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Reverse Nodes in Even Length Groups

Medium

Amazon

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Topics:

Linked Lists

You are given the head of a singly linked list. The nodes in the linked list are sequentially assigned to non-empty groups whose lengths form the sequence of the natural numbers (1, 2, 3, 4, ...). The length of a group is the number of nodes assigned to it. In other words:

The 1st node is assigned to the first group.
The 2nd and the 3rd nodes are assigned to the second group.
The 4th, 5th, and 6th nodes are assigned to the third group, and so on.

Note that the length of the last group may be less than or equal to 1 + the length of the second to last group.

Reverse the nodes in each group with an even length, and return the head of the modified linked list.

For example:

Input: head = [5,2,6,3,9,1,7,3,8,4] Output: [5,6,2,3,9,1,4,8,3,7]

Explain your approach, time complexity, and space complexity. Handle edge cases such as an empty list or a list with only one node.

Reverse Nodes in Even Length Groups

Medium

Amazon

2 views

Topics:

Linked Lists

The 1st node is assigned to the first group.
The 2nd and the 3rd nodes are assigned to the second group.
The 4th, 5th, and 6th nodes are assigned to the third group, and so on.

Note that the length of the last group may be less than or equal to 1 + the length of the second to last group.

Reverse the nodes in each group with an even length, and return the head of the modified linked list.

For example:

Input: head = [5,2,6,3,9,1,7,3,8,4] Output: [5,6,2,3,9,1,4,8,3,7]

Explain your approach, time complexity, and space complexity. Handle edge cases such as an empty list or a list with only one node.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the nodes in the linked list?
Can the linked list be empty, or contain only one node?
If the last group has an even length but fewer nodes than the intended group size based on the earlier groups, should I reverse only those nodes in the last group?
Should I modify the existing linked list in-place, or create a new one?
What should I return if the input linked list is null?

Brute Force Solution

Approach

We are given a linked list and need to reverse groups of nodes with even lengths. The brute force approach involves exhaustively trying all possible group sizes, checking if they are even, and reversing them if they are.

Here's how the algorithm would work step-by-step:

Start with a group size of 1, then a group size of 2, then 3, and so on until the end of the list.
For each group size, check if the number of nodes in the current group is even.
If the group size is even, reverse the order of the nodes in that group. If it is odd, leave it as is.
Move to the next group of nodes, repeating the process of checking the group size and reversing if necessary.
Continue until you have processed all the nodes in the linked list.
The final result is the modified linked list with even-length groups reversed.

Code Implementation

def reverse_even_length_groups_brute_force(head):
    if not head:
        return None

    dummy_node = ListNode(0)
    dummy_node.next = head
    previous_node = dummy_node
    group_size = 1

    while head:
        group_head = head
        group_tail = None
        group_length = 0

        # Determine group size and tail node
        for _ in range(group_size):
            if not head:
                break
            group_length += 1
            group_tail = head
            head = head.next

        # Check if the length of the group is even
        if group_length % 2 == 0:
            # Reverse the current group if the length is even
            current_node = group_head
            previous_node_in_group = None

            for _ in range(group_length):
                next_node = current_node.next
                current_node.next = previous_node_in_group
                previous_node_in_group = current_node
                current_node = next_node

            # Correct next pointer of previous node
            previous_node.next = previous_node_in_group

            # Correct next pointer of the group head
            group_head.next = head
            previous_node = group_head

        else:
            # Do nothing if the length of the group is odd
            previous_node.next = group_head
            previous_node = group_tail

        group_size += 1

    return dummy_node.next

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the linked list, grouping nodes with increasing group sizes (1, 2, 3, ...). For each potential group, the algorithm must traverse the list to determine its length, an operation proportional to the current group size. If a group's length is even, the algorithm reverses the nodes within that group, again an operation proportional to the group size. Since the group size increments up to a maximum value related to the total nodes, this results in nested operations and time complexity approximating O(n * (n/2)), which simplifies to O(n²).

