hearing "hey guys i'm still unemployed so let's solve a leetcode problem today" and knowing you became a software engineer at google is so real and inspiring bro, thanks
"hey guys i am still unemployed so lets solve a problem " this one cracked me up because right now all your efforts paid off and look back, we all have been there..ah what a journey!!!
I've been watching your solutions for a while now and I'm just so happy that I finally started to solve problems like the way you do. Thank you so much!!
So simple, thank you. I was casting the linked list values to strings and reversing them, then casting them back to ints and adding them... needless to say I was making spaghetti. Your example is very clean!
@@prudvi01 I used an if statement inside a while loop, would that be wrong even if it works, if not what's wrong with it ? As in does the solution have to be a linked list.
I did the same. I'm not convinved it's any worse than the solution presented in the video. looping through two linked lists (and a third for the output) is O(n) time complexity.
Excellent! Clear and plain explanations, and it's so much cleaner when hand mark-up is by pen instead of by mouse. :) This is linked in LeetCode as "optimized solution explained" on the problem itself, but it'd be faster to take: carry = val // 10 val = val % 10 ...and make it instead: carry = 1 val = val - 10 The modulus approach is fantastic for expanding our algorithm to support adding up three or more arrays, but in our use case carry will always be 0 or 1.
@@inspisiavideos8542 You divide by 10 so that you can add it to the next number which will be in the next 10's place. For example: if you took 25 + 26 if you add 5 + 6 and then divide by 10 you'll get 1. So i take my one and add it to the next 10's place which is 2 + 2 + 1 that i got from carrying the number which will total 51.
@@danielwilczak7833 Thank you so much!!! I got an internship during this Summer as QA and just finished it but I asked for an extension and if I could get it this time as Dev. They said yes. I have coded before with courses on how to make websites but I have never done code questions. Basically this next week is my first week. It’s so hard to get a job so they told me that they don’t have entry level positions and senior positions are more than likely ending up on people who are already in the company. So basically I’m learning as much as I can during my spare time to then perhaps they let me stay after the internship extension ends by the end of the year.
@@inspisiavideos8542 Congrats on your internship! Calling out 10 as a "magic number" is really great - bare numbers with no evident explanation can cause a lot of confusion in code, so 10 could be a constant called NUMBER_BASE since we're just using decimal notation here. For extra fun, make it not a constant, and adapt the algorithm to work in number bases other than 10!
@@philhamlin7935 hey thanks! My internship was supposed to end in August but was extended to the end of December. I’ve been applying to other companies but nothing yet even though I have projects on my portfolio. My internship manager told me to stay alert when an opening occurs and apply. 2 positions opened and I applied there this week. Hopefully they’ll let me know and I be ready for a test code. The company has many teams so probably the person who sees my resume doesn’t know me yet.
I think it's because we want the logic to be "the next thing is ___" for every iteration, so we make a dummy list node and immediately start with our logic (the next thing for the dummy node is ___). If you use dummy.val, you have the first step do the logic of "the value here is ___" and then the later steps are "the next thing is ___" which is more inconvenient to write. That's how I thought of it, not sure if that's the right way to think about it.
