Transform linked list to sorted array

Question

Convert the given array to tree structure following the specified rules.

Codemia · Accepted Answer

To transform a linked list into a sorted array, we need to recognize that the problem involves extracting elements from a linear data structure (linked list) and organizing them into a different structure (array) while ensuring they are sorted. The most applicable pattern here is the straightforward traversal of the linked list to gather elements, followed by sorting the resultant array. Since linked lists do not allow random access, we cannot directly index into them; therefore, we will traverse the list to collect elements sequentially.

1. **Initialization**: Start by creating an empty list to hold the values extracted from the linked list.
2. **Traversal**: Use a pointer to traverse the linked list from the head until we reach the end (null).
3. **Collecting Values**: During traversal, append each node's value to the list.
4. **Sorting**: After collecting all values in the list, use Python's built-in sorting functionality (which is based on Timsort, O(n log n)).
5. **Return the Result**: Finally, return the sorted list.

**Example**: Given a linked list with values 4 -> 2 -> 1 -> 3, we would first collect [4, 2, 1, 3] and then sort it to get [1, 2, 3, 4].

```python
class ListNode:
    def __init__(self, value=0, next=None):
        self.value = value
        self.next = next

def linked_list_to_sorted_array(head):
    values = []
    current = head
    while current is not None:
        values.append(current.value)
        current = current.next
    values.sort()
    return values
```

The time complexity of this approach is O(n log n), where n is the number of elements in the linked list. This is due to the sorting step after collecting the values. The space complexity is O(n) as well, because we are storing all the values in a separate list before sorting them. This analysis holds under the assumption that the linked list can be traversed in O(n) time.

1. **Empty List**: If the linked list is empty (head is None), we should return an empty array `[]`.
2. **Single Element**: If the list contains only one node, the output should simply be an array with that one element.
3. **Already Sorted List**: If the linked list is already sorted, the output should remain the same after sorting.
4. **Duplicates**: The algorithm should correctly handle linked lists with duplicate values, ensuring that duplicates are preserved in the sorted output.

Transform linked list to sorted array

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