Subarray Sum Equals K - Prefix Sums - Leetcode 560 - Python

[@beeramrahulreddy11]

u forgot to add 2,1 to the hashmap

[@TobiasWheaton]

Good catch, i was thinking about that too bc its easier to see the pattern if its consistent on when u add (even if its not needed for this example)

[@prasad9012]

I just spent like 10 minutes rewinding and forwarding the video trying to understand why 2 wasn't added. Thankfully, I found your comment now.

[@dopark1027]

Actually, it is just not {2:1}. He forgot to add the current sum to the hash table when it is not found in the table.

[@amritkumar8876]

Yup .. I was also thinking about that ..

[@pwnweb5734]

haha yeep. my OCD went brrrr... I was like wheres the 2 :P ..

[@kimstuart7989]

You know I remember first learning CompSci, I thought the specific algorithm concepts (BFS, DFS, Dijkstra, etc.) were the hard ones to learn. Now that I am more experiences, I realize that the what I thought then were 'basic' data structures, i.e. arrays, strings, etc. are actually the harder ones to make sure you understand because those concepts has many dedicated algorithms to them, such as Kadane, prefix/suffix array, KMP etc. that you need to know to truly understand that data structure. So while I glossed over arrays because I knew their functions, I am now revisiting to make sure I understand the algorithms embedded in them. Learning truly never does stop!

[@fatcat22able]

Honestly same here. I figured that arrays would be relatively simple, and they are simple on the surface, but there are so many algorithms and different types of problems that involve them that it feels like they're an entire rabbit hole unto themselves.

[@servantofthelord8147]

Facts man

[@davidnwaekwu9019]

factssss, at a point i thought smth is wrong with the way i think .

[@suvajitchakrabarty]

If you are having difficulty understanding this or wrapping your head around the solution, maybe this might be a good way to think about it. The question asks how many subarray sum equals to k. For a subarray to sum to k, you need a subarray, as in a part of the array from index 'a' to index 'b' to have a sum equal to k. But when we reach any index 'b', we obviously do not know, if there was a subarray from index 'a' to index 'b' equal to k. However we do have the sums from 0 to index 'a' in our hash map, because we have been storing all sums starting from index '0' to every single index till now, and the count of them as the value of the key. Now obviously sum_0_to_a + sum_a_to_b = total sum so far (curSum). If we go to the original ask, which is we need a prefix that is sum_a_to_b to be equal to k. For that to hold true, replace sum_a_to_b with 'k'. Hence, sum_0_to_a + k = curSum. Hence curSum - k = sum_0_to_a. And then since we have been storing all possible values of sum_0_to_a so far in the hashmap, curSum - k must exist in the hashmap as a key, and we can simply add the value from the hashmap to add number of prefixes from 0 to any index which equalled to curSum - k .

[@meraki___]

Bro this comment is better than the video lmao. Thanks a lot for the explanation.

[@dhairyakhara783]

You made the problem 1000 times easier to understand. Thank you sir!

[@dishade7094]

Thanks for the perfect summary. Awesome bro !

[@brendan5878]

Thank you brother. Simple and clear explanation. Brilliant.

[@Slothieify]

Amongst every explanation I've come across, this was the only one that actually worked for me. Thank you!

[@Question418]

I like your attempt to explain it, even though I don't fully understand it. I appreciate it.

[@codetrooper9279]

U should

[@ToastFrench24]

lmao

[@apoorvwatsky]

1:55 I needed to hear that, I wanted to know why sliding window won't guarantee correct answer for these constraints. That cleared it up.

[@EchoZeeDot]

Thanks, NeetCode for these solutions and explanations, It would be really helpful to make a video about general techniques that you use to solve these problems... Because we wan to know how to approcah these kinds of problems by ourselves. Thanks again.

[@darkwhite2247]

08:21 - i have no idea how we are supposed to figure out during the coding interview that (0,1) has to be added into the hashmap by default. How is anyone solving this without memorizing during interviews? May be i'm dumb to not understand this even though the instructor is explaining it. Please help me understand the intution behind adding (0,1) in hashmap by default.

[@AndreiSokolov-k7j]

bcs if you find that n-k = 0 means that you dont need to remove anything so basically it be 0, 1 by default

[@timzhang6651]

Well explained, and it takes effort to make a video as good as this. Thank you!

[@josembass]

Great explanation. How are we supposed to come up with a solution like this in 30m without having seen it before!

[@darshana5254]

We cannot. These are classic problems asked during interviews. Whoever come across this problem would easily solve it. Other would go with bruteforce approach which doesn't work on interviews.