is it bad to convert them to ints and add them then make a new linked list? 💀 def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: n1 = "" n2 = "" while l1: n1 += str(l1.val) l1 = l1.next while l2: n2 += str(l2.val) l2 = l2.next total = int(n1[::-1]) + int(n2[::-1]) total = str(total)[::-1] res = dummy = ListNode() for n in total: new = ListNode(n) dummy.next = new dummy = dummy.next return res.next
"Hey guys, I'm still unemployed.." Many kisses for you neet! you make me laugh.. And relax... To the best ASMR Channel OUT-THERE!!!! Please go back to this microphone and keyboard, and relaxed voice!!!!!!!!! damn I enjoy this
i solved it using more memory but it eliminates thinking about edge cases: nums1, nums2 = [], [] cur = l1 while cur: nums1.append(cur.val) cur = cur.next cur = l2 while cur: nums2.append(cur.val) cur = cur.next
nums1 = nums1[::-1] nums2 = nums2[::-1]
num1 = int(''.join(map(str, nums1))) num2 = int(''.join(map(str, nums2))) vals = [] number = str(num1 + num2) for c in number: vals.append(int(c)) dummy = ListNode(0) cur = dummy while vals: cur.next = ListNode(vals.pop()) cur = cur.next
# Definition for singly-linked list. # class ListNode(object): # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution(object): def addTwoNumbers(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ def reverse_in_place(head): before = None temp = head while temp is not None: after = temp.next temp.next = before #flip pointer before = temp #cross 'gap' temp = after # move temp over return before #before is the new head of reversed LL l1 = reverse_in_place(l1) l2 = reverse_in_place(l2) temp_l1 = l1 l1_val = "" temp_l2 = l2 l2_val = "" while temp_l1 is not None: l1_val += str(temp_l1.val) temp_l1 = temp_l1.next while temp_l2 is not None: l2_val += str(temp_l2.val) temp_l2 = temp_l2.next total = int(l1_val) + int(l2_val) total = str(total) temp_out = ListNode(int(total[0])) new_LL = temp_out for i in range(1, len(total)): if temp_out.next is None: temp_out.next = ListNode(int(total[i])) temp_out = temp_out.next new_LL = reverse_in_place(new_LL) return new_LL This is my brute force method for this problem but i dont think its THAT bad on time and space complexity LOL what do you guys thing
I was able to solve this one, but did not think of dividing the current sum by 10 and using Math.floor() to get the first digit as the carry number, and then using modulo to get the second number aka the current sum. So darn clever! Instead I converted the sum to a string and grabbed the values by index.
I cant figure out how does cur work. If I print id(cur) == id(dummy) it's true, but if i print cur and dummy inside or after while cycle the output is different. Tell me please what's going on or What should I google to clarify that ?
I was really close to the solution on this one. I had an extra node at the end because I was creating my node in cur instead of cur.next(). Slowly getting better at these
We are treating dummy.next as the head node. Dummy.next.val contains the value of the head node, and dummy.next.next contains the next node and so on. Basically we created the dummy node to avoid setting head node = firstvalue and then doing head.next for the others
Can Some please explain why the dummy is still getting updated when after the first iteration cur is updated to a different object. newNode = ListNode(columnSum % 10) curr.next = newNode curr = newNode
At the start of the loop cur = dummy. On the first iteration you are assigning cur.next (aka dummy.next) with the first digit of the solution. So now dummy.next equals the head node of the solution so at the end we return dummy.next. dummy.next will always point to this head node no matter what cur gets changed to.
I don't understand what's going on with carry. Wouldn't always be a left over number there by the time two get to the second digit in the list nodes. First time through: val = 2 + 5 + 0 carry = 7 // 10 Doesn't that leave carry as .7 when we get to the next digits? I'm very confused.
In Python, we can perform floor division (also sometimes known as integer division) using the // operator. This operator will divide the first argument by the second and round the result down to the nearest whole number So, 7 // 10 = 0
Can anybody explain me? In the begin we wrote "dummy = cur". After that we updating "cur" in the while-loop. But we don't update "dummy". So how "dummy" is updating together with "cur" if we don't redefine "dummy = ..."
For some reason it was extremely easy problem for me, first time that I just saw the solution instantly in full (unlike all “easy” ones before) and was so surprised of everyone’s animosity to this problem, lol.
it's a bit compicated to explain it in words. I would rather suggest you to watch the videos on how to create linklist object and other related problems ( which aren't directly given into the function like this leetcode problem)
Bruh, my dumb ass thought transforming the f*cking linked list into a number and than adding both together and making another linked list of that was easier "Error - found cycle" disagree...
That also works! I solved it like that first. But this is a better approach lol. Removes the pain of converting linked list to integer and then back to linked list.
