LeetCode 55. Jump Game (Algorithm Explained)

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#coding #programming #softwareengineering

[@azfarzaidi513]

This is simply beautiful! You've really taught me something on how to think about a solution approach! Thank you Nick!

[@ravishekhawat5489]

That was some clean explanation Nick. I was aware of only the O(n2) solution. Keep it up.

[@yashsheth3033]

Absolutely loved the explanation and how you broke it down into smaller sub problems! Superb stuff!

[@nimanizky]

That algorithm explanation section helps so much

[@Harjotse]

i was strugling with the sol from last 3 days but you made it clear in just 10 min kudos for your efforts

[@satyamgupta6030]

nick my brother you don't know how much ur videos help me in building approach and solving a problem. Your approach are always better than others please keep on making such videos.

[@aditimangs]

@nick, Do you have a solution to Leetcode 45 with similar approach, where you start iterating from the right end of Index

[@vivekgr3001]

love the way you explain the code using dry runs

[@jamesgeng7815]

This is the best and easy-understanding solution I can tell. Thank you dude, excellent

[@mrhelloimstupid]

Hey, do you also have a video for jump game 2?

[@seshafermi5776]

Thank a bunch Nick, it was so much clearer after your explaining.

[@NickWhite]

how am i still covering the code...

[@user-sj3fp2xq2m]

Rookie mistake! Jk, no worries, still helpful.

[@sitaramakella8922]

best explanation I have seen for this problem !! thanks Nick !

[@thegreatsilence1081]

Hi Nick ..I did not get why you have comapre i+nums[i] >= lastGoodIndexPosition....why not just to compare nums[i]

[@palak2708]

Probably your best explanation! Great video.

[@bulianxile]

Love your video and the way you are talking dude!

[@xxsanyxx1]

Great explanation and beautiful code! Thanks Nick!

[@fadi.casual3796]

Can you explain the minimum number of jumps (aka jump game ii) required to reach the end. @Nick

[@Fapados123]

Now that's a simple and efficient solution, combined with good explanation. I didn't even think of going backwards, just tried to brute-force it recursively.

[@cindrasenareddy1929]

then why is leetcode is not accepting the solution . its giving TLE

[@69_kalesh]

cuz recursive approach TS is in exponential form , which is uh.. really bad.. you can see the question's constraints.@@cindrasenareddy1929

[@BadriBlitz]

Well the approach is so good and explained really well.Thanks Nick.

[@urvashichaudhuri9516]

Thanks for the explanation Nick. Nice approach. In case you have solved Leetcode 45 too (Jump Game II), could you share your solution?

[@DurgeshYadav-bc5nm]

This is actually quite neat. It only does whatever the questions asks.. nothing more. We don't really need to explore all possibilities when we only need to return whether or not it's possible to reach the last index. Cool solution this

[@rosonerri-faithful]

Nick since the return type asked is boolean,keep the return values as true and false,other than i and 0. good explaination. understood it

[@satang500]

thanks for the video, but your small rectangle bottom-left corner covers some of the code part/writing when you're explaining.

[@sarthakpatidar1099]

The explanation was awesome and you made it look easy. Thanks

[@satyamgupta6030]

another great solution, sweet and simple.

[@wigcrafter]

really love your video, how is ur interview going so far?

[@kirtipalve1772]

Best explanation, clean and concise. thanks!

[@t3techtilltoday861]

you are amazing bro, short and on-point. Perfect video :-)

[@ricardongh72]

Great strategy bro ! Sometimes those simple solutions can be hard to think of

[@hhjdkdnchidnd]

super easy to understand after watching your video!!

[@sumanmallipeddi5950]

Nice Explanation. But technically I don't think we have to start from last Index as we are moving from behind and we dont care whats the value in the last index as we have reached there. (On line 4, we can start from lastGoodIndexPosition-1 (which is num.length-2))

[@arishsheikh3000]

Last good index means smallest index from which path to last index is possible

[@sayoojvalsan4879]

why i + nums[i] >=. What is ">" doing here. Sorry could not understand that part

[@CricketThomas]

good explanation! you should make a video about how you may not need leetcode for a dev job

[@gu3566]

Love your explanation!

[@BronToLearnBornToTeach]

wow... the explaination, now this problem look really easy

[@coordinator3039]

Really having trouble with all the interview problems. I need an explanation of how I can understand these algorithms

[@dishagupta7446]

amazing explaination!!

[@duongvuhai1222]

the last index's value does not matter so we can skip it by doing nums.length -2 instead of nums.length -1 in the loop

[@shubhangishinde5544]

Will you make one video on Jump Game 2 ?

[@nileshpatil3710]

clean, neat and easy to understand.

[@johnnywilson3071]

This method seems a lot easier than trying to do it via DFS/Backtracking. You could also start the loop from "last_good_index_position -1" or nums.size()-2 as long as LGIP is initialized as nums.size()-1. It would mean one less loop required which probably makes minimal difference in terms of speed.

[@kellenstuart4698]

Yeah that's what I was thinking

[@rajeshkartha147]

Wow.. Thanks a lot. Very clear explanation.

