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Head of Career Skills Development & Coaching

*Based on past data of successful IK students

Given a linked list, reverse every group of `k`

nodes. If there is a remainder (a group of less than `k`

nodes) in the end, reverse that last group, too.

```
{
"head": [1, 2, 3, 4, 5, 6],
"k": 3
}
```

Output:

```
[3, 2, 1, 6, 5, 4]
```

Input list consists of two whole groups of three. In the output list the first three and last three nodes are reversed.

```
{
"head": [1, 2, 3, 4, 5, 6, 7, 8],
"k": 3
}
```

Output:

```
[3, 2, 1, 6, 5, 4, 8, 7]
```

There are two whole groups of three and one partial group (a remainder that consists of just two nodes). Each of the three groups is reversed in the output.

- The function has two parameters: head of the given linked list and
`k`

. - Return the head of the linked list after reversing the groups of nodes in it.

Constraints:

- 1 <= number of nodes in the list <= 100000
- -2 * 10
^{9}<= node value <= 2 * 10^{9} - 1 <=
`k`

<= number of nodes - Cannot use more than constant extra space

We have provided one solution for this problem.

The "constant extra space" constraint rules out using recursion because recursion uses the call stack space (up to O(n) if `k = n`

).

We are forced to use an iterative approach.

This solution demonstrates how that can be implemented.

O(n).

O(1).

O(n).

```
/*
Asymptotic complexity in terms of length of given linked list \`n\`:
* Time: O(n).
* Auxiliary space: O(1).
* Total space: O(n).
*/
// Helper function: Reverse singly linked list in O(length) time and O(1) extra space.
void reverse_linked_list(LinkedListNode *cur)
{
LinkedListNode *prev = NULL;
LinkedListNode *next;
while (cur)
{
next = cur->next;
cur->next = prev;
prev = cur;
cur = next;
}
}
LinkedListNode *reverse_linked_list_in_groups_of_k(LinkedListNode *head, int k)
{
/*
Input:
list: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> NULL
k: 3
Output:
3 -> 2 -> 1 -> 6 -> 5 -> 4 -> 8 -> 7-> NULL
Groups to be reversed are (1 -> 2 -> 3), (4 -> 5 -> 6) and (7 -> 8).
We will call reverse_linked_list function when (start = 1 and stop = 3),
(start = 4 and stop = 6) and (start = 7 and stop = 8).
*/
// Points to previous start node.
LinkedListNode *prev_of_start = NULL;
LinkedListNode *start = head;
LinkedListNode *stop = head;
int count = 0;
// Traverse whole linked list.
while (stop)
{
count++;
/*
If we have covered k nodes in between start
and stop (inclusive) or we are at the last node.
*/
if (count == k || stop->next == NULL)
{
// Points to next node of start.
LinkedListNode *next_of_stop = stop->next;
/*
We want to reverse start to stop nodes,
set stop->next = NULL so we know where to stop.
*/
stop->next = NULL;
// Reverse start to stop nodes.
reverse_linked_list(start);
if (prev_of_start == NULL)
{
// Head will change when we are reversing the linked list first time.
head = stop;
}
else
{
/*
We have reversed start to stop nodes, hence now stop will be next node of
prev_of_start.
*/
prev_of_start->next = stop;
}
/*
We have reversed start to stop nodes, hence next_of_stop will be next node of start.
*/
start->next = next_of_stop;
/*
In the above example, after we have reversed first k nodes list will be:
3 -> 2 -> 1 -> 4 -> 5 -> 6 -> 7 -> 8 -> NULL,
start will point to 1, next_of_stop will point to 4.
Now we will set start and stop to point at 4.
And prev_of_start should be previous of 4 that is 1.
*/
prev_of_start = start;
start = next_of_stop;
stop = next_of_stop;
// Reset counter.
count = 0;
}
else
{
stop = stop->next;
}
}
return head;
}
```

We hope that these solutions to reverse a linked list in groups of k problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

If you are preparing for a tech interview at FAANG or any other Tier-1 tech company, register for Interview Kickstart's FREE webinar to understand the best way to prepare.

Interview Kickstart offers interview preparation courses taught by FAANG+ tech leads and seasoned hiring managers. Our programs include a comprehensive curriculum, unmatched teaching methods, and career coaching to help you nail your next tech interview.

We offer 18 interview preparation courses, each tailored to a specific engineering domain or role, including the most in-demand and highest-paying domains and roles, such as:

- Back-end Engineering Interview Course
- Front-end Engineering Interview Course
- Full Stack Developer Interview Course

To learn more, register for the FREE webinar.

Note: Input and Output will already be taken care of.

