House Robber

[@KevinNaughtonJr]

HOW DOPE IS MY SKI MASK

[@jlecampana]

Love the Mask! and well, this one is a very nice and do-able DP problem. I was wondering if you could do #276 - Paint Fence? I couldn't do it for the life of me, and it's labelled as "Easy", so maybe I'm missing something and you can enlighten me. Thanks in Advance Kevin!

[@KevinNaughtonJr]

@@jlecampana Haha thanks and don't pay attention to the labels they're all subjective! I actually think I already have a video on that problem if you wanna check it out! And anytime, thank YOU for your support!!!

[@jlecampana]

@@KevinNaughtonJr Oh I see it, you have done a video for LeetCode #256 - Paint House, How could I have missed it! I will see if I can actually solve the problem before watching your video. But I do think you should definitely check out #276, I could be wrong but I think it's Harder than most DP regulars. Have a nice day!

[@KevinNaughtonJr]

@@jlecampana I'll check it out thanks for the suggestion!!!

[@PrathamMantri]

Thanks Kevin for the best explanation of DP problem. I could use this example as a start for solving DP problems.

[@aydasu]

love the dp questions. i feel like i am starting to get it. thanks Kevin!

[@KevinNaughtonJr]

So so so so SO happy to hear that!

[@alperozdamar517]

I saw this question in real interview and be able to do it. Thank you Kevin. :)

[@KevinNaughtonJr]

No way that's amazing anytime :)

[@bingo7137]

Purely Awesome! i'm kinda struggling with understanding the dynamic programming problems and this is the best explanation out here. Thank you once again for all your help.

[@ahasunos5914]

Kevin, I've been following your tutorials since past couple of weeks. You don't need to doubt why do you have subscribers. Thank you for all the concepts.

[@priyanshmathur3010]

Bro, all those memes and explanations were lit!! Great Video :D

[@jaysaxena185]

Excellent Solution Kevin, cleared the concept of dp, through this explanation. Too Good, short, and on-point.

[@jaidevsingh1009]

The clearest video on this problem, thank you!

[@svddwd]

Very nice explanation and approach . Thank you.

[@mohithguptakorangi1766]

now, that's how you explain "SHIT" in a "BEAUTIFUL" way

[@mkhanyisigamedze4807]

This took me 2 hours and I'm reminded of how understanding key concepts is so efficient. Love your videos

[@KevinNaughtonJr]

thank you!

[@alammahtab08]

Very well explained. Below is the code in case if someone wants to try out. Both, with the dp[] array and without dp array. With dp[] array : O(n) Space class Solution { public int rob(int[] nums) { if(nums == null || nums.length == 0) return 0; if(nums.length == 1) return nums[0]; if(nums.length == 2) return Math.max(nums[0], nums[1]); int[] dp = new int[nums.length]; dp[0] = nums[0]; dp[1] = Math.max(nums[0], nums[1]); for(int i = 2; i < nums.length; i++) { dp[i] = Math.max(nums[i] + dp[i-2], dp[i-1]); } return dp[nums.length-1]; } } Without dp[] array : O(1) Space class Solution { public int rob(int[] nums) { if(nums == null || nums.length == 0) return 0; if(nums.length == 1) return nums[0]; if(nums.length == 2) return Math.max(nums[0], nums[1]); int maxBeforeTwoHouse = nums[0]; int maxBeforeOneHouse = Math.max(nums[0], nums[1]); int maxAtI = Math.max(nums[0], nums[1]); for(int i = 2; i < nums.length; i++) { maxAtI = Math.max(maxBeforeTwoHouse+ nums[i] , maxBeforeOneHouse); maxBeforeTwoHouse = maxBeforeOneHouse; maxBeforeOneHouse = maxAtI; } return maxAtI; } } github.com/eMahtab/house-robber

[@selinadu1579]

Awesome solution. Plz plz keep doing these tutorials. Love them!