Thanks for making this. Trying this problem is the first time I've ever tried using a Linked List, so this helped solidify the solution LeetCode provides.
Hi. Can someone help me understand how and why there is l1.val? I thought l1 is a list which is [2, 4, 3]? How does l1.val calling integer 2? Or it's from ListNode class? I'm very confused.
if you see the ListNode class there is a val, which is used to access the int values of the nodes so it is used as node.val (in this case l1.val) hope it helps
Hi @@smilenpassion thank you for your respond, but I am still confused. Class listnode is not inheritance under class solution. How can class solution call any object from class listnode? Also, even it can call the object from listnode, shouldn't be v1 = self.val if l1? Or set the self.val to l1.val? Thank you again for helping me to understand it
Both dummy and cur are both pointers to the same list, but serve different purposes: We are using dummy to point to the start of the list and cur will be used to add additional nodes + move to the latest node in the list.
curr is what you use to traverse the dummy linked list, while, "dummy" will always point at the head of the linked list. Since you're returning "dummy.next" you'll be returning the head of the resultant linked list.
0:00 Lol now you're at Google. Here I am just trying to get Amazon or Microsoft now. Awesome solution, I managed to actually solve this one myself but wanted to see your way of doing it, much cleaner. Thank you! Liked!
@@tsunningwah3471 dummy is initialized to default value of 0 for the purpose of having an initial node to attach everything else to. dummy.next is the real start of the summed list.
At the start of the loop cur = dummy. On the first iteration you are assigning cur.next (aka dummy.next) with the first digit of the solution. So now dummy.next equals the head node of the solution so at the end we return dummy.next. dummy.next will always point to this head node no matter what cur gets changed to.
I quote your response to ask the question further. As I understand, cur = dummy means let cur be dummy, but they are not the same one. I still don't understand whey cur.next will change dummy.next. Did I miss anything? Thank you so much!
@@alicia8170 hi yeah i'm wondering the same thing - although we didn't reference dummy at all or change its value, it gets built on as we change cur and cur.next - is this a python binding or pointer issue?
This was hard argh I had no idea how to implement carry. Thank you for explaining! I don't understand the dummy node part. can we not use a dummy node? Why not use curr and return curr?
You could, but in my experience the dummy node generally makes building the linked list easier since you can loop through it with less conditional statements. If you don't make the dummy node, you would either have to have a separate if statement to check that our starting node is not null or find some other way to do a one-off initialization of the list. With the dummy node, you can process the nodes each time without having to worry about if it's your starting node or not.
🚀 neetcode.io/ - A better way to prepare for Coding Interviews
Seems like the code is not working. Showed me the wrong result when compile.
hearing "hey guys i'm still unemployed so let's solve a leetcode problem today" and knowing you became a software engineer at google is so real and inspiring bro, thanks
Bruh he left job at amazon, and decided to voluntarily be unemployed. Dude's been a genius all along.
"hey guys i am still unemployed so lets solve a problem " this is savage bro
he earns fro youtube tho
😂
TH-cam hired him
Now he is at google
💯😂
that intro hit me different
lol
hhhhhhhhhhh me too
I get so inspired/motivated when you mention unemployment in your old videos because of where you ended up
"hey guys i am still unemployed so lets solve a problem " this one cracked me up because right now all your efforts paid off and look back, we all have been there..ah what a journey!!!
I've been watching your solutions for a while now and I'm just so happy that I finally started to solve problems like the way you do. Thank you so much!!
If the person teaching these is unemployed what chances do we have...
That beginning took me by surprise, instantly made my evening better. Thanks
So simple, thank you. I was casting the linked list values to strings and reversing them, then casting them back to ints and adding them... needless to say I was making spaghetti.
Your example is very clean!
lol that was my initial solution as well! got dirty real quick
I did same, thanks
@@prudvi01 I used an if statement inside a while loop, would that be wrong even if it works, if not what's wrong with it ? As in does the solution have to be a linked list.