Space Complexity

O(1) – The provided plain English solution iterates through the linked list, potentially reversing sections. The reversal process in place would not require extra data structures that scale with the input size N (number of nodes). No additional lists, hashmaps, or significant data structures are created, implying the space used is constant regardless of the linked list's length. Therefore, the space complexity is O(1).

Optimal Solution

Approach

The problem involves reversing groups of nodes in a linked list, but only if the group has an even number of nodes. The optimal approach is to iteratively process the linked list, identifying groups, checking their size, and reversing them in place when required, without using extra space proportional to the list's size.

Here's how the algorithm would work step-by-step:

Start at the beginning of the list.
Identify the next group of nodes to consider. The size of the group increases with each iteration (1, 2, 3, ...).
Count the number of nodes in this identified group.
Check if the count of nodes in the group is an even number.
If the count is even, reverse the order of nodes within that group directly in the list. This means changing which node points to which, without creating new nodes.
If the count is odd, leave the group as it is, maintaining the original order of nodes.
Move on to the next group of nodes in the list, increasing the group size, and repeat the process from step 2 until the end of the list is reached.
Return the head of the modified linked list.

Code Implementation

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def reverse_nodes_in_even_length_groups(head):
    dummy_node = ListNode(0)
    dummy_node.next = head
    previous_node = dummy_node
    group_length = 1

    while previous_node.next:
        # Identify the start of the current group
        current_node = previous_node.next
        tail_node = current_node
        group_size = 0

        # Count the nodes in the current group
        while tail_node and group_size < group_length:
            tail_node = tail_node.next
            group_size += 1

        # If group has even length, reverse it
        if group_size % 2 == 0:
            # These operations keep track of reversing the sublist
            next_group_start = tail_node
            current_node_for_reverse = previous_node.next
            previous_node_for_reverse = None

            for _ in range(group_size):
                temp_node = current_node_for_reverse.next
                current_node_for_reverse.next = previous_node_for_reverse
                previous_node_for_reverse = current_node_for_reverse
                current_node_for_reverse = temp_node

            # Stitch the reversed group back
            previous_node.next = previous_node_for_reverse
            current_node.next = next_group_start
            previous_node = current_node

        else:
            # Advance previousNode to the end of odd group
            for _ in range(group_size):
                previous_node = previous_node.next

        group_length += 1

    return dummy_node.next

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the linked list in groups, where the size of each group increases by one in each iteration. The sum of the group sizes will approximate the total number of nodes (n) in the list. Reversing each even-length group takes time proportional to the group's size, but this is still bounded by the overall iteration through the list. Because each node is visited and potentially reversed only once, the overall time complexity is O(n).

Space Complexity

O(1) – The algorithm operates directly on the linked list, modifying node pointers in place. It utilizes a few constant space variables such as pointers to track the current group and potentially a few variables for reversing the nodes within a group. The amount of extra memory required does not scale with the size N of the linked list, hence the space complexity is O(1).

Edge Cases

Case	How to Handle
Empty linked list	Return null immediately as there are no nodes to reverse.
Linked list with only one node	Return the original list as a group of size 1 is odd.
Linked list with two nodes (an even group of size 2)	Reverse the two nodes and return the new head.
Linked list with a length that is not a perfect triangular number (e.g., 1, 3, 6, 10...)	Handle remaining odd-length group at the end without reversing.
Linked list with a very large number of nodes (potential stack overflow with recursion)	Use an iterative approach to avoid stack overflow issues.
Linked list with all identical values	The reversal logic should still function correctly regardless of node values.
Memory constraints with extremely long linked list	Ensure we are not creating unnecessary copies of the list and operate in place where possible.
Reversal within a group leads to circular references	Carefully manage pointers during reversal to avoid circular linked list structures.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the nodes in the linked list?
Can the linked list be empty, or contain only one node?
If the last group has an even length but fewer nodes than the intended group size based on the earlier groups, should I reverse only those nodes in the last group?
Should I modify the existing linked list in-place, or create a new one?
What should I return if the input linked list is null?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start with a group size of 1, then a group size of 2, then 3, and so on until the end of the list.
For each group size, check if the number of nodes in the current group is even.
If the group size is even, reverse the order of the nodes in that group. If it is odd, leave it as is.
Move to the next group of nodes, repeating the process of checking the group size and reversing if necessary.
Continue until you have processed all the nodes in the linked list.
The final result is the modified linked list with even-length groups reversed.

