets with sum=6, How can we achieve this?so we can tell ok i have reached at index 2 by adding element of index 1 or not both case has been added ------ means dp[i-1] we need only bcoz we are need of last index decision only nothing more than that so this why we are using a huge 2D arrayjust store our running state and last state that's it.1.Time Complexity:- O(N*val)2.Space Complexity:- O(Val)where val and n are targetSum and number of element.`

C++

 `#include ` `using` `namespace` `std;`   `int` `CountSubsetSum(vector<``int``>& arr, ``int` `val, ``int` `n)` `{` `    ``vector<``int``> PresentState(val + 1, 0),` `        ``LastState(val + 1, 0);` `    ``// consider only last and present state we dont need the` `    ``// (present-2)th state and above and we know for val to` `    ``// be 0 if we dont pick the current index element we can` `    ``// achieve` `    ``PresentState[0] = LastState[0] = 1;` `    ``if` `(arr[0] <= val)` `        ``LastState[arr[0]] = 1;` `    ``for` `(``int` `i = 1; i < n; i++) {` `        ``for` `(``int` `j = 1; j <= val; j++)` `            ``PresentState[j]` `                ``= ((j >= arr[i]) ? LastState[j - arr[i]]` `                                 ``: 0)` `                  ``+ LastState[j];` `        ``// this we will need in the next iteration so just` `        ``// swap current and last state.` `        ``LastState = PresentState;` `    ``}` `    ``// Note after exit from loop we will having a present` `    ``// state which is nothing but the laststate itself;` `    ``return` `PresentState[val]; ``// or return` `                              ``// CurrentState[val];` `}` `int` `main()` `{` `    ``vector<``int``> arr = { 3, 3, 3, 3 };` `    ``cout << CountSubsetSum(arr, 6, arr.size());` `}`

Java

 `import` `java.io.*;` `import` `java.lang.*;` `import` `java.util.*;`   `class` `GFG {`   `    ``static` `int` `subsetSum(``int` `arr[], ``int` `n, ``int` `val)` `    ``{`   `        ``int``[] LastState = ``new` `int``[val + ``1``];` `        ``// consider only last and present state we dont need` `        ``// the (present-2)th state and above and we know for` `        ``// val to be 0 if we dont pick the current index` `        ``// element we can achieve`   `        ``LastState[``0``] = ``1``;` `        ``if` `(arr[``0``] <= val) {` `            ``LastState[arr[``0``]] = ``1``;` `        ``}` `        ``for` `(``int` `i = ``1``; i < n; i++) {` `            ``int``[] PresentState = ``new` `int``[val + ``1``];` `            ``PresentState[``0``] = ``1``;` `            ``for` `(``int` `j = ``1``; j <= val; j++) {`   `                ``int` `notPick = LastState[j];` `                ``int` `pick = ``0``;` `                ``if` `(arr[i] <= j)` `                    ``pick = LastState[j - arr[i]];`   `                ``PresentState[j] = pick + notPick;` `            ``}` `            ``// this we will need in the next iteration so` `            ``// just swap current and last state.` `            ``LastState = PresentState;` `        ``}` `        ``// Note after exit from loop we will having a` `        ``// present state which is nothing but the laststate` `        ``// itself;` `        ``return` `LastState[val]; ``// or return` `                               ``// CurrentState[val];` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `n = ``4``;` `        ``int` `a[] = { ``3``, ``3``, ``3``, ``3` `};` `        ``int` `sum = ``6``;`   `        ``System.out.print(subsetSum(a, n, sum));` `    ``}` `}`   `// This code is contributed by Sanskar`

