🚀 neetcode.io/ - I created a FREE site to make interview prep a lot easier, hope it helps! ❤
@dineshkumarkb1372 Жыл бұрын
All your videos are a treasure . Every single one is worth rewatching during interviews. Never ever delete these videos or stop uploading new ones.
@naive-fleek74209 ай бұрын
during interviews??
@dineshkumarkb13729 ай бұрын
@@naive-fleek7420 I meant during prep for interviews. It’s implied dude.
@allen7243 жыл бұрын
This is a great explanation. I like this method better than LeetCode's published solution! It is more intuitive for me. Thanks and keep up these videos.
@lakshmanprabhu6722 Жыл бұрын
Same here!. Went through a lot of solutions, but this is just gold.
@ghdshds189910 күн бұрын
this explanation was actually dog
@matthewtang14903 жыл бұрын
Please don't stop making videos :) I just binged your DP playlist in 2 days
@NeetCode3 жыл бұрын
Wow, I bet you would nail any interview now! Thanks for the kind words
@crimsonghoul89835 ай бұрын
2 DAYS????
@aniketyadav6247 Жыл бұрын
The below code also works : 1.) Traverse the entire nums array. On each i-th iteration, update the farthest_jump to the max of current value of farthest_jump and i + nums[i] 2.) If i is equal to the current jump we have completed the current jump and can now prepare to take the next jump (if required). So we increment the jump by 1 and set curr_jump to farthest jump. 3.) If that's not the case then do not update the jumps variable and the curr_jump variable since we haven't yet completed the current jump. 4.) In the end of the traversal you will get the minimum jumps. Hope this helps :) def jump(self, nums: List[int]) -> int: farthest_jump = 0 jump = 0 curr_jump = 0 for i in range(len(nums)-1): # Find the Farthest Jump farthest_jump = max(farthest_jump, i + nums[i]) # it means we have made the jump if i == curr_jump: # Point curr jump to the farthest jump curr_jump = farthest_jump jump += 1 return jump
@venkateshnaidu21332 жыл бұрын
Amazing solution, loved it! Here is a minor tweak to handle an edge case (no possible path) int minJumps(int arr[], int n){ // Your code here int l=0, r=0; int jumps = 0; while(r < n-1) { int farthest = 0; for(int i = l; i r) return -1; jumps++; } return jumps; }
@akshatsamdani Жыл бұрын
Thanks for posting this. I was thinking about the same
@PorkBoy695 ай бұрын
The test cases are generated such that you can reach nums[n - 1].
@PRAVEENKUMAR-fi1wu2 ай бұрын
No need to do this. It is said in question. "We can always reach to the end" @@akshatsamdani
@acala1273 жыл бұрын
This is the most elegant and clear explanation I have seen for this problem. Thank you!
@karthikinamanamelluri720810 ай бұрын
Great explanation!! Even after this I was confused why the while condition is r < len(nums)-1, and not r < len(nums). I thought why can't we can change it to r < len(nums), and return res-1. This explains the algorithm better, since the result we are finding from the loop is the no. of blocks, and the no. of jumps is one less than no. of blocks. But this solution is not working for all the cases.
@prashantgupta61603 жыл бұрын
please continue this series, you are born to teach coding to other people.
@heroicrhythms8302 Жыл бұрын
if we come to the middle of the array and can't reach the end anymore, that means we have encountered a 'zero'. then we should return -1. implement a check saying (if l>r: return -1)
@licokr Жыл бұрын
Crazy. This channel explains coding solutions in the easiest way. It saves my life.
@ng.manisha9 ай бұрын
You are literally a savior! I have my google interview lined up soon and all your videos actually teach me tricks how to think when faced with a problem!
@iamnoob7593Ай бұрын
did u make it into Google?
@protyaybanerjee50513 жыл бұрын
Man, we need more videos. Great production quality :)
@ma_sundermeyer Жыл бұрын
Nice simplification of the BFS.. I had a similar idea but stayed with the conventional deque implementation: q = deque([(0,0)]) max_i = 1 while q: i,num_j = q.popleft() if i >= len(nums) - 1: return num_j for j in range(max_i, nums[i] + i + 1): q.append((j, num_j+1)) max_i = max(max_i, nums[i] + i + 1)
@soumyajitganguly25939 күн бұрын
Great explanation man. Easy to generalize when jumping to last value is not possible.
@denshaSai2 жыл бұрын
whats the reason that it is "while r < len(nums) - 1:" not just "while r < len(nums) :"?