I did the same. I'm not convinved it's any worse than the solution presented in the video. looping through two linked lists (and a third for the output) is O(n) time complexity.
def addTwoNumbers(l1,l2):
l1.reverse()
l2.reverse()
l1s= str(l1)
l2s= str(l2)
l1s= "".join(l1s)
l2s= "".join(l2s)
l1 = ""
l2 = ""
for i in range(1, len(l1s), 3):
l1 +=l1s[i]
for i in range(1, len(l2s), 3):
l2 +=l2s[i]
l1s=int(l1)
l2s=int(l2)
res=l1s+l2s
res= str(res)
res= list(res)
res.reverse()
return res
Excellent! Clear and plain explanations, and it's so much cleaner when hand mark-up is by pen instead of by mouse. :)
This is linked in LeetCode as "optimized solution explained" on the problem itself, but it'd be faster to take:
carry = val // 10
val = val % 10
...and make it instead:
carry = 1
val = val - 10
The modulus approach is fantastic for expanding our algorithm to support adding up three or more arrays, but in our use case carry will always be 0 or 1.
You would require 'if' statements.
if val >= 10:
carry = 1
val = val - 10
Which would reduce the elegance of code
Nothing about this problem is elegant or worth optimization.
I traversed through both linked list and got the actual numbers, added them, then created a linked list with the result's digits as nodes.
hi, Please let me know. Why is the value divided by 10? I'm learning how to code and I don't know where he got the number 10? (on minute 7:29)
@@inspisiavideos8542 You divide by 10 so that you can add it to the next number which will be in the next 10's place. For example: if you took 25 + 26 if you add 5 + 6 and then divide by 10 you'll get 1. So i take my one and add it to the next 10's place which is 2 + 2 + 1 that i got from carrying the number which will total 51.
@@danielwilczak7833 Thank you so much!!! I got an internship during this Summer as QA and just finished it but I asked for an extension and if I could get it this time as Dev. They said yes. I have coded before with courses on how to make websites but I have never done code questions. Basically this next week is my first week. It’s so hard to get a job so they told me that they don’t have entry level positions and senior positions are more than likely ending up on people who are already in the company. So basically I’m learning as much as I can during my spare time to then perhaps they let me stay after the internship extension ends by the end of the year.
@@inspisiavideos8542 Congrats on your internship! Calling out 10 as a "magic number" is really great - bare numbers with no evident explanation can cause a lot of confusion in code, so 10 could be a constant called NUMBER_BASE since we're just using decimal notation here. For extra fun, make it not a constant, and adapt the algorithm to work in number bases other than 10!
@@philhamlin7935 hey thanks! My internship was supposed to end in August but was extended to the end of December. I’ve been applying to other companies but nothing yet even though I have projects on my portfolio. My internship manager told me to stay alert when an opening occurs and apply. 2 positions opened and I applied there this week. Hopefully they’ll let me know and I be ready for a test code. The company has many teams so probably the person who sees my resume doesn’t know me yet.
A complete beginner question, how to call and declare the inputs for addTwoNumbers function,
just to test the program !
"hey guys i am still unemployed so lets solve a problem" mans representing his entire audience 😂😂😂
This code only passes 1365/1568 test cases. Update needed.
Congratulations that you are now employed!! Happy for you!
@NeetCode Could you please explain the time complexity of the problems you solve? It will be really useful for us.
For this one it should be O(n + m), where n is length of first linked list and m is length of second linked list.
@@wlockuz4467 no bro... Either O(n) or O(n+1) where n is the length of linked list which has more nodes
Can you please explain what will be the time complexity and space complexity for this piece of code in example?
@@alok7485 Time complexity will be O(max(n, m)) and Space complexity will be O(max(m,n))
Hopefully you found a job!
Dude, thankyou for this solutions. They are helping me alottttt! Also, if you could please make a video on intersection of two linked lists
Can you please explain what will be the time complexity and space complexity for this piece of code in example?
Why are these things so difficult 😣?
