Delete Nodes From Linked List Present in Array - Leetcode 3217 - Python
Summary
TLDRIn this coding tutorial, the focus is on solving the problem of deleting nodes from a linked list that are present in an array. The video demonstrates the use of a hash set for efficient lookup, which reduces the time complexity to O(n+m). A dummy node is introduced to simplify the removal process, and a two-pointer approach is explained to iterate through the list and remove the nodes. The tutorial also covers a one-pointer alternative for deletion. The video concludes with a live coding session, showcasing the efficiency of the solution.
Takeaways
- 📝 The video tutorial focuses on solving the problem of deleting nodes from a linked list based on values present in an array.
- 🔍 The presenter suggests using a hash set for constant time lookups to determine if a node's value exists in the given array.
- 🔑 The idea is to iterate through each node of the linked list and remove it if its value is found in the hash set.
- 🔁 The tutorial introduces the concept of a 'dummy node' to simplify the process of node deletion in a linked list.
- 💡 The presenter explains that with a dummy node, you can handle edge cases like deleting the head of the list more easily.
- 🔭 The video demonstrates a two-pointer approach to traverse and modify the linked list, keeping track of the current and previous nodes.
- ⏱️ The time complexity of the algorithm is discussed as O(n + m), where n is the length of the linked list and m is the size of the array.
- 💾 Space complexity is noted as O(n) due to the additional hash set data structure used for lookups.
- 👨💻 The presenter provides a step-by-step coding example in Python, emphasizing the importance of not skipping nodes during deletion.
- 🔄 An alternative single-pointer approach is also discussed, showing adaptability in solving the problem with different methods.
- 📢 The tutorial ends with a call to action for viewers to subscribe for more content on data structures, system design, and related topics.
Q & A
What problem is the video tutorial addressing?
-The video tutorial is addressing the problem of deleting nodes from a linked list that are present in a given array.
What is the expected outcome if all nodes in the linked list are removed?
-If all nodes in the linked list are removed, the head of the remaining list could be null, indicating an empty list.
Why does the tutorial suggest using a hash set for lookup?
-The tutorial suggests using a hash set for lookup because it provides constant time complexity for lookups, making the process more efficient compared to linear scans.
What is the role of a dummy node in linked list problems?
-A dummy node simplifies the process of modifying a linked list, especially when dealing with edge cases such as removing the head of the list or nodes from the beginning.
How does the tutorial propose to remove a node from the linked list?
-The tutorial proposes using a two-pointer approach where one pointer keeps track of the previous node and the other of the current node, adjusting the pointers to effectively remove the current node if its value is in the hash set.
What is the time complexity of the algorithm described in the tutorial?
-The time complexity of the algorithm is O(n + m), where n is the number of nodes in the linked list and m is the size of the array.
What is the space complexity of the algorithm?
-The space complexity of the algorithm is O(n), which is due to the additional space used by the hash set to store the values from the array.
Why does the tutorial mention not shifting the previous pointer when a node is deleted?
-Shifting the previous pointer when a node is deleted could lead to skipping a node in the list, which is incorrect behavior for the problem at hand.
What alternative method does the tutorial suggest for solving the problem?
-The tutorial suggests an alternative method using only one pointer, which simplifies the code by always moving the pointer one step ahead.
How does the tutorial ensure that the original order of the nodes is maintained after deletions?
-The tutorial ensures that the original order is maintained by only adjusting the pointers to skip over the nodes that are to be deleted, without changing the order of the remaining nodes.
What additional resources does the tutorial mention for further learning?
-The tutorial mentions a free newsletter and upcoming system design videos as additional resources for further learning on data structures, system design, and more.
Outlines
👨💻 Solving the Delete Nodes from Linked List Problem
This paragraph introduces a coding challenge to delete nodes from a linked list based on values present in an array. The speaker proposes using a hash set for efficient lookups to determine if a node's value should be removed. The approach involves iterating through the linked list and removing nodes whose values exist in the hash set. A dummy node is suggested to simplify the removal process, especially for nodes at the beginning of the list. The speaker explains the use of two pointers, one to keep track of the previous node and another for the current node, to efficiently delete nodes without skipping any. The time complexity is discussed as O(n+m), where n is the length of the linked list and m is the size of the array, and the space complexity is O(n) due to the hash set.