[@OlsZn]

I love how you have the perfect camera window to block the previous tries of this question ;)

[@loirobin]

@Kevin your solution explanation is simple and crystal clear, thanks a lot

[@objectivesworld31]

Hi Kevin nice explanation of dp. but we are sequentially processing so there is no need of dp array. if we change logic like this int first = nums[0]; int second = Math.max(nums[0],nums[1]); int ans = 0; for(int i=2; i

[@hookedbeans]

This is brilliant. Thank you!

[@studyaccount794]

We can improve this solution even further. We can use 3 variables instead of using a complete new array. Just like we do in fibonacci. Code- public int rob(int[] nums) { int loot1 = 0; int loot2 = 0; for (int i = 0; i

[@wheresthebeach0138]

Hey Kevin, thank you for this! IMO, it's your best video to date :) I think you took the perfect amount of time to explain the logic before diving into the code; really clicked for me. Thank you!

[@NehaJoshi-x5n]

very intuitive, thanks!

[@SaumyaSharma007]

Thank you so much Sir 🌟 Awesome explanation out there. Huge love and Respect from India 🌟🤗

[@ENGCS_JaiSaxena]

that is unbelivable , how are u able to solve so easily dude.great work.

[@wakita]

Thanks for doing the dynamic programming problems, I've been trying various difficulty ones for a while, and one of the things thats helped me the most has been watching your Leetcode videos, along with BackToBack SWE and Tushar Roy. I did something pretty similar to you in my Javascript solution, but did it in place, modifying the original input array. I think this could also be done in constant space and not modifying the input by using a variable to represent max up to i - 1 and max up to i - 2. Javascript solution ===== var rob = function(nums) { if (nums.length === 0) return 0; if (nums.length === 1) return nums[0]; for (let i=1; i

[@georgiossamaras5063]

You don't need the if statement for the 2 houses case. Nice explanation, thanks!

[@killersdeat0]

Great intuition! A lot of people only keep 2 numbers (value if you robbed the house vs value if you didn't) and alternate saving the variables but I think that building the dp table is a lot more intuitive. We can also simplify the space complexity of the solution to be O(1) once we see the concept of building the table out in a real interview scenario

[@KevinNaughtonJr]

Thanks and definitely very good point!!!

[@vaishnavipamulapati9920]

i was struggling with this problem for a day 😔 but your explanation was very intuitive! thank you!

[@hyonsoo79]

Very easy to understand. Thank you!

[@KevinNaughtonJr]

Anytime Kevin!

[@mp0157]

Thank you Kevin!! Good and simplified explanation! :) This helped!

[@AmolGautam]

Thank you for the explaination.

[@economicriskcapitalmodelpr9849]

This is my first visit to this channel and it's really impressive.Thanks for sharing your knowledge !! I am going to watch at least one video everyday.

[@KevinNaughtonJr]

Anytime and thanks for the support!

[@DhruvPatel-kg5ut]

Awesome bro. This is the 3rd video I saw for this problem and now I actually could figure out the solution. Thanks.

[@KevinNaughtonJr]

Anytime Dhruv happy to hear the video was helpful :)

[@ichdiegross]

You did a great job at explaining the concept properly!

[@KevinNaughtonJr]

Thanks Pranav!

[@crimsoncad3230]

This is the 1st dp problem for which I paused the video and got the DP logic on my own. This should be the 1st video to watch when you start learning DP. Initially I thought of 2 independent loops that calculates the sum of all odd and even plce elements, then take the max out of it. But this seems more optimized and useful for new concept.

[@vaichegodas]

you cant use 2 for loops to solve the problem

[@shredchic]

I tried the odd/even thing as well first, but it will miss the max in some cases. Consider [8, 1, 9, 2, 5, 20] With checking odd vs. even, the max would be 23. However, the best robbery is 37. :)

[@hitec1691]

Nicely explained. Thank you

[@Slumpicus]

You're killing it, dude. Keep it up!