Can someone explain why we don't use the first dummy list node? Why not use dummy.val and assign?
I think it's because we want the logic to be "the next thing is ___" for every iteration, so we make a dummy list node and immediately start with our logic (the next thing for the dummy node is ___). If you use dummy.val, you have the first step do the logic of "the value here is ___" and then the later steps are "the next thing is ___" which is more inconvenient to write.
That's how I thought of it, not sure if that's the right way to think about it.
is it bad to convert them to ints and add them then make a new linked list? 💀
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
n1 = ""
n2 = ""
while l1:
n1 += str(l1.val)
l1 = l1.next
while l2:
n2 += str(l2.val)
l2 = l2.next
total = int(n1[::-1]) + int(n2[::-1])
total = str(total)[::-1]
res = dummy = ListNode()
for n in total:
new = ListNode(n)
dummy.next = new
dummy = dummy.next
return res.next
Here after when you got job in Google.... congratulations:)
Better explanation than the official solution explanation.
nulll and zero are different, you should instead append the while statement to while l1 or l2 or carry>0
In python, 0 evaluates to its Boolean value (False) in a conditional clause. So it's fine to just have carry.
@jaspreetsahota1840 good to know!
"Hey guys, I'm still unemployed.." Many kisses for you neet! you make me laugh.. And relax... To the best ASMR Channel OUT-THERE!!!! Please go back to this microphone and keyboard, and relaxed voice!!!!!!!!! damn I enjoy this
I have interview on 21st March, there is no way I am gonna pass :(
This level of programming is going above my head
kaya hua? Pass kiya tumne?
Did u pass?
how did it goes?
i solved it using more memory but it eliminates thinking about edge cases:
nums1, nums2 = [], []
cur = l1
while cur:
nums1.append(cur.val)
cur = cur.next
cur = l2
while cur:
nums2.append(cur.val)
cur = cur.next
nums1 = nums1[::-1]
nums2 = nums2[::-1]
num1 = int(''.join(map(str, nums1)))
num2 = int(''.join(map(str, nums2)))
vals = []
number = str(num1 + num2)
for c in number:
vals.append(int(c))
dummy = ListNode(0)
cur = dummy
while vals:
cur.next = ListNode(vals.pop())
cur = cur.next
return dummy.next
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
def reverse_in_place(head):
before = None
temp = head
while temp is not None:
after = temp.next
temp.next = before #flip pointer
before = temp #cross 'gap'
temp = after # move temp over
return before #before is the new head of reversed LL
l1 = reverse_in_place(l1)
l2 = reverse_in_place(l2)
temp_l1 = l1
l1_val = ""
temp_l2 = l2
l2_val = ""
while temp_l1 is not None:
l1_val += str(temp_l1.val)
temp_l1 = temp_l1.next
while temp_l2 is not None:
l2_val += str(temp_l2.val)
temp_l2 = temp_l2.next
total = int(l1_val) + int(l2_val)
total = str(total)
temp_out = ListNode(int(total[0]))
new_LL = temp_out
for i in range(1, len(total)):
if temp_out.next is None:
temp_out.next = ListNode(int(total[i]))
temp_out = temp_out.next
new_LL = reverse_in_place(new_LL)
return new_LL
This is my brute force method for this problem but i dont think its THAT bad on time and space complexity LOL what do you guys thing
By far the best explanation of this problem I’ve seen 👏 Thanks for sharing
Are you still unemployed?
I was able to solve this one, but did not think of dividing the current sum by 10 and using Math.floor() to get the first digit as the carry number, and then using modulo to get the second number aka the current sum. So darn clever! Instead I converted the sum to a string and grabbed the values by index.
I cant figure out how does cur work. If I print id(cur) == id(dummy) it's true, but if i print cur and dummy inside or after while cycle the output is different. Tell me please what's going on or What should I google to clarify that ?
A little confused,how are we adding in reverse order?
we're just adding v1 and v2 which is normal addition isn't it?
