# Bijective function between two nested lists python

When working with nested lists in Python, it can be useful to find a bijective function that maps elements from one nested list to another. In this article, we will explore three different ways to solve this problem.

## Option 1: Using a Recursive Function

One way to solve this problem is by using a recursive function. We can define a function that takes two nested lists as input and returns a bijective function between them. Here’s an example:

``````def bijective_function(list1, list2):
if len(list1) != len(list2):
return None

mapping = {}

for i in range(len(list1)):
if isinstance(list1[i], list) and isinstance(list2[i], list):
mapping.update(bijective_function(list1[i], list2[i]))
else:
mapping[list1[i]] = list2[i]

return mapping

# Example usage
list1 = [1, [2, 3], 4]
list2 = ['a', ['b', 'c'], 'd']

result = bijective_function(list1, list2)
print(result)``````

This recursive function checks if the lengths of the input lists are equal. If they are not, it returns None. Otherwise, it initializes an empty dictionary to store the mapping between elements. It then iterates over the elements of the lists and checks if they are nested lists. If they are, it recursively calls the function on the nested lists and updates the mapping. If they are not nested lists, it adds the mapping between the elements to the dictionary. Finally, it returns the mapping.

## Option 2: Using List Comprehension

Another way to solve this problem is by using list comprehension. We can define a function that takes two nested lists as input and returns a bijective function between them. Here’s an example:

``````def bijective_function(list1, list2):
if len(list1) != len(list2):
return None

mapping = {list1[i]: list2[i] for i in range(len(list1))}

for i in range(len(list1)):
if isinstance(list1[i], list) and isinstance(list2[i], list):
nested_mapping = bijective_function(list1[i], list2[i])
if nested_mapping is None:
return None
mapping.update(nested_mapping)

return mapping

# Example usage
list1 = [1, [2, 3], 4]
list2 = ['a', ['b', 'c'], 'd']

result = bijective_function(list1, list2)
print(result)``````

This approach uses list comprehension to create the initial mapping between elements. It then iterates over the elements of the lists and checks if they are nested lists. If they are, it recursively calls the function on the nested lists and updates the mapping. If the nested mapping is None, it returns None. Finally, it returns the mapping.

## Option 3: Using a Stack

A third way to solve this problem is by using a stack. We can define a function that takes two nested lists as input and returns a bijective function between them. Here’s an example:

``````def bijective_function(list1, list2):
if len(list1) != len(list2):
return None

stack = [(list1, list2)]
mapping = {}

while stack:
l1, l2 = stack.pop()

if isinstance(l1, list) and isinstance(l2, list):
if len(l1) != len(l2):
return None

for i in range(len(l1)):
stack.append((l1[i], l2[i]))
else:
mapping[l1] = l2

return mapping

# Example usage
list1 = [1, [2, 3], 4]
list2 = ['a', ['b', 'c'], 'd']

result = bijective_function(list1, list2)
print(result)``````

This approach uses a stack to keep track of the nested lists. It starts by pushing the input lists onto the stack. It then pops elements from the stack and checks if they are nested lists. If they are, it iterates over the elements and pushes them onto the stack. If they are not nested lists, it adds the mapping between the elements to the dictionary. Finally, it returns the mapping.

After exploring these three options, it is clear that the best solution depends on the specific requirements of the problem. The recursive function is the most straightforward and intuitive approach, but it may not be the most efficient for large nested lists. The list comprehension approach offers a concise solution, but it may be harder to understand for beginners. The stack-based approach provides a more iterative solution, which can be useful for handling large nested lists efficiently. Ultimately, the choice of the best option depends on the specific needs of the problem at hand.

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### 8 Responses

1. Oaklyn says:

Option 3: Using a Stack seems like the most interesting and challenging approach. Lets stack it up! 🥞

2. Gabriela Ochoa says:

Option 1: Recursive function FTW! Its like a never-ending maze of code, but damn, it works!

3. Arian says:

Option 3: Using a Stack sounds like a fun and challenging approach! Cant wait to give it a try!

1. Jeremy says:

I tried using a Stack once and it was a disaster. Its a convoluted mess that only adds unnecessary complexity. Stick to more straightforward methods, trust me.

4. Remi says:

Option 1 seems like a fun challenge, but List Comprehension in Option 2 is more concise. Thoughts?

5. Rylan says:

Option 1 sounds like a rabbit hole, option 2 seems concise, but option 3 has me intrigued.

6. Tiana Gonzalez says:

Option 3: Using a Stack? More like Option 3: Using a Snack! Whos with me? 🍿🙋‍♂️

7. Blair says:

Option 2 is like having a magic wand for nested list bijective functions! 👌🔮