🔍 Optimizing with a Single Pointer Technique
The second paragraph explores an alternative solution to the linked list node deletion problem using only a single pointer. The speaker refactors the code to eliminate the need for a 'current' pointer, relying solely on a 'previous' pointer to manage node deletions. The approach involves checking if the next node in line should be deleted and, if so, adjusting the 'previous' pointer to skip over it. If not, the 'previous' pointer is simply moved forward. The speaker runs the optimized code to demonstrate its effectiveness and mentions that, while this single-pointer technique might not be more efficient, it is a valid alternative. The paragraph concludes with a teaser for upcoming content, including a free newsletter and system design videos.
Mindmap
Keywords
💡Linked List
💡Array
💡Hash Set
💡Dummy Node
💡Two-pointer Technique
💡Time Complexity
💡Space Complexity
💡Node Removal
💡Constant Time
💡Algorithm
💡Linear Scan
Highlights
Introduction to solving the problem of deleting nodes from a linked list based on values in an array.
Explanation of the need to remove nodes from a linked list if their values exist in a given array.
Discussion on the potential outcomes of the linked list after node deletion, including it being empty or null.
Proposing the use of a hash set for efficient lookup to determine if a node's value exists in the array.
Algorithm outline for iterating through the linked list and removing nodes based on the hash set lookup.
Introducing the concept of a dummy node to simplify linked list modifications.
Demonstration of how to handle edge cases when removing nodes from the beginning of the linked list.
Trick for using two pointers to keep track of the previous and current nodes for deletion.
Explanation of why not to shift the previous pointer when a node is deleted to avoid skipping nodes.
Code implementation of the algorithm using a hash set and two-pointer technique.
Time complexity analysis of the algorithm, considering the size of the linked list and array.
Space complexity discussion, noting the use of a hash set and its impact on space usage.
Alternative solution using only one pointer for node deletion, simplifying the code.
Final code review and demonstration of the solution's efficiency.
Announcement of upcoming educational resources, including a free newsletter and system design videos.
Transcripts
hey everyone welcome back and let's
write some more neat code today so today
I'll solve the problem delete nodes from
linked list present in an array so I'll
keep this one nice and short the idea is
we're given a linked list that might
look like this we're also given an array
of numbers if any of these nodes has a
value that exists within that array we
want to remove that node from the linked
list and so in this case we would remove
all three of these and then return the
remaining linked list it could could be
empty I mean it could technically be
null it could be non-empty and if it's
non-empty of course we want to return
the original notes we want to return the
head of the remaining list how do we go
about solving this problem well the
first thing that you need to do is how
do we do the lookup how do we know if
this value exists in there we can do a
linear scan we can go through that
entire thing either by looping over it
or using a built-in function but since
it's an array it's going to be linear
time lookups though are quicker when
hash data structures whether it's a hash
map or in this case a hash set is going
to be sufficient so that's exactly what
we're going to do the lookup in that
case will go from oven to constant time
so the algorithm is going to be this go
through each node if the value exists in
the hash set we will convert this into a
hash set then remove the node okay so
now you might be wondering how do we go
about removing a node well the first
thing I want to tell you is that there
is a trick with link list problems you
almost always want to have a dummy node
when you're modifying a linked list it
makes the code just a little bit easier
let me show you what I mean like this
example here if we were updating this
link list we'd say that okay well right
now we have a pointer this is the head
of the link list and we are now trying
to delete it okay so since this is the
head we should probably change the head
pointer to now be at the next node but
if we were removing a node from the
middle suppose this one down here like
this is our head right now we're not
removing the head we're removing the
node after the head so in that case we
just want to take this pointer and then
shift it over there so we can remove the
difference between both of these by
doing something called a dummy node The
Edge case comes from the fact that we
could remove from the beginning of the
list so let's just create a random node
doesn't really matter what the value is
I'll say it's zero and let's set the
next pointer to the head of the existing
link list now we're going to do the
exact same thing as before this is a one
so let's remove it how do we do that
well one possible way is to use two