The digits are in reverse order.
I'm getting memory limit exceeded. Did something change on Leetcode's end?
what's the function of next and val do?
I got some errors when I ran your exact code. Is your solution missing the case when the highest position creates a carry of 1?
you might have missed the "+ carry" bit when summing up the "val"
while l1 or l1 or carry:
The first medium LL problem I could solve with optimized code.
I was really close to the solution on this one. I had an extra node at the end because I was creating my node in cur instead of cur.next(). Slowly getting better at these
I did 'print(Solution().addTwoNumbers(l1,l2))' and got error "list object has no attribute 'val'... how do I fix it?
I'm really proud of mylsef. I finally managed to solve a medium problem by myself, although my solution was slow as hell xd
To deal with edge case just before . return dummy.next write
If carry==1: cur.next= ListNode (1)
can someone please explain why dummy.next gives the full listnode in return? ( if i go dummy.next.val it gives y the value of the second node)
We are treating dummy.next as the head node. Dummy.next.val contains the value of the head node, and dummy.next.next contains the next node and so on. Basically we created the dummy node to avoid setting head node = firstvalue and then doing head.next for the others
Thank you
Can Some please explain why the dummy is still getting updated when after the first iteration cur is updated to a different object. newNode = ListNode(columnSum % 10)
curr.next = newNode
curr = newNode
At the start of the loop cur = dummy. On the first iteration you are assigning cur.next (aka dummy.next) with the first digit of the solution. So now dummy.next equals the head node of the solution so at the end we return dummy.next. dummy.next will always point to this head node no matter what cur gets changed to.
Hi Sajith, Here a good explanation: stackoverflow.com/questions/58759348/when-does-a-pointer-to-a-linked-list-change-the-actual-list
@@ai4sme Thanks for this. TIL python variables are essentially references/pointers to objects i.e. class instances.
You are best. I always pause ads blocker for you.
"you probably do but you might think you don't".
Haha, thanks for the encouragement.
Why do you write in the while loop just "l1 or l2 or carry"
Awesome, I love how you dealt with all the edge cases without cluttering the code
Did you find a job now?
could have reduced these statements within single block, otherwise awesome explanation thank you
if(l1 != null){
v1 = l1.val;
l1 = l1.next;
}
if(l2 != null){
v2 = l2.val;
l2 = l2.next;
}
Why my code is not working can any one correct this code.
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode head = null;
public ListNode tail = null;
public void addNode(int data) {
ListNode newNode = new ListNode(data);
if(head == null) {
head = newNode;
tail = newNode;
}
else {
tail.next = newNode;
tail = newNode;
}
}
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int num1=0;
int num2=0;
int a;
int i=1;
ListNode temp1 = l1;
ListNode temp2 = l2;
while(temp1 != null && temp2 !=null){
//coverting this to number
num1 = num1+((temp1.val)*i);
num2 = num2+((temp2.val)*i);
i=i*10;
temp1 = temp1.next;
temp2 = temp2.next;
}
System.out.println(num1+" "+num2);
// adding two numbers
int num3 = num1+num2;
while(num3>0){
//insert the nodes
a = num3%10;
if(a==0)
addNode(0);
else
addNode(a);
// here the zero is not inserting
num3 = num3/10;
}
return head;
}
}
Wow this is so clean✅✅
I brute- forced my way to this problem and it's functional but really ugly 😀
NC: You do probably remember how to add two numbers together
Me: You severely overestimate my abilities
I don't understand what's going on with carry. Wouldn't always be a left over number there by the time two get to the second digit in the list nodes.
First time through: val = 2 + 5 + 0
carry = 7 // 10
Doesn't that leave carry as .7 when we get to the next digits? I'm very confused.
In Python, we can perform floor division (also sometimes known as integer division) using the // operator. This operator will divide the first argument by the second and round the result down to the nearest whole number
So, 7 // 10 = 0
its interesting how you build the answer after the dummy node and return the answer. Its not like the test cases can detect the dummy node
This intro make this video my favorite one 😂. Could be a useful cue to remember the solution in an interview even.
