Excellent explanation, you explain the algorithms so easy and well. Observation: You can write the last line of code as: return goal == 0. regards!

@pankajjatav6448

Жыл бұрын

Also you can start loop from second last element: for i in range(len(nums)-2, -1, -1):

@yonavoss-andreae4952

10 ай бұрын

@@pankajjatav6448 that triggers an edge case in which two elements returns false ie [1,2]

@gordonwu8117

7 ай бұрын

@@yonavoss-andreae4952 if length is 0 then "i" will start on index 0, which works as intended

@learnwithujjwal_

4 ай бұрын

or simply return not goal

we can start the loop from the len(nums) - 2 position since goal is already set at the last position when we declare it.

@shashanksrivastava7262

Жыл бұрын

I am acturally surprised how his code actually worked like, wouldn't i+nums[i] be always greater than goal ?

@anupamkolay193

Жыл бұрын

@@shashanksrivastava7262 Yes I'm also thinking about that too.

@quanghuyluong1249

Жыл бұрын

@@shashanksrivastava7262 Consider [3,2,1,0,4]: i+nums[i] would not be greater and the goal will never be shifted

@experiment8924

11 ай бұрын

@@shashanksrivastava7262 Yes, i+nums[i] would be greater than goal for the first iteration, which would satisfy the if statement but the goal will be the same, i.e the last index and the program would move on the next iteration and work as usual.

@shravankumar3717

10 ай бұрын

Thats why in this example we cannot reach last element. Algorithm works

12:00 jaw dropping intuition, I could have never thought about it. Thanks for the explanation.

that explanation for greedy part was just awesome🔥🔥

Man excellent job. I went through many videos but wasn't able to understand. U made us all feel that it is very simple. Thanks a-lot again.

This channel is so much underrated. This video is just amazing!

Thank you for these videos, I found you recently and you are the channel I have been looking for

That greedy solution is ingenius. I am in awe!! Thank you.

There is no need to start from the end and move backwards. You can naturally progress from the beginning, like this: int reach = 0; for (int i = 0; i < nums.length && i = nums.length-1;

@ma_sundermeyer

Жыл бұрын

yeah, also started from the beginning, it's more intuitive. You can also stop early at zeros that you can't cross: max_index = -1 for i in range(len(nums)-1): if nums[i] == 0 and max_index

@PippyPappyPatterson

Жыл бұрын

@@ma_sundermeyer Why does everyone start from the end? Does it help with other problems that use a similar solution (that can't be implemented front-forwards)? From the beginning is a million times easier to remember.

@eku333

Жыл бұрын

@@PippyPappyPatterson I disagree. NeetCode's solution is easier to understand imo.

@PippyPappyPatterson

Жыл бұрын

@@eku333 do u normally iterate across ur arrays in reverse? or forwards?

@eku333

Жыл бұрын

@@PippyPappyPatterson iterating over an array in reverse is not a rarely used dp technique.

These are the questions/solutions that make me fall in love with algorithms :) many many thanks @NeetCode

The tricky part of the greedy is to prove that it actually works; we clearly have multiple options for moving the goal and we always pick the first one. How do we guarantee that we are not going to get stuck using this approach?

@sivaprakashkkumar9691

2 жыл бұрын

how it becomes greedy solution please explain it.

@mangalegends

2 жыл бұрын

This is what I have a little trouble wrapping my brain around. We always pick the first option, but what if the first option is 0? Then we're stuck lol. I guess you could add extra logic to handle that but the greedy solution doesn't seem to have include that

@VipulDessaiBadGAMERbaD

2 жыл бұрын

but this prob is not to find optimal path but to find only if we can reach the destination, that is why its okay to select the first elment.