@arunavbhattacharya33532 жыл бұрын
I fail to intuitively understand why do we need to iterate till n-1 instead of n
@eternalmeme79202 жыл бұрын
@@arunavbhattacharya3353 It's because, 1) All no. are positive, therefore if we touch the n-2 element, i.e. r=n-2, then we are iterating from l to r(inclusive), if we iterate at r=n-2, then since all no. positive, farthest will definitely be greater than >=n-1, therefore r
@visase20362 жыл бұрын
One of the main reasons for this is , Since we are accounting the 1st jump at position 0 , we are not considering the last element to calculate the answer ie [2,3,1,1,4] We re incrementing ans+1 by taking 2 into account , which is not necessary if you work logically and since we are accounting that as one jump we are ignoring the last element!
@pruthvirajpatil7798 Жыл бұрын
Simple - Because if r is at len(nums)-1, we would have achieved the goal. No need to proceed ahead.
@megavoltism2 жыл бұрын
It's funny how he always colors-in his arrow heads Anyway, really cool insight about BFS!
@ge_song57 ай бұрын
question: why do we have to stop by the last_index - 1? while r < len(nums) - 1:
@ssvivek48 ай бұрын
I don't think if I'll ever see a better explanation to this problem. Kudos man!
@ekanshsharma1309 Жыл бұрын
why we write r < nums.size() - 1..... not just r< nums.size()??
@RiteshThakur-o6r3 ай бұрын
Because if r is at nums.size, then it has to terminate because it has reached the final index.
@pranavmaiya438626 күн бұрын
hey did you get placed?
@sniper3243 жыл бұрын
Your videos and explanations are some of the best on KZbin, really great stuff man, keep it up!
@adityagoyal42996 ай бұрын
The content of this channel is priceless. Been binge watching your videos
@darkexodus6404 Жыл бұрын
Your explanation is too good! Understood it clearly.
@loia5tqd0012 ай бұрын
9:28 Why while r < len(nums) - 1 not while r < len(nums) ?
@abibhavankumar2522Ай бұрын
The array positions starts from 0 to length of array - 1. If you go till len(nums) then you will be considering an extra element that is not present in the array. Hope that clears your doubt!
@divyanshkhetan Жыл бұрын
This is a great approach. I especially liked how you related it to the concept of BFS. It helped in visualizing the approach so much!
@ahmad38233 жыл бұрын
I am new to serious coding but great job! this took me some time and now way close to this neatness level!
@akshatgupta1073 жыл бұрын
Please please man, I love your channel so much, you have never disappointed me. Make a list of important greedy problems please
@kashishsharma68092 жыл бұрын
Oh man I unnecessarily used queue like in bfs. 🤧 I implemented exact bfs to find hops.. But using the interval instead of queue was awesome 😎
@commandernorton882 жыл бұрын
Drinking game: take a sip when he says "Right?"
@bahaaiman85883 жыл бұрын
This is the only one that I understood. Thanks a ton !
@NeetCode3 жыл бұрын
Happy to help!
@sagarnair90212 жыл бұрын
one line should be added after updating j i.e if j
@yathartharana7956 Жыл бұрын
Searching the whole day and find this solution the best one 🙌🏼
@sauravchandra1017 күн бұрын
Loved the analogy to BFS.
@stith_pragya11 ай бұрын
Thank You So Much for this wonderful video................🙏🙏🙏🙏🙏🙏
@Gnaneshh3 жыл бұрын
One hell of an explanation ! Thank you
@arunraj25272 жыл бұрын
I love his patience and way of talking through the problem.
@VarunMittal-viralmutant2 жыл бұрын
The standard solution for this question is like the LIS variant which is O(N**2). And that gives TLE on LeetCode I think the solution described above works only when it is guaranteed that the end can be reached, else it fails. Correct ? Modified to take care of unreachable cases: def find_shortest_jump_path(jumps): l, r = 0, 0 i = 0 res = 0 while l = len(jumps)-1 else -1
@God0fSloth7 ай бұрын
Thank you very much! Your code solved my problem. But what's the variable ( i ) used for? why are we increasing it?
@VarunMittal-viralmutant7 ай бұрын
@@God0fSloth That looks like some junk code, not used in final solution :)
@venkatsaireddy14129 ай бұрын
why left is re-initialised with only right +1, it can be moved to more than right +1?
@user-pn9sr2rq3z6 ай бұрын
great explanation and video! Espcially coloring. Unfortuntatelly does not work on leetcode any longer , giving timeout exceeded error.
@chandrachurmukherjeejucse58167 ай бұрын
Your explanation and drawing is just awesome.
@dennisg36836 ай бұрын
Why does the first time iterating through the array the for loop starts at 0 and goes to just the first index (0 + 1)?
