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Python | Find the number of unique subsets with given sum in array
• Last Updated : 17 Sep, 2019

Given an array and a sum, find the count of unique subsets with each subset’s sum equal to the given sum value.

Examples:

```Input :
4 12 5 9 12
9
Output :
2
(subsets will be [4, 5] and )

Input :
1 2 3 4 5
10
Output :
3
```

We will use dynamic programming to solve this problem and this solution has time complexity of O(n2). Below is `dp[][]` used in the code –

```   _ | 4 12  5  9 12
0 |[1, 1, 1, 1, 1]
1 |[0, 0, 0, 0, 0]
2 |[0, 0, 0, 0, 0]
3 |[0, 0, 0, 0, 0]
4 |[1, 1, 1, 1, 1]
5 |[0, 0, 1, 1, 1]
6 |[0, 0, 0, 0, 0]
7 |[0, 0, 0, 0, 0]
8 |[0, 0, 0, 0, 0]
9 |[0, 0, 1, 2, 2]```

Below is the Python Code :

 `# Python code to find the number of unique subsets ` `# with given sum in the given array ` ` `  `def` `help``(a,s): ` `     `  `        ``dp ``=` `[[``0` `for` `i ``in` `range``(``len``(a))] ``for` `j ``in` `range``(``0``,s``+``1``)] ` `        ``for` `i ``in` `range``(``len``(a)): ` `            ``dp[``0``][i] ``=` `1` `        ``for` `i ``in` `range``(``1``,s``+``1``): ` `            ``if` `i ``=``=` `a[``0``]: ` `                ``dp[i][``0``] ``=` `1` `        ``for` `i ``in` `range``(``1``, s``+``1``): ` `            ``for` `j ``in` `range``(``1``, ``len``(a)): ` `                ``if` `a[j]<``=``i: ` `                        ``dp[i][j] ``=` `dp[i][j``-``1``] ``+` `dp[i``-``a[j]][j``-``1``] ` `                ``else``: ` `                        ``dp[i][j] ``=` `dp[i][j``-``1``] ` `        ``return` `dp[s][``len``(a)``-``1``] ` `  `  `# driver code ` `a ``=` `[``4``, ``12``, ``5``, ``9``, ``12``] ` `s ``=` `9` ` `  `print``(``help``(a,s)) `

Output :

```2
```

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