[@KevinNaughtonJr]

Thanks Grayson I really appreciate it and don't worry I'm not going anywhere! New video is uploading now so get ready to check it out. Thanks so much for your support!

[@ridhwaans]

the confusing part is how the dp array carries over the largest sum without overlapping sums and maintaining non adjacency

[@humblecoder9119]

Super explanation. Good to understand DP like this.

[@AshishSingh-dn8wb]

We actually don't need to check if nums.length==2 in the beginning. We are taking care of that in the loop itself. Anyway, loved the explanation. Love your videos man!

[@tirthjayswal9895]

Really Good Explentation

[@AliMehrpour-Volcano]

Crystal clear and neat explanation, good job as always 👍

[@mp0157]

Very well explained intution! :) Thanks, this was of immense help!

[@KevinNaughtonJr]

Mihir Phatak thanks Mihir! Happy to hear the video was helpful :)

[@vashi1989]

simple and nice explanation.. i dont think we need an extra array.. i tried with the provided nums arrays and it worked.

[@lazycoder1910]

The explanation was really amazing

[@ChocolateMilkCultLeader]

Bro this one was genius. Hats off for a great solution

[@KevinNaughtonJr]

devansh thanks dude

[@rajdipdas9413]

this problem can be solved by including first number from first index or excluding first number from first index and continue for the next indices.

[@datajunkie3427]

The question starts with: You are a professional robber planning to rob houses along a street... That's a good start

[@kairu86]

Very helpful. Thanks!

[@renetchi]

Why do you have subscribers? lollll because you're awesome!

[@Dorin-Baba]

I think you got a dude moving in the background =)

[@adityapaithon6499]

yea creeped me out at 5:18

[@oneautumnleaf47]

Any referrence for dp? Where can I learn that? They don't teach it at school. lol

[@saulgoodman980]

My Python solution: def rob(nums): dp = [0, 0] # having 0s instead of nums[0] and max(nums[:2]) handles cases where len(nums) is 0 or 1 or 2 for num in nums: dp.append(max(dp[-1], num + dp[-2])) return dp[-1]

[@yv6358]

Oh! Hey when did you start programming?

[@saulgoodman980]

@@yv6358 It's been 5 years since I started (in school), but I've become serious only from the last 1.5 years

[@yv6358]

@@saulgoodman980 What happened to your career in law?

[@saulgoodman980]

@@yv6358 Oh xD Well, you'll know in BCS S06 ;)

[@yv6358]

@@saulgoodman980 waiting for it

[@manulscode]

The robber is already sleeping in the background.:D

[@KevinNaughtonJr]

hahahah

[@theanonymousfoodie8906]

this question was asked in my interview and I'm not able to solve that, now it looks very simple to me. Dammmmm!!!

[@ecea044gauravgogoi2]

so nicely explained

[@Vikasslytherine]

How to identify if a certain problem's going to need dynamic problem to solve?

[@cam5862]

1. great understandable DP explanation 2. love your freeze frame gold esp at 1:22 (that's some field theory sh*t I had to learn for cryptography 🤣) thanks for your helping me prep for my Google interview next week!

[@Dorin-Baba]

So, i guess you're a googleer now? :) Hope yes

[@cam5862]

@@Dorin-Baba unfortunately not :( I'm an Amazonian tho!

[@peterdinklage3856]

Awesome explanation mate!! thanks a lot!!

[@KevinNaughtonJr]

Swapnil Shukla anytime happy it was helpful!

[@seal0118]

how do results not overlap, is there proof to this?