Can anybody explain me? In the begin we wrote "dummy = cur". After that we updating "cur" in the while-loop. But we don't update "dummy". So how "dummy" is updating together with "cur" if we don't redefine "dummy = ..."
Exactly my doubt!!!
For some reason it was extremely easy problem for me, first time that I just saw the solution instantly in full (unlike all “easy” ones before) and was so surprised of everyone’s animosity to this problem, lol.
Can you explain how can I call the l1 and l2 listnodes in the addtwonumber function to test?
it's a bit compicated to explain it in words. I would rather suggest you to watch the videos on how to create linklist object and other related problems ( which aren't directly given into the function like this leetcode problem)
@@crosswalker45 hi, thanks a lot man.
Couldn’t have solved this without you. Thank you so much
Can you please explain what will be the time complexity and space complexity for this piece of code in example?
Bruh, my dumb ass thought transforming the f*cking linked list into a number and than adding both together and making another linked list of that was easier
"Error - found cycle" disagree...
That also works!
I solved it like that first. But this is a better approach lol. Removes the pain of converting linked list to integer and then back to linked list.
Can someone explain the dummy listnode and how changing in cur changes dummy?
I am so happy that LC decided to give us reversed lists for this problem
The Leetcode dislike count was a useful metric
BRO! UNIQUE..!! "Hey Guys, I am still UNEMPLOYED so let's solve a problem!" Lmao!😂
Legendary Introduction!
Why can't you reverse both list and just add the two reversed list together?
Love how you started the video! xD You made my life easier by posting these videos. Great work! :) :D
i hate the problem already simply because i didn't understand the solution
doesn't dummy.next return the head of the list?
Why can't we just return cur.next ??
I mean the cur is pointing to dummy right? So just print it right idk if I'm right
i think that will skip first value
Explanation was crystal clear and helped me solve the problem super quick. Thanks!
Why can;t we simply reverse two lists first, sum them and then reverse an output?
Thanks for making this. Trying this problem is the first time I've ever tried using a Linked List, so this helped solidify the solution LeetCode provides.
Can you please explain what will be the time complexity and space complexity for this piece of code in example?
When was dummy head ever updated in the code? Please help!
The intro slapped me back into reality. As always, thanks for the awesome explanations 😄
This is the way I learnt about pointer, I can't believe it
i'm amazed at your ability. it's the nuances in the code that's the hardest to figure out. kudos to you.
Wow, this was very well explained!
Hi. Can someone help me understand how and why there is l1.val?
I thought l1 is a list which is [2, 4, 3]? How does l1.val calling integer 2?
Or it's from ListNode class? I'm very confused.
if you see the ListNode class there is a val, which is used to access the int values of the nodes so it is used as node.val (in this case l1.val) hope it helps
Hi @@smilenpassion thank you for your respond, but I am still confused.
Class listnode is not inheritance under class solution. How can class solution call any object from class listnode?
Also, even it can call the object from listnode, shouldn't be v1 = self.val if l1? Or set the self.val to l1.val?
Thank you again for helping me to understand it
so do we have a first node that is just null in all of these
Man I was super stressed out solving these leetcodes, but the " Hey I am still unemployed" made my day! LMAO
Stupid question but how does "dummy" get updated when all code changes are done to "cur"?
Both dummy and cur are both pointers to the same list, but serve different purposes: We are using dummy to point to the start of the list and cur will be used to add additional nodes + move to the latest node in the list.
Can someone explain what happens inside the stack with the statements "cur = dummy" and "return dummy.next"?
curr is what you use to traverse the dummy linked list, while, "dummy" will always point at the head of the linked list. Since you're returning "dummy.next" you'll be returning the head of the resultant linked list.
0:00 Lol now you're at Google. Here I am just trying to get Amazon or Microsoft now.
