Python|查找数组中具有给定总和的唯一子集的数量

给定一个数组和一个总和，找出每个子集的总和等于给定总和值的唯一子集的计数。

例子：

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

Input :
1 2 3 4 5
10
Output :
3

我们将使用动态规划来解决这个问题，这个解决方案的时间复杂度为 O(n ² )。下面是代码中使用的dp[][] –

_ | 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]

以下是Python代码：

# 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))

输出 ：

2