Jobs

Interviews

Discussions

Reverse Nodes in k-Group

Hard

Reverse Nodes in k-Group

Hard

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 `k` and the values within the linked list nodes themselves? Are negative values possible?
What should I return if the input `head` is null or if the linked list is empty?
If the number of nodes is less than `k`, should I reverse the entire list or leave it unchanged?
Is `k` guaranteed to be a positive integer greater than 0?
Are there any memory constraints I should be aware of when modifying the linked list in-place?

Brute Force Solution

Approach

Imagine you have a train of linked train cars, and you want to reverse small groups of them. The most straightforward way is to just try reversing every possible group of the right size, one at a time. We check each possible reversed configuration to see if it's the one we want.

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

Start at the beginning of the train of cars.
Look at the first 'k' cars as a group.
Reverse the order of these 'k' cars.
Now, look at the next 'k' cars after the reversed group.
Reverse the order of this new group of 'k' cars.
Keep doing this for all the groups of 'k' cars you can find in the train.
If you get to the end and don't have enough cars for a full group of 'k', leave those cars alone.
That's your reversed train!

Code Implementation

def reverse_nodes_in_k_group_brute_force(head, k):
    if not head:
        return None

    current_node = head
    group_head = head
    previous_group_tail = None
    
    while current_node:
        # Check if there are at least k nodes left
        test_node = current_node
        count = 0

        while test_node and count < k:
            test_node = test_node.next
            count += 1

        if count < k:
            break

        # Reverse the k-group
        previous_node = None
        current_group_node = current_node
        next_node = None
        group_count = 0

        while current_group_node and group_count < k:
            next_node = current_group_node.next
            current_group_node.next = previous_node
            previous_node = current_group_node
            current_group_node = next_node
            group_count += 1

        # Adjust head pointer if it's the first group
        if current_node == head:
            head = previous_node

        # Connect the reversed group with the previous group
        if previous_group_tail:
            previous_group_tail.next = previous_node

        # Update previous group tail and current node
        previous_group_tail = current_node
        current_node.next = current_group_node
        current_node = current_group_node

    return head

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the linked list, processing it in chunks of size k. Reversing a group of k nodes takes O(k) time. Since we iterate through the entire list of n nodes (or until there are fewer than k nodes remaining), and each node is part of at most one reversed group, the total time complexity is proportional to n. Therefore, the overall time complexity is O(n).

Space Complexity

O(1) – The algorithm operates in place by reversing groups of k nodes within the linked list. It uses a few constant extra variables, such as pointers to keep track of the current, previous, and next nodes during the reversal process. The space used does not depend on the length of the linked list (N), therefore the auxiliary space complexity is constant.

Optimal Solution

Approach

The core idea is to reverse small chunks of the linked list at a time. We identify groups of 'k' nodes, reverse them, and then connect them back into the original list. This avoids unnecessary back-and-forth movements and keeps things organized.

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

First, check if there are at least 'k' nodes remaining in the list. If not, don't reverse anything and just leave it as is.
If there are enough nodes, isolate the first 'k' nodes as a group to be reversed.
Reverse the order of the nodes within that isolated group.
Connect the reversed group to the node that originally came *before* the group.
Connect the end of the reversed group to the node that originally came *after* the group. Now the reversed group is correctly linked into the list.
Move on to the next group of 'k' nodes in the list and repeat the process from step 1 until you reach the end of the list.
By processing 'k' nodes at a time and carefully reconnecting them, we efficiently reverse the entire list in segments.

Code Implementation

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

def reverse_k_group(head, k):
    def reverse_list(head, tail):
        previous_node = None
        current_node = head

        while current_node != tail:
            next_node = current_node.next
            current_node.next = previous_node
            previous_node = current_node
            current_node = next_node

        return previous_node

    if not head or k == 1:
        return head

    dummy_node = ListNode(0)
    dummy_node.next = head
    group_previous = dummy_node

    while True:
        current_node = group_previous.next
        tail_node = group_previous

        # Check if there are at least k nodes remaining
        for _ in range(k):
            tail_node = tail_node.next
            if not tail_node:
                return dummy_node.next

        group_next = tail_node.next

        # Isolate the k-group by setting the tail's next pointer to None
        head_of_next_group = current_node
        tail_node.next = None

        # Reverse the k-group
        new_head = reverse_list(head_of_next_group, group_next)

        # Connect the reversed k-group to the previous group
        group_previous.next = new_head

        # Connect the tail of the reversed group to the next group
        head_of_next_group.next = group_next

        # Update group_previous to point to the tail of the reversed group
        group_previous = head_of_next_group

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the linked list, processing it in groups of k nodes. Reversing each group of k nodes takes O(k) time, but since each node is only reversed once and the groups are processed sequentially, the total time complexity is determined by the number of nodes in the linked list (n). The outer loop effectively visits each node once, and the reversal within the k-group is proportional to the number of nodes visited, so the time complexity is O(n) because k is a fixed constant with respect to n.

Space Complexity

O(1) – The algorithm primarily manipulates pointers within the existing linked list structure. It uses a few constant space variables to keep track of current, previous, and next nodes during the reversal process within each k-group. No additional data structures like arrays or hash maps are created that scale with the input size N (number of nodes in the linked list). Therefore, the auxiliary space used is constant and independent of the input size.