Awesome solution, I managed to actually solve this one myself but wanted to see your way of doing it, much cleaner. Thank you! Liked!
Only Platform I choose for any coding explanation
can anyone tell me why we're returning the dummy.next and not dummy itself? wont we be able to access every other element if we just return dummy?
yes i have the same question too. I thought dummy itself is the real list
@@tsunningwah3471 dummy is initialized to default value of 0 for the purpose of having an initial node to attach everything else to. dummy.next is the real start of the summed list.
@@m0sh3 thanks got it！
who the he are you
🕵🕵🕵🕵🕵🕵
Great video and great channel I am very excited to discover this channel I hope I benefit from this channel a lot .Thanks a lot😊
Haha, now you are so employed! Thanks for these great contents!
At the start of the loop cur = dummy. On the first iteration you are assigning cur.next (aka dummy.next) with the first digit of the solution. So now dummy.next equals the head node of the solution so at the end we return dummy.next. dummy.next will always point to this head node no matter what cur gets changed to.
I quote your response to ask the question further. As I understand, cur = dummy means let cur be dummy, but they are not the same one. I still don't understand whey cur.next will change dummy.next. Did I miss anything? Thank you so much!
@@alicia8170 hi yeah i'm wondering the same thing - although we didn't reference dummy at all or change its value, it gets built on as we change cur and cur.next - is this a python binding or pointer issue?
this solution gives TLE (Time Limit Exceeded) any other solution?
Most likely you have an infinite loop somewhere, are you updating your pointers?
This was hard argh I had no idea how to implement carry. Thank you for explaining! I don't understand the dummy node part. can we not use a dummy node? Why not use curr and return curr?
You could, but in my experience the dummy node generally makes building the linked list easier since you can loop through it with less conditional statements.
If you don't make the dummy node, you would either have to have a separate if statement to check that our starting node is not null or find some other way to do a one-off initialization of the list. With the dummy node, you can process the nodes each time without having to worry about if it's your starting node or not.
"I am still unemployed" Not anymore!
I hope that you are employed now bro
I am very happy for you that you are employed at Google now. 😀
#Here is the full code to run on your personal machine:
#Add two numbers:
class LinkedList:
def __init__(self):
self.head = None
class ListNode:
def __init__(self,val=0,next=None):
self.val=val
self.next=next
def addTwoNumbers(l1,l2):
dummy=ListNode()
cur=dummy
carry=0
while l1 or l2 or carry:
v1=l1.val if l1 else 0
v2=l2.val if l2 else 0
#newdigit
val=v1+v2+carry
carry=val//10
val=val%10
cur.next=ListNode(val)
cur=cur.next
l1=l1.next if l1 else None
l2=l2.next if l2 else None
return dummy.next
first_node = ListNode(2)
second_node = ListNode(4)
third_node = ListNode(3)
l1 = LinkedList()
l1.head = first_node
first_node.next = second_node
second_node.next = third_node
first_nodeb = ListNode(5)
second_nodeb = ListNode(6)
third_nodeb = ListNode(4)
l2 = LinkedList()
l2.head = first_nodeb
first_nodeb.next = second_nodeb
second_nodeb.next = third_nodeb
string_repr = ""
node = l1.head
while node is not None:
# string_repr += node.val
# string_repr += " -> "
string_repr += str(node.val)
string_repr += " -> "
node = node.next
string_repr += "None"
print(string_repr)
string_repr = ""
node = l2.head
while node is not None:
# string_repr += node.val
# string_repr += " -> "
string_repr += str(node.val)
string_repr += " -> "
node = node.next
string_repr += "None"
print(string_repr)
lresult=addTwoNumbers(l1.head,l2.head)
string_repr = ""
node = lresult
while node is not None:
# string_repr += node.val
# string_repr += " -> "
string_repr += str(node.val)
string_repr += " -> "
node = node.next
string_repr += "None"
print(string_repr)
LOVE you my GURU your awesome!!!
wonderful explanation
im still unemployed :(