[@user-us7rg4cd6p]

This is genius! Thanks a ton

[@SaifUlIslam-di5xv]

I had a gut feeling there was some approach I could solve this starting at the last index, but I was unsure of that.

[@stith_pragya]

Thank You So Much for this wonderful video...................🙏🙏🙏🙏🙏🙏

[@josephwu9455]

the 'i + nums[i] >= lastGoodIndexPosition' part is hard to grasp initially until using a test case '2 0 0', now it clicks. thanks

[@harshthakkar7503]

Nicely explained bro!!!

[@willturner3440]

How to know minimum number of steps

[@milenitrivedi7561]

nice explanation !

[@cocoandcairo]

Really good Explanation

[@ashishbarai1425]

Awesome 👍 Best in you tube one can find

[@saumya1singh]

awesome Nick

[@himanshusingh694]

I have solved 70 easy questions and 5 medium questions on leetcode.. still can't think of these solutions.. born with a freezed mind! :/

[@abhishekrajgeek]

Amazing! Thannks!!

[@etherioussanjudraganeel3163]

Thnx Jonathan

[@alokesh985]

Amazing stuff. This is why I love solving problems. What I thought was a pretty straightforward backtracking question turned out to be a reverse linear traversal

[@anonymousreviewer3816]

Thank You!!

[@myvinbarboza3038]

awesome Thank you

[@tejasnakhate]

Please do Jump game II and Jump game III

[@hamicartoon]

I think this will work only for 1st index, will not work if i want jump from other index. Ex a[]={2, 0, 1, 1, 4}, the code will true. where as it should fail when will start from index 1

[@pauljustin9583]

The question is you have to start from 1st index.

[@suzy1107]

it's == at the return statement, but overall great explanation!

[@carpettunnel8837]

Did you come up with your solution on the fly during the video or did you solve the problem ahead of time?

[@NickWhite]

I’ve solved this problem in a Coursera course I took before and it said I submitted a solution 10 months ago as well so I was pretty familiar with the problem

[@carpettunnel8837]

Nick White Felt slightly less dumb after reading your comment as my initial thinking while watching was to start from the beginning of the array...feeling dumb again after finding my previous comment was directly addressed at the end of the video. 🤣 Maybe try putting the video of yourself on screen right as it decreases the probability of it covering a line of code.

[@benh2908]

​@@NickWhite Hey, by any chance, do you remember the name of that Coursera course? And/or do you have any other courses you recommend? Thanks!

[@sachinshrestha2538]

@@NickWhite which coursera course did you took

[@tweissin]

Yes what was the Coursera course?

[@marq_8976]

I still don't understand how it works for the FALSE example.

[@Hav0c1000]

I hate this question... It looks like Graph Search, It looks like back tracking... and then the answer is just reverse linear sweep...

[@mohammedsadiq5729]

Nick Spiral Matrix lll 🙏

[@catmium7974]

Danke!

[@roycrxtw]

Brilliant.

[@sarscov9854]

Damn. I thought the question was asking that you can ONLY jump the number of steps at nums[i]. But it actually means you can jump nums[i] or less. The solution for my former knocks out almost all test cases, but it is the wrong answer. Apparently, you have to do backtracking.

[@kriswright5112]

your algorithm was O(N), but your rambling goodbye was O(N^2)

[@rupaldesai7098]

I din get how nums[i]+i is handling the logic

[@NickWhite]

nums[i] is how many indices you can jump forward and i is the current index so it basically just tells you the farthest index you can jump from your current position

[@rupaldesai7098]

@@NickWhite Thanks! Appreciate the brief explanation. Understood now!

[@0anant0]

@@rupaldesai7098 Sometimes it helps to use descriptive local variables int curMaxJump = nums[i]; if (i + curMaxJump >= lastGoodIndex) Sometimes, you can give descriptive names to the 'magic numbers' in code: final int EMPTY = 0, WALL = 1; final int X = 0, Y = 1; if (cell[X] == dest[X] && cell[Y] == dest[Y]) return true; ... if (board[newRow][newCol] == WALL) continue; This will help when you are reviewing your own code at a later date! :-)

[@madanmohanpachouly6135]

nice

[@shivamshukla9277]

Eminem - Im the fastest rapper alive Nick - Hey nick white here fndfsojdbfldmf.........

[@upol92]

This will not work for many cases including [2,0,0]

[@DurgeshYadav-bc5nm]

It works

[@prathyushagade3501]

you're too good at explaining it!

[@abhilashgoyal2234]

You can start the loop from "nums.length - 2" also. That will also be logically correct

[@Hav0c1000]

not if nums[-2] is 0

[@vaidehiyadav5507]

what about the corner case then .where there is single entity in the array.

[@100bands]

Never felt more stupid...damn it

[@vishalagrawal246]

please dry run this testcase [2,0,0] with your algo... thanks :)

[@crapid5252]

returns True

[@Sanyat100]

Comon bro. Use JavaScript ! I

[@LordLobov]

This problem is not intuitive at all :((

[@koownmyo]

the hate I have for this problem is unreal

[@krishnavar7219]

srikrishna

[@krishnavar7219]

6:42

[@mannana8550]

Please dont cover text with your video. Thanks.

[@aronblanche]

return nums[nums.length-2]!=0; why not this one liner?