[@scionOfIkshvaku1]

@@darshana5254 thanks bro, you’ve raised a very important point. Only if one has encountered such problems, only then will he be able to solve the problem cuz he knows how to optimum approach.

[@mobaisi13]

That's the neat part. You don't.

[@vineethsai1575]

I got this question in the interview and I did not know the optimal solution. However I was able to produce the n^2 solution quickly and the interviewer started giving me hints to use Hashmap to reduce it to O(N). I scrambled here and there but was able to talk about the "repeated work" being done and how I could use hashmap to save the work. In the end I was not able to code it up. Interviewer still passed me though.

[@calebmitcler8760]

@@vineethsai1575 what company

[@mangalegends]

I'm still trying to wrap my head around this EDIT: Came back a few weeks later and it makes 100% sense to me now

[@OusmanBah-f8v]

Me right here

[@CliffordFajardo]

Initially I thought it was a sliding window problem, Trying to wrap my head around this ...patience, time...ahhh. Thanks for the video!

[@ziomalZparafii]

I thought about it as a sliding window problem until I found out there can be negative values. If we would only have positive ones, you would increment right pointer while sum is less than k, then increment left while sume is more than k. That would be my approach (not tested).

[@aleksandr_t]

@@ziomalZparafii the main problem behind sliding window approach is some edge cases. For example: [1,-1,0], k = 0 If we will use two pointers approach (or dynamic sliding window), then we will return 2 instead of 3. 1) sum=1, count=0, left=0, right=0 2) sum=0, count=1, left=1, right=1 3) sum=0, count=2, left=2, right=2

[@ziomalZparafii]

@@aleksandr_t you have a negative value in your array. Read my comment once again :-)

[@matom2473]

in a real interview we can not make this mistake, it will take you 20 mins to find out sliding window will not work

[@RuslanZinovyev]

@@ziomalZparafii you still can use a sliding window with negative numbers, it depends on requirements and edge cases, so your first comment is not entirely accurate.

[@scottloewke5199]

Good explanation and, fortunately, the omission of the [key = 2, count = 1] map entry did not affect the result. However, it would be good if NeetCode could add text at that part in the video to make it clear what happened.

[@AndreiSokolov-k7j]

It's not about a bug in the code, it's about a bug in the video. 9:55 Here you should have added key 2 and value 1 to the hashmap and that's it.

[@Wes-Tyler]

You explained it so perfectly that I didn't even need to see the code to be able to code it myself :) thank you!!

[@arunraman6630]

A slightly cleaner solution using a defaultdict class Solution: def subarraySum(self, nums: List[int], k: int) -> int: prefixSums = defaultdict(int, { 0 : 1 }) curSum = 0 res = 0 for n in nums: curSum += n res += prefixSums[curSum - k] prefixSums[curSum] += 1 return res

[@thndesmondsaid]

the idea that we're expected to come up with this solution on the fly in an interview is bonkers to me

[@Adam-tz6gk]

This is by far the best explanation in the web for this problem, thank you so much NeetCode!

[@X3Maverick]

00:40 The brute force approach is O(n^3), not O(n^2). There are ~n^2 subarrays but summing a subarray is O(n).

[@spacetime_wanderer]

It is n^3 if you calculate sum from 0 to n everytime. It is n^2 if you maintain a sum variable for the innerloop and simply add num[i] to it.

[@tempregex8520]

i appreciated you explaining this. But, it would have been more helpful(esp to people like me) where you would have taken the time to explain more about the use of HashMap for maintaining prefixSum -> Count to me it felt like the explanation went "too fast" as soon as you discussed the brute force solution.But, in any case a great attempt indeed.

[@cranos666]

Crazy good explanation. Subscribed. Thank you man !

[@smonkey001]

Godlike explanation to a looked easy problem. Hats off my man

[@unwarysage05w32]

This is a way better explanation than leetcode premium’s explanation thank you!

[@raghav28489]

Let me try to break down the explanation since it took quite some time for me to wrap around the head on this one: For a subarray starting at index A and ending at index I, the sum of that subarray is: sum(indexA to indexI) = sum(0 to indexI) - sum(0 to indexA) = prefixSum(indexI) - prefixSum(indexA) From the problem statement, we want sum(indexA to indexI) = k. Replacing it in the above equation gives k = prefixSum(indexI) - prefixSum(indexA) Rearranging above equation gives, prefixSum(indexA) = prefixSum(indexI) - k At any point in ith index of the iteration we already have prefixSum(indexI), which is basically the currentSum we've maintained so far. So in the above equation we already have the RHS value. This tells us that if we’ve seen a previous prefix sum equal to currentSum - k, then there exists a subarray ending at index I whose sum equals k. To efficiently find these previous prefix sums, we use a hashmap that tracks: Keys: All prefixSum values encountered so far. Values: The number of times that prefixSum has occurred (i.e., how many potential starting points indexA exist for a subarray). Whenever the target: currentSum - k is found in the hashMap, it means there are one or more subarrays ending at index I whose sum equals k. We add the count of such starting points to our result.