@meghaldarji5983 жыл бұрын
The best solution there is for this problem. I am saying after watching all other videos on this problem.🙌🙌
@vincenttung-or3ns10 күн бұрын
This is freaking genius!
@rohanb95122 жыл бұрын
Nicely explained the intuition. Exactly wat i was looing for. Probably the best explanation in YT
@ohmegatech6666 ай бұрын
Just want to clarify that this is still a dynamic programming solution. That's because this solution is a moving window algorithm which are examples of dynamic programming. That is because they involve breaking the problem down into sub-problems and finding an optimal answer to those sub-problems, thus finding the optimal answer to the main problem. In this case the main problem is optimizing the fewest jumps to get to the end. The smaller sub-problem is finding the max jump within the current window.
@themagickalmagickman Жыл бұрын
I used a Dijkstra's approach to solve the problem, but this is a simpler and quicker answer... wow.
@PrathamMittal-t8m Жыл бұрын
Best channel for explaining the leetcode problems to a dumbo like me
@tobito__3 жыл бұрын
How is this solution a O(n)? Isnt there a for loop inside a while loop which makes it a O(n^2)?
@abhinav-lq9msАй бұрын
I think it's O(n) since we are always updating l to be r+1, so it won't again traverse the same value
@krishnavamsichinnapareddy2 жыл бұрын
You simply nailed it. Love from India ♥️
@hwang16075 ай бұрын
how would you come up with this in an interview, I could come up with the DP solution, but its n^2 complexity
@pixelbyteee3 ай бұрын
isn't this a O ( n^2 ) solution? as there are two loops? ( u mentioned it is an O ( n ) earlier )
@697Alok3 жыл бұрын
What an explanation. Loved it :)
@kevintran61022 жыл бұрын
This explanation is crystal clear. Thank you!
@MrExamer2 жыл бұрын
great explanation on the BFS mind behind the problem
@aravinda1595 Жыл бұрын
You are so good I just need to watch the explanation part and boom ! i can write my own code
@amarrajeev2903 Жыл бұрын
I think the edge cases also must be handled in the code. Suppose if the Algorithm could not find the minJumps it should return -1 as such. Thoughts on this?
@markolainovic Жыл бұрын
No need, in the description it says that the solution existence is guaranteed.
@kewtomrao2 жыл бұрын
Awesome explanation!! I did the regualr BFS and got stuck in a MLE error. Now i know my mistakes.
@xinniu31452 жыл бұрын
Thank you for sharing. We can put farthest=0 before the while loop right?
@shantanukumar40812 жыл бұрын
Great Explanation !!!
@ardhidattatreyavarma53377 ай бұрын
what a beautiful piece of code
@shashijais7892 жыл бұрын
Again Superb solution, which I was looking for! Thanks for this explanation.
@huatsai433 жыл бұрын
Thank you for the clear drawing explanation!
@NeetCode3 жыл бұрын
Thanks, happy it was helpful 🙂
@VY-zt3ph Жыл бұрын
Hello Neet!! You didn't mention the time coplexity here. Can uou please explain what was the time complexity? It seems it is n2 because of two nested loops. But I am still not sure.
@111Apurva11 ай бұрын
it's not n2 it is O(n) after calculating farthest you are not considering the element again (starting from r+1), In worst case of all 1's it would still go just once over each element
@aishwaryaranghar33853 жыл бұрын
i tried this code but it somehow throws tle. would be glad if you let me know a little updation in the code. thanks
@sauravsingh71573 жыл бұрын
It's giving TLE because this code goes in infinite loop in cases where we can't reach the end of the array. Ex- 1 2 0 0 6 7 - for this test case loop will never break. Here is a working code : def minJumps(self, arr, n): res = 0 l= r = 0 flag = 0 while r < len(arr) -1: farthest = 0 for i in range(l,r+1): farthest = max(farthest,i + arr[i]) l=r+1 r= farthest if l > r: flag =1 break res+=1 if flag ==1: return -1 return res
@shantanushende63 жыл бұрын
really really good! Felt like one comment did not do justice to the level of simplicity!
@vishalmishra70183 жыл бұрын
Really neat code! Nicely done and explained.
@Cld1363 жыл бұрын
Excellent explanation. Much Thanks!
@rabbyhossain615010 ай бұрын
Memory Complexity: O(n) class Solution: def jump(self, nums: List[int]) -> int: steps = 0 visited, q = set(), collections.deque() q.append(0) visited.add(0) while q: length = len(q) for _ in range(length): idx = q.popleft() if idx == len(nums) - 1: return steps for pos in range(1, nums[idx] + 1): if (idx + pos) not in visited: q.append(idx + pos) visited.add(idx + pos) steps += 1
@rohands9195Ай бұрын
Why is the time complexity of DP step = O(n^2)?