Given a linked list, reverse every group of `k`

nodes. If there is a remainder (a group of less than `k`

nodes) in the end, reverse that last group, too.

```
{
"head": [1, 2, 3, 4, 5, 6],
"k": 3
}
```

Output:

```
[3, 2, 1, 6, 5, 4]
```

Input list consists of two whole groups of three. In the output list the first three and last three nodes are reversed.

```
{
"head": [1, 2, 3, 4, 5, 6, 7, 8],
"k": 3
}
```

Output:

```
[3, 2, 1, 6, 5, 4, 8, 7]
```

There are two whole groups of three and one partial group (a remainder that consists of just two nodes). Each of the three groups is reversed in the output.

- The function has two parameters: head of the given linked list and
`k`

. - Return the head of the linked list after reversing the groups of nodes in it.

Constraints:

- 1 <= number of nodes in the list <= 100000
- -2 * 10
^{9}<= node value <= 2 * 10^{9} - 1 <=
`k`

<= number of nodes - Cannot use more than constant extra space

We have provided one solution for this problem.

The "constant extra space" constraint rules out using recursion because recursion uses the call stack space (up to O(n) if `k = n`

).

We are forced to use an iterative approach.

This solution demonstrates how that can be implemented.

O(n).

O(1).

O(n).

```
/*
Asymptotic complexity in terms of length of given linked list \`n\`:
* Time: O(n).
* Auxiliary space: O(1).
* Total space: O(n).
*/
// Helper function: Reverse singly linked list in O(length) time and O(1) extra space.
void reverse_linked_list(LinkedListNode *cur)
{
LinkedListNode *prev = NULL;
LinkedListNode *next;
while (cur)
{
next = cur->next;
cur->next = prev;
prev = cur;
cur = next;
}
}
LinkedListNode *reverse_linked_list_in_groups_of_k(LinkedListNode *head, int k)
{
/*
Input:
list: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> NULL
k: 3
Output:
3 -> 2 -> 1 -> 6 -> 5 -> 4 -> 8 -> 7-> NULL
Groups to be reversed are (1 -> 2 -> 3), (4 -> 5 -> 6) and (7 -> 8).
We will call reverse_linked_list function when (start = 1 and stop = 3),
(start = 4 and stop = 6) and (start = 7 and stop = 8).
*/
// Points to previous start node.
LinkedListNode *prev_of_start = NULL;
LinkedListNode *start = head;
LinkedListNode *stop = head;
int count = 0;
// Traverse whole linked list.
while (stop)
{
count++;
/*
If we have covered k nodes in between start
and stop (inclusive) or we are at the last node.
*/
if (count == k || stop->next == NULL)
{
// Points to next node of start.
LinkedListNode *next_of_stop = stop->next;
/*
We want to reverse start to stop nodes,
set stop->next = NULL so we know where to stop.
*/
stop->next = NULL;
// Reverse start to stop nodes.
reverse_linked_list(start);
if (prev_of_start == NULL)
{
// Head will change when we are reversing the linked list first time.
head = stop;
}
else
{
/*
We have reversed start to stop nodes, hence now stop will be next node of
prev_of_start.
*/
prev_of_start->next = stop;
}
/*
We have reversed start to stop nodes, hence next_of_stop will be next node of start.
*/
start->next = next_of_stop;
/*
In the above example, after we have reversed first k nodes list will be:
3 -> 2 -> 1 -> 4 -> 5 -> 6 -> 7 -> 8 -> NULL,
start will point to 1, next_of_stop will point to 4.
Now we will set start and stop to point at 4.
And prev_of_start should be previous of 4 that is 1.
*/
prev_of_start = start;
start = next_of_stop;
stop = next_of_stop;
// Reset counter.
count = 0;
}
else
{
stop = stop->next;
}
}
return head;
}
```

We hope that these solutions to reverse a linked list in groups of k problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

If you are preparing for a tech interview at FAANG or any other Tier-1 tech company, register for Interview Kickstart's FREE webinar to understand the best way to prepare.

Interview Kickstart offers interview preparation courses taught by FAANG+ tech leads and seasoned hiring managers. Our programs include a comprehensive curriculum, unmatched teaching methods, and career coaching to help you nail your next tech interview.

We offer 18 interview preparation courses, each tailored to a specific engineering domain or role, including the most in-demand and highest-paying domains and roles, such as:

- Back-end Engineering Interview Course
- Front-end Engineering Interview Course
- Full Stack Developer Interview Course

To learn more, register for the FREE webinar.

**Attend our free webinar to amp up your career and get the salary you deserve.**

Hosted By

Ryan Valles

Founder, Interview Kickstart

- Designed by 500 FAANG+ experts
- Live training and mock interviews
- 17000+ tech professionals trained

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