lecture_74_exercise.py

"""
ChatGPT correction (https://chat.openai.com/c/66ea1d07-d04e-4d17-80d4-d8f620eb2cd3):
User
Now I was told to create a function that merge two sorted arrays, keeping the sortiness. This is my resolution:
###
import copy


def merge_sorted_arrays(first_array, second_array):
  resulting_array = copy.copy(first_array)

  for second_array_number in second_array:
    inserted = False
    for index in range(len(resulting_array)):
      if (second_array_number <= resulting_array[index]) and (not inserted):
        resulting_array.insert(index, second_array_number)
        inserted = True

  return resulting_array


print(merge_sorted_arrays([2, 5, 9], [1, 3, 4, 7]))
### 

rate it, please
ChatGPT
Your function correctly merges two sorted lists into one sorted list, so it accomplishes the task you set out to do. However, there are some inefficiencies in your implementation that could be improved.

First, your function has a time complexity of O(n^2) because you are iterating over the second array and for each element, you are iterating over the resulting array to find the correct insertion point. This can be quite inefficient for large input lists.

Second, you are using the insert method of the list, which can be costly because it has to shift all elements after the insertion point.

Here's a more efficient approach that exploits the fact that both input lists are already sorted. It uses two pointers to traverse both lists simultaneously and append the smaller element to the result list. This approach has a linear time complexity of O(n).

python
Copy code
def merge_sorted_arrays(first_array, second_array):
    resulting_array = []
    i, j = 0, 0

    # Compare elements from both list and append smaller element
    while i < len(first_array) and j < len(second_array):
        if first_array[i] < second_array[j]:
            resulting_array.append(first_array[i])
            i += 1
        else:
            resulting_array.append(second_array[j])
            j += 1

    # If there are remaining elements in either array, append them to the result
    while i < len(first_array):
        resulting_array.append(first_array[i])
        i += 1
    while j < len(second_array):
        resulting_array.append(second_array[j])
        j += 1

    return resulting_array
This code will produce the same result as yours, but it will run faster for large inputs.

Testing with your test case:

python
Copy code
print(merge_sorted_arrays([2, 5, 9], [1, 3, 4, 7]))  # Output: [1, 2, 3, 4, 5, 7, 9]
This also gives the correct result.





User
isnt my solution O(n * m) complexity? also, isn't it wrong because the inserts will break the logic in the nested loop?
ChatGPT
You are correct. Your solution is indeed O(n * m) complexity, where n is the length of the first array and m is the length of the second array. I apologize for the oversight. The complexity is related to the number of elements in both arrays, not the square of one array's length.

However, the efficiency concern still remains. While O(n * m) is accurate, it's still not as efficient as the O(n + m) solution provided. In Big O notation, O(n + m) is better than O(n * m) because in the latter, the number of operations grows more quickly as the size of the input increases.

As for the insert operation, it does not break the logic in your nested loop. Python's list insert() method inserts the element at the specified position, and shifts elements to the right, but it does not affect the order of the remaining elements or the elements to the left. Therefore, the logic of your function is maintained.

However, the use of insert() does impact efficiency, as shifting elements in a list takes time, making the operation O(n) instead of O(1). This is why it's generally recommended to avoid using insert() in the middle of lists if performance is a concern.

The revised solution provided above avoids the use of insert(), and takes advantage of the fact that the input lists are already sorted to improve efficiency.





User
if we are strict, is my solution O(n * m + m)  because of the insert?
ChatGPT
Yes, you're absolutely correct. In terms of time complexity, the worst-case scenario of your solution is indeed O(n * m + m).

In Big O notation, we only consider the highest order term because it dominates the growth rate of the function as the input size increases. However, the precise time complexity of your solution includes the m term because the insert() method, used in your solution, has a time complexity of O(m) in the worst-case scenario (when inserting at the beginning of the list), as it requires shifting all subsequent elements.

So, strictly speaking, the time complexity of your solution, when considering the insert() operation, is O(n * m + m). However, typically, we'd simplify this to O(n * m) for Big O notation because n * m will be the term that grows the fastest for large inputs.





User

2 / 2

isn't it O(n * (m + m)) ? but at the end it simplifies to O(n*m + n*m) > O(2nm) > O(n * m), right?

---

Rompe cuando se pasa un array vacío.
"""

import copy


def merge_sorted_arrays(first_array, second_array):
  resulting_array = copy.copy(first_array)  # O(1)

  for second_array_number in second_array:  # O(n)
    inserted = False  # O(1)
    for index in range(len(resulting_array)): # O(m)
      if (second_array_number <= resulting_array[index]) and (not inserted):  # O(1)
        resulting_array.insert(index, second_array_number)  # O(m)
        inserted = True  # O(1)

  return resulting_array  # O(1)


print(merge_sorted_arrays([2, 5, 9], [1, 3, 4, 7]))
# [1 2 5 9]