如何从元组创建所有的组合和？(面谈)

Question

我在面试中遇到一个问题，这是很有挑战性的。本课题以决策分析为基础。问题是，让我们假设我们有一个元组；

(15, 8, 8, 3)

我们想要创建所有组合的所有和一个一个，而不重复和求和相同的数字，如这个输出；

[(23, 8, 3), (18, 8, 8), (15, 11, 8)]

另一个例子；

(6, 5, 3, 8)

产出如下：

[(11, 3, 8), (9, 5, 8), (14, 5, 3), (6, 8, 8), (6, 13, 3), (6, 5, 11)]

注意:订单是灵活的。

我真的很想知道答案，所以如果有人对这个编码挑战感兴趣，可以帮助我改进我的思维结构。

Answer 1

In[2]: from itertools import combinations
  ...: 
  ...: 
  ...: def solution(nums):
  ...:     result = []
  ...:     seen = set()
  ...:     for p in combinations(range(len(nums)), r=2):
  ...:         dex_1, dex_2 = p
  ...:         if nums[dex_1] == nums[dex_2]:
  ...:             continue
  ...:         current = []
  ...:         for i, elem in enumerate(nums):
  ...:             if i == dex_1:
  ...:                 current.append(elem + nums[dex_2])
  ...:             elif i != dex_2:
  ...:                 current.append(elem)
  ...:         sorted_current = tuple(sorted(current))
  ...:         if sorted_current not in seen:
  ...:             result.append(tuple(current))
  ...:         seen.add(sorted_current)
  ...:     return result
  ...: 
In[3]: solution((15, 8, 8, 3))
Out[3]: [(23, 8, 3), (18, 8, 8), (15, 11, 8)]
In[4]: solution((6, 5, 3, 8))
Out[4]: [(11, 3, 8), (9, 5, 8), (14, 5, 3), (6, 8, 8), (6, 13, 3), (6, 5, 11)]

Answer 2

您可以尝试这种方法：

import itertools

final_=[]
for m in list(itertools.permutations(order,r=4)):
    if m[:2][0]==m[:2][1]:
        pass
    else:
        final_.append(tuple(sorted((sum(m[:2]),)+m[2:])))

print(set(final_))

产出：

when order=(15, 8, 8, 3)

产出：

{(3, 8, 23), (8, 8, 18), (8, 11, 15)}

当order=(6, 5, 3, 8)

产出：

{(5, 8, 9), (3, 8, 11), (3, 5, 14), (6, 8, 8), (5, 6, 11), (3, 6, 13)}