@Sim0000n

Жыл бұрын

@@mangalegends Let's consider the array 3105. Goal is 3, i is 3. I becomes 2, goal stays the same (as it is 0.) I becomes 1, Goal stays the same as 1+1 is not greater or equal than 3. i becomes 0, but as 0+3 ≥ 3, we're good. So no reason to be stucked

@jorgemejia1586

Жыл бұрын

Even if you hit a 0 during the loop, that’s fine. The goal post will be left at an actual reachable/moveable location (aka an index where the value is not zero)

greedy approach is brilliant. love it

Outstanding illustration with the diagrams. I just tried out the same code looping i from len(nums) - 2 to -1 because anyway the first run didn't alter the goal's position and it got accepted :) thank you so much!

amazing explanation. It really showcase why greedy approach can work wonders sometimes

Great Explanation! Thx for making it clearer for greedy algorithm

thats an awesome solution and explination!! I thought of an answer where we only stop our iteration once we come upon a zero. then we just check if we can if the zero can be jumped over by previous elements. the only problem is the edge case where nums= [0],where you got check for it. var jump = function (nums) { if (nums.length === 1) return true; let prevPosition, prevValue; let passFlag = true; for (let i = 0; i if (nums[i] === 0) { passFlag = false; prevPosition = i - 1; while (prevPosition >= 0) { prevValue = nums[prevPosition]; if (prevPosition + prevValue > i) { passFlag = true; break; } prevPosition--; } if (!passFlag) return false; } } return true; };

Nice. I practiced the dp solution and got stuck on greedy. I was able to make the connection to the last and second before last elements , but couldn’t think of moving the goalpost as you say. Nice solution.

Love when I can implement someone's explanation without directly looking at the implementation. Thank you so much brotha

@NeetCode

Жыл бұрын

Nice! 💪

Oh my god I cannot believe I can finally understand, design the steps and write the correct code! Finally my work paid off!!

Your explanation works like magic! Thank you so much 😊

Thank you so much sir! Your logics and way of explaining is really impressive and make attention to the solution.

your explanations are really easy to understand. I always look for your videos. I was looking for buy and sell stocks III and IV videos from your channel but did'nt find them. Watched other channel videos but they were not as easy to understand.

I love how simple the solution is. I was sketching out a very complex one :D

Spent couple of hours with this task :( after watching first 3 minutes of the video, was able to write the solution. You're God!

The greedy approach is simply mind-blowing!

This is the best greedy solution I have seen till now!

Awesome solution man! Loved it

You and Tech Dose are my go to for leetcodes.

@NeetCode

3 жыл бұрын

Yeah, Tech Dose is really good at explaining.

@MrYp-ds7sz

3 жыл бұрын

I found neetcode more clear and easy.

@rajdeepchakraborty7961

3 жыл бұрын

@@MrYp-ds7sz neetcode>techdose

@adithyagowda4642

2 жыл бұрын

@@rajdeepchakraborty7961 neetcode==techdose

really nicely explained man! thank you so much

Thanks. You're very talented in explanation.

Thanks mate, helped me a lot. Easy and very quick

Never stop making these please lol

Explanation 💯 clean !

Thanks buddy. You're literally the best

I think u can also solve it in O(nlogn) using a BIT or segment tree , where u will start from the end and see whether the current node range (l = i , r = i+nums[i]) summation is greater than 0 or r >= nums.size() , if so then update the current node to be one and continue

excellent video. I am learning a lot from your videos. Great work

You can also solve this problem by finding zeroes in the array and check if there are numbers before zeroes long enough to jump over zero. For example, if you see [3,2,1,0,...], you can instantly tell you're stuck at 0.

@tunepa4418

Жыл бұрын

that makes sense but this is still technically O(n^2) right?. consider this example [2,3,4,5,6,7,8,8,0,0,0,0,0,0,0,0,0,0,9]. For every non zero element, we have to do ~n/2 work to check if it crosses over all the zeros (i.e constant amount of work for every non zero element)

@archiecalder5252

11 ай бұрын

Working backwards, if we record the index of the first encountered 0, then the work required to check if an element crosses is constant. @@tunepa4418

@jffrysith4365

10 ай бұрын

@@tunepa4418 no, I think it would be the same way, except we just only start checking once we find a zero. I don't think it'd be any faster though... because you'd still have to check if each value is a zero...

@kavabanger88

10 ай бұрын

​@@tunepa4418if we are going from the end of list and found a zero, and then found a long enough jump to jump over it, we dont care about zeroes between first zero and jump position

@856dtrad9

8 ай бұрын

@@tunepa4418 O(n) class Solution: def canJump(self, nums: List[int]) -> bool: obstacle = -1 for pos in range(len(nums) - 2, -1, -1): if nums[pos] == 0 and obstacle == -1: obstacle = pos if pos + nums[pos] > obstacle: obstacle = -1 return obstacle == -1

awesome solution and explanation, thank you !