Code Implementation

def reverse_even_length_groups_brute_force(head):
    if not head:
        return None

    dummy_node = ListNode(0)
    dummy_node.next = head
    previous_node = dummy_node
    group_size = 1

    while head:
        group_head = head
        group_tail = None
        group_length = 0

        # Determine group size and tail node
        for _ in range(group_size):
            if not head:
                break
            group_length += 1
            group_tail = head
            head = head.next

        # Check if the length of the group is even
        if group_length % 2 == 0:
            # Reverse the current group if the length is even
            current_node = group_head
            previous_node_in_group = None

            for _ in range(group_length):
                next_node = current_node.next
                current_node.next = previous_node_in_group
                previous_node_in_group = current_node
                current_node = next_node

            # Correct next pointer of previous node
            previous_node.next = previous_node_in_group

            # Correct next pointer of the group head
            group_head.next = head
            previous_node = group_head

        else:
            # Do nothing if the length of the group is odd
            previous_node.next = group_head
            previous_node = group_tail

        group_size += 1

    return dummy_node.next

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start at the beginning of the list.
Identify the next group of nodes to consider. The size of the group increases with each iteration (1, 2, 3, ...).
Count the number of nodes in this identified group.
Check if the count of nodes in the group is an even number.
If the count is even, reverse the order of nodes within that group directly in the list. This means changing which node points to which, without creating new nodes.
If the count is odd, leave the group as it is, maintaining the original order of nodes.
Move on to the next group of nodes in the list, increasing the group size, and repeat the process from step 2 until the end of the list is reached.
Return the head of the modified linked list.

Code Implementation

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def reverse_nodes_in_even_length_groups(head):
    dummy_node = ListNode(0)
    dummy_node.next = head
    previous_node = dummy_node
    group_length = 1

    while previous_node.next:
        # Identify the start of the current group
        current_node = previous_node.next
        tail_node = current_node
        group_size = 0

        # Count the nodes in the current group
        while tail_node and group_size < group_length:
            tail_node = tail_node.next
            group_size += 1

        # If group has even length, reverse it
        if group_size % 2 == 0:
            # These operations keep track of reversing the sublist
            next_group_start = tail_node
            current_node_for_reverse = previous_node.next
            previous_node_for_reverse = None

            for _ in range(group_size):
                temp_node = current_node_for_reverse.next
                current_node_for_reverse.next = previous_node_for_reverse
                previous_node_for_reverse = current_node_for_reverse
                current_node_for_reverse = temp_node

            # Stitch the reversed group back
            previous_node.next = previous_node_for_reverse
            current_node.next = next_group_start
            previous_node = current_node

        else:
            # Advance previousNode to the end of odd group
            for _ in range(group_size):
                previous_node = previous_node.next

        group_length += 1

    return dummy_node.next

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty linked list	Return null immediately as there are no nodes to reverse.
Linked list with only one node	Return the original list as a group of size 1 is odd.
Linked list with two nodes (an even group of size 2)	Reverse the two nodes and return the new head.
Linked list with a length that is not a perfect triangular number (e.g., 1, 3, 6, 10...)	Handle remaining odd-length group at the end without reversing.
Linked list with a very large number of nodes (potential stack overflow with recursion)	Use an iterative approach to avoid stack overflow issues.
Linked list with all identical values	The reversal logic should still function correctly regardless of node values.
Memory constraints with extremely long linked list	Ensure we are not creating unnecessary copies of the list and operate in place where possible.
Reversal within a group leads to circular references	Carefully manage pointers during reversal to avoid circular linked list structures.