Edge Cases

Case	How to Handle
Empty list (head is null)	Return null immediately as there are no nodes to reverse.
k is 1	Return the original list as reversing groups of size 1 has no effect.
k is greater than the list length	Return the original list as there aren't enough nodes to form a group of size k.
List length is a multiple of k	The entire list is reversed in groups of k, ensuring all nodes are processed.
List length is not a multiple of k	The last group of nodes (less than k) should remain in their original order, and the algorithm should correctly link the reversed portion with the remaining unreversed nodes.
List with only one node	Return the original list immediately as there are no nodes to reverse.
k is a large number that can potentially cause integer overflow if not handled carefully during calculations	Use appropriate data types (e.g., long) to prevent overflow when manipulating k or related values if needed.
k is zero or negative	Throw an exception or return the original list, depending on the desired behavior for invalid input.

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 `k` and the values within the linked list nodes themselves? Are negative values possible?
What should I return if the input `head` is null or if the linked list is empty?
If the number of nodes is less than `k`, should I reverse the entire list or leave it unchanged?
Is `k` guaranteed to be a positive integer greater than 0?
Are there any memory constraints I should be aware of when modifying the linked list in-place?

Brute Force Solution

Approach

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

Start at the beginning of the train of cars.
Look at the first 'k' cars as a group.
Reverse the order of these 'k' cars.
Now, look at the next 'k' cars after the reversed group.
Reverse the order of this new group of 'k' cars.
Keep doing this for all the groups of 'k' cars you can find in the train.
If you get to the end and don't have enough cars for a full group of 'k', leave those cars alone.
That's your reversed train!

Code Implementation

def reverse_nodes_in_k_group_brute_force(head, k):
    if not head:
        return None

    current_node = head
    group_head = head
    previous_group_tail = None
    
    while current_node:
        # Check if there are at least k nodes left
        test_node = current_node
        count = 0

        while test_node and count < k:
            test_node = test_node.next
            count += 1

        if count < k:
            break

        # Reverse the k-group
        previous_node = None
        current_group_node = current_node
        next_node = None
        group_count = 0

        while current_group_node and group_count < k:
            next_node = current_group_node.next
            current_group_node.next = previous_node
            previous_node = current_group_node
            current_group_node = next_node
            group_count += 1

        # Adjust head pointer if it's the first group
        if current_node == head:
            head = previous_node

        # Connect the reversed group with the previous group
        if previous_group_tail:
            previous_group_tail.next = previous_node

        # Update previous group tail and current node
        previous_group_tail = current_node
        current_node.next = current_group_node
        current_node = current_group_node

    return head

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, check if there are at least 'k' nodes remaining in the list. If not, don't reverse anything and just leave it as is.
If there are enough nodes, isolate the first 'k' nodes as a group to be reversed.
Reverse the order of the nodes within that isolated group.
Connect the reversed group to the node that originally came *before* the group.
Connect the end of the reversed group to the node that originally came *after* the group. Now the reversed group is correctly linked into the list.
Move on to the next group of 'k' nodes in the list and repeat the process from step 1 until you reach the end of the list.
By processing 'k' nodes at a time and carefully reconnecting them, we efficiently reverse the entire list in segments.

Code Implementation

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

def reverse_k_group(head, k):
    def reverse_list(head, tail):
        previous_node = None
        current_node = head

        while current_node != tail:
            next_node = current_node.next
            current_node.next = previous_node
            previous_node = current_node
            current_node = next_node

        return previous_node

    if not head or k == 1:
        return head

    dummy_node = ListNode(0)
    dummy_node.next = head
    group_previous = dummy_node

    while True:
        current_node = group_previous.next
        tail_node = group_previous

        # Check if there are at least k nodes remaining
        for _ in range(k):
            tail_node = tail_node.next
            if not tail_node:
                return dummy_node.next

        group_next = tail_node.next

        # Isolate the k-group by setting the tail's next pointer to None
        head_of_next_group = current_node
        tail_node.next = None

        # Reverse the k-group
        new_head = reverse_list(head_of_next_group, group_next)

        # Connect the reversed k-group to the previous group
        group_previous.next = new_head

        # Connect the tail of the reversed group to the next group
        head_of_next_group.next = group_next

        # Update group_previous to point to the tail of the reversed group
        group_previous = head_of_next_group

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty list (head is null)	Return null immediately as there are no nodes to reverse.
k is 1	Return the original list as reversing groups of size 1 has no effect.
k is greater than the list length	Return the original list as there aren't enough nodes to form a group of size k.
List length is a multiple of k	The entire list is reversed in groups of k, ensuring all nodes are processed.
List length is not a multiple of k	The last group of nodes (less than k) should remain in their original order, and the algorithm should correctly link the reversed portion with the remaining unreversed nodes.
List with only one node	Return the original list immediately as there are no nodes to reverse.
k is a large number that can potentially cause integer overflow if not handled carefully during calculations	Use appropriate data types (e.g., long) to prevent overflow when manipulating k or related values if needed.
k is zero or negative	Throw an exception or return the original list, depending on the desired behavior for invalid input.