Excellent explanation and great video! Thank you!

@NeetCode

3 жыл бұрын

Glad it was helpful!

Please keep posting. Also I have a recommendation. Please add python in your title and thumbnail. You will surely reach more people. Ex. Jump game leetcode python solution #55 - Greedy approach explained Thank you for making these videos.💯❤️

nice video! loved the explanation

Thanks for your explanation!!

bro is a genius for coming up with that greedy solution THANK YOUU FOR THIS VIDEO!

thank you for clearly explanation this question!!!

Amazing explanation!

Really good video, I was stuck trying to do it in a DP way, but this is really clean! Also, the last line can be simplified to "return goal == 0" as this returns a boolean.

@bryanleow5024

Жыл бұрын

this is the dp solution in O(n^2), but only passes 77/170 on LC due to TLE. I think they really want you to use the greedy approach class Solution(object): def canJump(self, nums): dp = [0 for _ in range(len(nums))] # goal can be reached from itself dp[-1] = 1 for i in range(len(nums)-2, -1, -1): for j in range(1, nums[i]+1): # as long as one of your descendants (who u can reach from your current spot) can reach the goal, you can too if (dp[i+j] == 1): dp[i] = 1 # for better efficiency break; return dp[0]

I have started to binge watch Neet ode recently, 2024 is going to be awesome ❤

Fantastic approach love you needcode

One small nitpick in the implementation would be, that you could just return goal==0 (which evaluates to boolean value). Some could say using the ternary / if...else is more readable, but it's a matter of opinion I guess. Nevertheless, great video as always. I truly appreciate the way you explain algo, it's very clear imo. :)

Absolutely amazing!!!!

God bless you! 😀 This helped a lot.

Amazing! Thank you so much!

thats a unique greedy soln.! awesome!

The if condition can be modified to If nums [i] >= goal - i i.e if the value at the current index is greater than or equal to the difference between the goal and the index For better understanding 😄 thank you

@mattgolden100

Жыл бұрын

This really helped me conceptualize. Thank you!

@amitdwivedi9951

Жыл бұрын

plz exlain hy nums [i] >= goal - i ?

@anjanobalesh8046

Жыл бұрын

@@amitdwivedi9951 i didn't get your question ??

@gunadeepakpallikonda

Жыл бұрын

Mann!!! That's really helpful 👏🏻 🙂

Starting at the end, we always have to traverse the entire length of the array. Starting at the beginning is more performant as we can return true as soon as the current value is > last index - n.

superb video, thanks a ton

youre so good! thank you so much

brilliant solution!

I could come up with the DP memorization solution by myself, but got TLE, the greedy solution is optimal but unintuitive.

@freindimania11

5 ай бұрын

Had the same experience, DP with memo got me TLE, however DP with tabulation got through - although still slower than other submissions.

Fantastic! I can't believe that is so simple...Amazing!

Thank you for the video.

You can skip the last index check by initiating for loop len(nums) -2

Best explanation ever!

Very good explaination , thanx

youre amazing keep it up

Amazing. You made this incredible video explaining the problem and you end it with "return True if goal == 0 else False" instead of just "return goal == 0".

@Rajmanov

5 ай бұрын

Due to its more comprehensive nature, simplicity doesn't always equate to superiority.

great video

You are awesome man

Superb explanation

Thank you so much!

Alternative: 1) Start on the right 2) Go to the left, skip all non-zeros 3) On each zero, find any position on the left that can skip it - i.e. larger than the distance from it 4) If there's none, return false 5) If there's any, continue from there, to the point 2) 6) Return true when reaching the left end.

@Marcelo-yp9uz

2 жыл бұрын

I think that's no longer O(n)

Brilliant and elegant!