[@PrashantNigam]

An improvement over this solution. Same as for Nth Fibonacci number. Without DP array. // T/S: O(n)/O(1) public int rob(int[] nums) { if (nums.length == 0) return 0; if (nums.length == 1) return nums[0]; int twoBack = nums[0]; int oneBack = Math.max(nums[0], nums[1]); int curr = b; for (int i = 2; i < nums.length; i++) { curr = Math.max(nums[i] + twoBack, oneBack); twoBack = oneBack; oneBack = curr; } return curr; }

[@jagrutijamba9093]

Hi Kevin, nice try. But your alogrithm would not return the correct answer in the following case : [2,1,3,4]. The answer should be 2+4 = 6. As both the locations are non-consecutive and lead to max loot. Hence, I tried solving this problem with recursion instead.

[@ayanagrawal]

Update : This is in medium now.

[@RohanDaDev]

Okay, im not sure if its this simple or not, but wouldn't the question devolve into checking whether the sum of the even numbered indices is greater than the odd? the only possible combos are all the even numbered indices and odd numbered indices. But i don' think its that easy

[@arpitverma8060]

Awesome explanation of this question by a bottom up approach ..Can you explain it by a Top down ?

[@shubhammishra1225]

Now this question become Leetcode medium. Still Dope mask and video.

[@sandeep-rai07]

Awesome explanation

[@bhargavlokinindi1895]

You are awesome man, Neat & clear, keep it up bro

[@Coolguylht]

I notice that you didn't use length + 1 for dp array this time, but for some other questions like number of coins and decoded ways all used length + 1 may I ask why this is otherwise?

[@FrankLi92]

Do we need to create a new array? Can we overwrite the current array for O(1) space?

[@brijeshtiwari1357]

Can we sort this array first and then start adding non adjacent elements from n-1 ?

[@christianrascioni8178]

If you sort the array first, you lose the original order and don't know what houses are adjacent anymore

[@shubhoch2368]

Hey, first of all I really like your videos the way you approach a dp problem makes it so soo simple. Thank you! Can you please make a video on Leetcode problem number 152, I just want to know how you will approach that problem in case you have time to do so!

[@Plopcorn02]

Kevin how many years of experienced have you had as a coder for using Java. Im intrested.

[@Matt-xq6ow]

I'm confused by the solution, can someone explain? I'm stuck understanding: dp[i] = Math.max(nums[i] + dp[i-2], dp[i-1]); Doesn't this mean we possibly have the chance of selecting an adjacent house instead of every other house?

[@ishikanagar3236]

loved it!

[@mayankdixit401]

My greatest achievement was that I was able to find that it is a dp problem. Now I can relate to the solution, thanks :)

[@KevinNaughtonJr]

Nice!!!

[@marlegagaming1274]

man, you made it look so easy...

[@vkRonaldo97]

can someone explain how this works with [4,1,1,4]

[@tomiwaolasoko7291]

Thanks, great solution!

[@yicai7]

Omg I really like ur bgm at the beginning!!

[@KevinNaughtonJr]

thanks Yi!

[@anurag1908]

Just WOW!!!

[@我願榮光歸香港-k5o]

Thank you! Really explained my struggles.

[@tavleenkaur6941]

Hey Kevin, your videos are very insightful and helping alot of people like me prepping up for the interviews. One thing I’d like to point in this question is that Is DP even necessary here? The first intuition that hit my mind when I read the question was to take the sum of the entire array (sum) and sum of all the alternate elements beginning from 0 (i.e. element at index 0 ,2, 4, till n-1 or n-2), lets call it X. The answer should be max(X , sum-X). Please let me know if there is something wrong in this approach ?

[@darkcaper703]

[2,1,1,2] -> u need to rob 1st and last to get maximum

[@sunittechjourney9640]

What did you do on line 13? int[] dp = new int[nums.length] ? What is that?

[@lazycoder1910]

It's java 😂

[@AhmadHassan-de9xo]

What if my sack where i put the money has a certain weight, and with each house's money there is a weight bounty i have to take into consideration, how would the program change ?