@supervince1102 жыл бұрын
You always have the simplest solution!
@Kushagra_212 жыл бұрын
one of the best solution in internet for this question. Thanks a lot!!
@ankurtiwari56642 жыл бұрын
i don't know python still i watch all ur videos Next level explanation of every approach
@venkatrampramod7978 Жыл бұрын
Your solutions are absolutely brilliant. I love the way you break down the solution with diagrams.
@gokulnaathbaskar98082 жыл бұрын
Nice, looking at BFS for the first time in an array.
@francescosacripante1661 Жыл бұрын
Correct me if I'm wrong, but couldn't we solve it by using Dijkstra algorithm? I mean we could create a graph and with this search for the shortest path to the end
@shreshtashetty29852 жыл бұрын
Nice solution, but how do you come up with this in an interview? In general, how do you tackle a greedy-solution problem?
@darshansimha21662 жыл бұрын
Nice video, I understand the solution. But I am having a tough time understanding the complexity for an array of all 1's wont the for loop inside run for every iteration in the array, so won't that make it O(n^2)?
@divyasri94322 жыл бұрын
I have got the same doubt forloop inside while loop will give time complexity O(n^2) how does this become optimal solution?
@bmusuko2 жыл бұрын
@@divyasri9432 I think the key here, we shift the left pointer to r+1 so we only inspect each element in the array once, it means it is o(n)
@minhthinhhuynhle91032 жыл бұрын
2 loops doesn't mean O(N^2)
@user-2802cvsfkj Жыл бұрын
this is nuts dawg, who allowed this to exist. im literally shaking and crying rn, wtf.
@amrutaparab49395 ай бұрын
Thank god for you!
@DivyaSingh-bl4cj2 жыл бұрын
Best explanation channels for python.
@amanshaukat11052 жыл бұрын
This is a great explanation. Very intuative. lets say it is not guaranteed that answer will exist, and we have to return -1 in such case. How could we handle this. Please help.
@allwell85709 ай бұрын
You simplified it!! Thanks
@algorithmo1343 жыл бұрын
@NeetCode Can we use a heap to find the maximum steps we can jump instead of looping over all the number of steps we can take and then finding the maximum?
@oogieboogie7028 Жыл бұрын
TC of heap is n.logn. This approach is O(n)
@thevagabond85yt Жыл бұрын
I programmed by modifying Greedy Approach of Jump Game(I) : ``` class Solution { public int jump(int[] nums) { int N= nums.length, goal= N-1, jumpCount= 0; while(goal >0) { for(int i=0; i= goal) { goal= i; jumpCount++; break; } // if }//for }//while return jumpCount; } } ```
@yashjain1011 Жыл бұрын
such a nice explanation. This video was so great. You earned a subscriber.
@avishekarora2 жыл бұрын
Best Solution explaination, thanks
@emekaobasi4765 Жыл бұрын
why till len(nums)-1, shouldn't it be len(nums) ?
@peng94310 ай бұрын
I don't get it neither... starting to hate myself when I can't understand something people don't think it is worthy to explain...
@dent18082 жыл бұрын
clearest explanation ever
@trantung20133 жыл бұрын
Really elegant solution.
@gulershad43752 жыл бұрын
def jump(nums): jump = 0 farthest = float('-inf') end = 0 for i in range(len(nums)): farthest = max(farthest, nums[i]) + 1 if i == end and i != len(nums) - 1: jump += 1 end = farthest return jump
@sunshine-mc2oi2 жыл бұрын
Thank you for the awesome video. It's super easy to understand. Is there any chance you can make a video about 1326. Minimum Number of Taps to Open to Water a Garden and Video Stitching, they seem relevant to this topic. Thank you so much.
@TheLaidia3 жыл бұрын
very clear explanation. Thank you!!
@alonalon87942 жыл бұрын
at the end of the algorithm r might point to an index which is lesser than the last index?
@anupamdubey57362 жыл бұрын
Best Explanation! Thanks
@vasumahalingam51629 ай бұрын
brilliant but I wouldnt be able to solve this by myself in a coding interview. Very clever indeed.
@LeCharles-bw4wp9 ай бұрын
Thanks.But I suppose the time should be O(n squares)
@MsSkip603 жыл бұрын
Awesome content as usual :)
@yifeicheng91022 ай бұрын
what's the time complexity of this solution?
@swathiayas3 жыл бұрын
Really good videos! Been watching alot of your videos lately! Thank you making such awesome videos!