Python3

 `def` `CountSubsetSum( arr, val, n):`   `    ``LastState``=``[``0``]``*``(val ``+` `1``);` `    ``# consider only last and present state we dont need the` `    ``# (present-2)th state and above and we know for val to` `    ``# be 0 if we dont pick the current index element we can` `    ``# achieve` `    ``LastState[``0``] ``=` `1``;` `    ``if` `(arr[``0``] <``=` `val):` `        ``LastState[arr[``0``]] ``=` `1``;` `    ``for` `i ``in` `range``(``1``,n):` `        ``PresentState``=``[``0``]``*``(val ``+` `1``);` `        ``PresentState[``0``]``=``1``;` `        ``for` `j ``in` `range``(``1``,val``+``1``):` `            ``if``(j >``=` `arr[i]):                ` `                ``PresentState[j] ``=` `LastState[j ``-` `arr[i]]``+` `LastState[j];` `            ``else``:` `                ``PresentState[j] ``=` `LastState[j]` `                `  `        ``# this we will need in the next iteration so just` `        ``# swap current and last state.` `        ``LastState ``=` `PresentState;` `    ``# Note after exit from loop we will having a present` `    ``# state which is nothing but the laststate itself;` `    ``return` `PresentState[val]; ``# or return` `                              ``# CurrentState[val];` `                              `  `arr ``=` `[ ``3``, ``3``, ``3``, ``3` `];` `print``(CountSubsetSum(arr, ``6``, ``len``(arr)));`

C#

 `// C# implementation of the approach` `using` `System;` `using` `System.Collections.Generic;            `   `class` `GFG` `{`   `  ``static` `int` `subsetSum(``int``[] arr, ``int` `n, ``int` `val)` `  ``{`   `    ``int``[] LastState = ``new` `int``[val + 1];` `    ``// consider only last and present state we dont need` `    ``// the (present-2)th state and above and we know for` `    ``// val to be 0 if we dont pick the current index` `    ``// element we can achieve`   `    ``LastState[0] = 1;` `    ``if` `(arr[0] <= val) {` `      ``LastState[arr[0]] = 1;` `    ``}` `    ``for` `(``int` `i = 1; i < n; i++) {` `      ``int``[] PresentState = ``new` `int``[val + 1];` `      ``PresentState[0] = 1;` `      ``for` `(``int` `j = 1; j <= val; j++) {`   `        ``int` `notPick = LastState[j];` `        ``int` `pick = 0;` `        ``if` `(arr[i] <= j)` `          ``pick = LastState[j - arr[i]];`   `        ``PresentState[j] = pick + notPick;` `      ``}` `      ``// this we will need in the next iteration so` `      ``// just swap current and last state.` `      ``LastState = PresentState;` `    ``}` `    ``// Note after exit from loop we will having a` `    ``// present state which is nothing but the laststate` `    ``// itself;` `    ``return` `LastState[val]; ``// or return` `    ``// CurrentState[val];` `  ``}`   `  ``// Driver code` `  ``public` `static` `void` `Main (String[] args)` `  ``{` `    ``int` `n = 4;` `    ``int``[] a = { 3, 3, 3, 3 };` `    ``int` `sum = 6;`   `    ``Console.WriteLine(subsetSum(a, n, sum));` `  ``}` `}`   `// This code is contributed by sanjoy_62.`

Javascript

 `function` `countSubsetSum(arr, val, n) {` `    ``let presentState = ``new` `Array(val + 1).fill(0); ` `    `  `    ``// consider only last and present state we dont need the (present-2)th ` `    ``// state and above and we know for val to be 0 if we dont pick the ` `    ``// current index element we can achieve` `    ``let lastState = ``new` `Array(val + 1).fill(0);` `    ``presentState[0] = lastState[0] = 1;` `    ``if` `(arr[0] <= val) {` `        ``lastState[arr[0]] = 1;` `    ``}` `    `  `    ``for` `(let i = 1; i < n; i++) {` `        ``for` `(let j = 1; j <= val; j++) {` `            ``presentState[j] = ((j >= arr[i]) ? lastState[j - arr[i]] : 0) + lastState[j]; ` `            ``// this we will need in the next iteration so just swap current and last state.` `        ``}` `        ``lastState = [...presentState];` `    ``}` `    `  `    ``// Note after exit from loop we will having a present ` `    ``// state which is nothing but the laststate itself;` `    ``return` `presentState[val]; ``// or return CurrentState[val];` `}`   `let arr = [3, 3, 3, 3];` `console.log(countSubsetSum(arr, 6, arr.length));`   `// This code is contributed by ritaagarwal.`

Output

```6
```

Time Complexity: O(sum*n), where the sum is the â€˜target sumâ€™ and â€˜nâ€™ is the size of the array.
Auxiliary Space: O(sum).Â

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