Python程序获取所有具有总和x的子集

我们得到一个包含 n 个数字和一个数字 x 的列表，任务是编写一个Python程序来找出列表中所有可能的子集，使得它们的总和为 x。

例子：

Input: arr = [2, 4, 5, 9], x = 15

Output: [2, 4, 9]

15 can be obtained by adding 2, 4 and 9 from the given list.

Input  : arr = [10, 20, 25, 50, 70, 90], x = 80

Output : [10, 70]

         [10, 20, 50]

80 can be obtained by adding 10 and 70 or by adding 10, 20 and 50 from the given list.

方法#1：

这是一种蛮力方法。找到给定列表的所有可能子集和并检查和是否等于 x。使用这种方法的时间复杂度为 O(2^n)，非常大。

Python3

# Python code with time complexity
# O(2^n)to print all subsets whose
# sum is equal to a given value
from itertools import combinations
  
  
def subsetSum(n, arr, x):
    
    # Iterating through all possible
    # subsets of arr from lengths 0 to n:
    for i in range(n+1):
        for subset in combinations(arr, i):
              
            # printing the subset if its sum is x:
            if sum(subset) == x:
                print(list(subset))
  
  
# Driver Code:
n = 6
arr = [10, 20, 25, 50, 70, 90]
x = 80
subsetSum(n, arr, x)

输出：

[10, 70]
[10, 20, 50]

方法#2：

中间相遇是一种将搜索空间划分为两个大小相等的部分，对这两个部分进行单独搜索，然后组合搜索结果的技术。使用这种技术，两次搜索可能需要比一次大搜索更少的时间，并将时间复杂度从 O(2^n) 变成 O(2^(n/2))。

蟒蛇3

# Efficient Python code to
# print all subsets whose sum
# is equal to a given value
from itertools import combinations
  
  
def subsetSum(li, comb, sums):
    # Iterating through all subsets of
    # list li from length 0 to length of li:
    for i in range(len(li)+1):
        for subset in combinations(li, i):
              
            # Storing all the subsets in list comb:
            comb.append(list(subset))
              
            # Storing the subset sums in list sums:
            sums.append(sum(subset))
  
  
def calcSubsets(n, arr, x):
    
    # Dividing the list arr into two lists
    # arr1 and arr2 of about equal sizes
    # by slicing list arr about index n//2:
    arr1, arr2 = arr[:n//2], arr[n//2:]
      
    # Creating empty lists comb1 and sums1
    # to run the subsetSum function and
    # store subsets of arr1 in comb1
    # and the subset sums in sums1:
    comb1, sums1 = [], []
    subsetSum(arr1, comb1, sums1)
      
    # Creating empty lists comb2 and sums2
    # to run the subsetSum function and
    # store subsets of arr2 in comb2
    # and the subset sums in sums2:
    comb2, sums2 = [], []
    subsetSum(arr2, comb2, sums2)
      
    # Iterating i through the indices of sums1:
    for i in range(len(sums1)):
          
        # Iterating j through the indices of sums2:
        for j in range(len(sums2)):
              
            # If two elements (one from sums1
            # and one from sums2) add up to x,
            # the combined list of elements from
            # corresponding subsets at index i in comb1
            # and j in comb2 gives us the required answer:
            if sums1[i] + sums2[j] == x:
                print(comb1[i] + comb2[j])
  
  
# Driver Code:
n = 6
arr = [10, 20, 25, 50, 70, 90]
x = 80
calcSubsets(n, arr, x)

输出：

[10, 70]
[10, 20, 50]