[@HarinathSrinivas]

Hey, the question you have solved is, "maximum sum in an array such that no 2 elements are adjacent". I was asked this question in an interview, tell me how to solve this. Given an array and an int value K, find Maximum sum of numbers, such that no 2 elements are adjacent and sum not greater than K.

[@RohanDaDev]

damm

[@CoolnesXcore]

If you want to save some extra space, we can just add the elements to the nums array. :)

[@masoomraza5155]

Thank you you are a life saver

[@ayeshaadhikari6123]

Thanks a tonne Kevin!

[@GURUYATHI]

Great Solution..!

[@daipayanhati2347]

strange that the the recursive overlaping subsets gives a heap overflow error; int Max_Rob(vector& nums,int i,int * dp){ if(i>=nums.size()){ return 0; } if(dp[i]!=0){ return dp[i]; } return dp[i] = max(Max_Rob(nums,i+2,dp)+nums[i],Max_Rob(nums,i+3,dp)+nums[i+1]); } int rob(vector& nums) { int dp[120] = {0}; return Max_Rob(nums,0,dp); }

[@ashmin3636]

Good stuff, blud!

[@KevinNaughtonJr]

thanks Ashhmin!

[@MyVegeta]

yes i have the same reaction when i understand DB

[@DanOhCaptainDaniel]

Every time you say "hopefully that's not too confusing" makes me start to assume the explanation you are about to make is going to be confusing haha. I really enjoy your videos, but I think the way you explain for some reason, it does not click for me. Maybe leetcode is still a little too complex for my programming level... I may just have to spend more time looking at the problems myself. Either way, thank you for having the solutions. I hope you answer every leetcode problem so we all have a reference solution for all the leetcode problems! Thank you again.

[@cocoarecords]

thanks wow well explained and ery clear

[@KevinNaughtonJr]

Thanks so much happy to hear the explanation made sense :)

[@atift5465]

Ughh this is why I hate DP Problems. They always have a gotcha moment. It either clicks with us or it doesnt. I always had intuition that this was a DP problem before solving it but for some reason went towards the Brute Force and used many edge cases. Its hard to explain what I did and the problem worked for many cases but it failed many other cases. Anyhow, thanks for sharing again Kevin. Great work. Can you explain how you typically come up with solutions or algorithms to DP problems?

[@KevinNaughtonJr]

That's ok Atif it just takes practice keep working hard! What I normally try and do for dp problems is kinda what I did in this video...I try to think abou the simplest case. So whenever approaching a dp problem (and using bottom-up processing), I try to think about the absolute simplest case and build from there!

[@atift5465]

@@KevinNaughtonJr yea, true. I just need to tackle them more often until I can pick up exactly how to break down each problem into sub-problems. And as you said, it comes easier with practice. As always, thank you so much for this content and for replying.

[@KevinNaughtonJr]

@@atift5465 Definitely, it's all about practice and anytime Atif thanks for your continued support :)

[@ankurgupta4696]

Nice Explanation:)

[@mostlyharmless161]

Epic thumbnail btw xD

[@nivedhapadmanabhan1794]

how does the 1st 3 condition making a diff, can we not directly start by setting dp[0] and dp[1]

[@shrutee11]

The first three conditions are called as base conditions where only those situations will hold true. As Kevin has coded, nums has either 0,1 or 2 houses. If your nums is having more than 2 houses, we're gonna use a dynamic programming array and save the intermediate house value at each step. So, we would have to start again from 0, 1, and so on. Hope this helps!

[@shilpaas2433]

It works without those base cases because we anyway have the values saved in the dp array. I was also wondering what made this dynamic programming problem different from the others I've seen from Kevin.

[@aashkas]

Wow finally a good explanation

[@KevinNaughtonJr]

thanks!

[@heeseung5]

Nice mask haha made me laugh

[@KevinNaughtonJr]

Haha thanks happy to hear it :)

[@wenzhongtan1358]

Great explanation! Just a question, we could probably shorten the last line to from dp[nums.length - 1] to dp[-1], right?