pointers keep track of the previous node
and keep track of the current node and
if the current node is something we want
to delete then take the previous next
pointer and set it to current. next
that's pretty much going to remove it
and then in this particular case the
previous pointer should stay here it
should not be shifted our current
pointer should only be shifted here
because if we shift both of the pointers
if we put the previous pointer over here
and the current pointer over here we're
g to consider removing this one we're
going to check okay is this a value we
want to remove and then we'll end up
removing it but we ended up skipping
this guy so that's why you don't want to
shift the previous poter if you deleted
the node okay but what if we don't
delete the node suppose we weren't
deleting this one well then we would
shift both of the pointers then we would
shift previous by one and we would shift
current by one and then you know do the
same thing here so that's more or less
how you go about solving the problem I
think we're actually ready to code this
up very quickly let's go over the time
and space we're given two inputs the
size of them is not necessarily going to
be the same let's say this is n let's
say this is M so overall the time
complexity we're going to convert this
into a hashset and we're going to
iterate over the entire link list that's
going to be n plus m space we are using
a data structure the hash set remember
so that's going to be Big O of n now
let's code it up so the first thing I'm
going to do is just take nums and
convert it into a hash set I'm able to
do this in Python because there's no
like static typing you might not be able
to do that in your own language you
might have to create a new variable name
um but now let's do the Demi node let's
create a random node and let's give it a
value of zero it doesn't really matter
what value we give it let's set the next
pointer though to the head of the given
linked list here and now let's basically
do that two-pointer thing I was talking
about having the previous pointer set to
uh the Demi node initially and then the
current pointer set to the head or we
could say dummy. next but it doesn't
really matter they're both the same and
now we say while current because current
is the node that we're considering to
delete and so we're going to check if
the current. Val is in the nums hash set
then we want to delete it and that's
very easy to do we just say previous do
next is equal to current. next and
remember in this case we don't shift the
previous pointer we don't want to do
that we don't want to skip a node we
only shift the current pointer so say
current is now going to be current. next
now the alternative is where we don't
delete the node in that case it's pretty
simple we just shift both of the pointer
so previous will be previous nextt
current will be current. nextt you might
notice that there is a duplicate line in
both of these conditions it doesn't
really matter in a real interview I
personally wouldn't care that much but
it does make the code a little bit
better so I'm going to go ahead and move
this line out of the if statement and
then delete this one lastly at the end
we are going to return
dummy. nextt if we ended up deleting
every single Noe in the linked list what
would dummy. nextext point to it would
probably point to null if we didn't
delete every single node it's going to
point to the new head of the link list
cuz even though we're deleting nodes
we're never changing the order of them
so this is the whole solution let's run
it as you can see it works it's pretty
efficient you can actually solve this
problem without keeping track of two
pointers you can get away with just
using one pointer so let's do that I
guess I'll leave it as previous and then
instead of current I'm going to say
previous do nextt because we know that
current was always one ahead of previous
anyway for current here I'm going to
change that again to previous do nextt
and to current here I'm also going to
change it to previous do nextt so now
this is just you know being shifted by
one here I'm going to change this to
previous do next and same thing here
actually I think I'm going a little bit
too much on autopilot because we don't
need to shift all of these so let me
delete this line and just reread what
we're doing here so if the next node is
one that we want to delete let's delete
it otherwise just shift the pointer I
think that's actually enough I don't
think we actually need anything else
since we only have one pointer anyway
and let me go ahead and run this and you
can see this one works as well I don't
think it's necessarily more efficient to
be honest um but the runtime says so but
if you found this helpful check out N.O
for a lot more I'm going to be launching
a free newsletter pretty soon it's going
to be covering data structures system
design and a lot more I'll also be
making a lot more system design videos
pretty soon I'm really excited about
that so subscribe if you want to see
those
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