Great videos, can you also explain how you get the time complexities? I am not how sure how you got the n^n and n^2

@robalexnat

2 жыл бұрын

n^n happens because as for each index number explored, he recursively explores the indices reachable by it, however because he isn't caching it essentially we end up with a lot of repeated work like he mentions. A more visual way of thinking about it: [3,2,1,0,4]. I start with the 3 at index 0, and recursively call (which is the same as a Depth First Search stack implementation) each of the indices reachable (0+1,0+2,0+3) = (1,2,3), I start with index 1. It has a value of 2, same as before, I then have another recursive call to index 2 with value 1, until we reach 0. Now when the stack unfolds, BECAUSE we did not cache, we still end up calling all the branches as before. Meaning when we roll back to index 1 (the one that had value 2), we only explored the first of (1+1,1+2) aka index 2, but we didn't make a call yet for index 3. With Caching this is reduced significantly, and essentially becomes a n + (n-1) + (n-2) +...+1 complexity problem which, when we drop lower terms, we get as O(n^2). Hope this is more clear.

I love you Mr. NeetCode.

I had a different solution that worked as well. We only can't jump to the end if the array have an element 0 in it and this element isn't the last one. So I loop through every element and check if it was 0, if so, I loop backwards to check if there is an element that can jump over the zero, if so, I'll continue the loop, if not, I'll break and return false

@RS-vu4nn

2 жыл бұрын

He is doing exactly the same thing but more efficiently

Thank you!!

Thanks!

Nice one🎉

You are the best!

You made problem soo easy Sir!!!

Holy Moly , what a piece of cake.

amazing content

What software do you used to draw? on this scene 9:23

hi if it you could only jump a fixed length (e.g. if the number was 2, and you could only jump 2 spaces, not 1), could you still start from the end? Or must yoi use DP?

Great explanation!! but wont the for loop start from len(nums)-2 ???

Just fabulous 🤩🤩

loved the soln.!!

For those who have the question of why is always first element chosen as the next goal in greedy approach: There are mainly two types of questions you might be facing: 1. Why always chose the first element as next goal post 2. What if I don't get a further route afterwards by choosing the first element. What if I would have chosen other element that time and I would have gotten answer ( In this case you must have thought what if I couldn't have reached 1 in any way, I would have missed the potential answer of keeping 3 as the next goal.....) Answer: According to the solution, we chose 1 as our goalpost. In the back of our mind we know it can also be reached by 3. you think that I might get stuck on further exploring the path with 1. ** Take a closer look, my friend, if you can reach 4 from 3, you will also definitely reach 1 from 3 ( because 4 is farther away from 1). So while choosing the first element you have the surety that if there is any other potential answer beyond that index, that index could also be reached with that potential answer(in this case 1 could also be reached by 3 as 4 was reachable by 3). And thus you know that you will get the answer by choosing the first element. ** Hope this clears your doubt....

@case6339

Жыл бұрын

Finally someone answered! Appreciated.

for the greedy explanation, what gives us the right to shift the goal post to the FIRST item that can reach it? There can be multiple items that can reach the goal post that are more "left"? ie in [1,3,2,1,5] we shift the goal from 5 to 1 immediately upon encountering 1, instead of looking further to the left such as 2....why is this greediness guaranteed to produce the correct result?

Hey Neetcode, do you have any videos on your routine when your were leetgrinding? Like how many questions per day u were doing and how long it took u to complete the 75 questions list?

@kneeyaa

2 жыл бұрын

I targeted 5 q each day to achieve all in 15 days

@haroldobasi2545

2 жыл бұрын

@@kneeyaa that’s crazy actually

Very good video

this is so good

Amazing!

you are a boss !!

super Explanation

You'r genius!

Greedy part was fire 🔥

Optimization: Notice that you can always jump forward unless you run into a zero. Therefore, just check each zero (if any), and make sure there's a number to the left that's big enough to jump over it. It's a little more code, but much faster, especially if the array is huge but only has a few zeros. Here's my solution: start = 0 while True: try: pos = nums.index(0, start, len(nums) - 1) except ValueError: # no more zeros found break for ptr in range(pos - 1, -1, -1): if nums[ptr] > pos - ptr: break else: return False start = pos + 1 return True

Thanks

O(n^2) DP solution gives TLE, thats pretty sad considering the greedy approach may not be apparent to many in an actual interview.