[@user-wx8hn3ng6b]

It's a good combination of two sum and prefix sum. And the intuition should be: 1. subarray sum -> prefix 2. equals k -> two sum The only difference is it's 'b - a = k' instead of 'b + a = k', and you just need to calculate the 'a' differently.

[@uzzwalpreetkaur3374]

I understood this explanation! Thanks!! I was trying with sliding window technique first and getting the wrong answer until I realised that we can have negative numbers. This is indeed is a neat.

[@Thewowhahahawow]

What a god of prefix sums. Great vid!

[@daylansit4537]

By far the best explanation to this problem

[@roman-berezkin]

Absolutely love this guy, not lessons but a masterpiece!

[@SonGoku-lc1sb]

Perfect explaination. The best one on youtube so far

[@quirkyquester]

just gotta keep practicing! Thank you Neetcode for making all these amazing videos!

[@lakcai4063]

This is probably the best explanation of this question on KZbin

[@MP-ny3ep]

Hands down the best coding youtube channel

[@edwardteach2]

Another way to think about why we would set our key value to {0:1} instead of memorizing, is to think what if that single element, itself is the target. For example: input = [1, 1, 4] , k = 4 occur = { 0: 1 } By just eyeballing, you'll see the output is 1, since the last element is a 4. When we iterate through the input, we'll do input[2] - k (4 - 4) = 0. Do we have a 0 in our hashmap/dictionary? Yes we do, it's value is 1.

[@alvjou]

except in this scenario wouldn't we be doing currSum - k (6 - 2) = 2, and we would be looking up the value for 2 in our hash map, which would be 1. I think your scenario only works if the single element that is the target is the first element in the input.

[@almasabdrazak5089]

so what you are saying is we have a map that shows how many ways we can come up with given sum so for sum 0 we can come up with 2 sequences, but these sequences have to be contiguous , how to prove that 0 has 2 only if we pop up contiguous elements from current sum ?

[@itspete2444]

Why are array and string problems 10x harder than graph, matrix and dp???

[@kaushik.aryan04]

yeah atleast graph matrix dp follow a single type of questions these are like so different from each other

[@benk2451]

Very well explained! Thank you so much!

[@aadil4236]

10:02 Why didn't we add 2 as a prefix there? Was that a mistake on your part? I'm confused.

[@andrewsowell3088]

Yep, should have been added

[@GdeVseSvobodnyeNiki]

Finally a great explanation what hashmap actually represents! Got the essentials of the algorithm on the first watch of this video, but was banging my head against the wall trying to figure it out on leetcode.

[@chenhaibin2010]

brilliant trick of adding a prefix 0 to count no prefix situation instead of another if statement

[@dharmvtiwari6286]

There is another edge case for adding 0->1in the starting , which is if the. nums[0] is 0. This intuition is quite tricky but great explanation.

[@mukeshmanral1327]

what a great explanation , saw many videos, amazing explanation of intuition.

[@hoatruong2474]

Very informative man. Keep up the good work.

[@jameszhang9422]

Very clear explanation! Even better on the second watch lol.

[@michelleyang1881]

really clear explanation. Thank you!

[@Nahwka]

Believe it or not this was the entire solution that actually killed me 😂

[@zj82119]

nice explanation with good diagram. really helpful.

[@soanonso]

Really good explanation. Thanks.

[@Mohith7548]

I don't know why it works, but still works! ;) Everytime I have to revisit the solution before interview. ;(

[@mikhassik]

Neetcode's videos are usually the ones that drive it home

[@dheepthaaanand3214]

The reason we can have multiple prefixes of a particular sum is because it could get negated by a subarray and then added back Like 1-1 1 -> there are 2 prefixes summing to 1

[@celinlu7274]

Your video is much better than the video in leetcode solution. Thank you.

[@phebos2228]

Bless you, O savior from heavens above!!

[@ShivangiSingh-wc3gk]

Thank you so much!! Amazing walkthrough

[@meylyssa3666]

thank you so much for you clear explanations!

[@AmirMairaj-f3x]

I watched the explanation then tried to code it on my own, def not as clean but I think its a little easier to understand: class Solution: def subarraySum(self, nums: List[int], k: int) -> int: prefix_sum = {} return_val = 0 prefix_sum[0] = 1 count = 0 for i in nums: count += i prefix = count - k if prefix in prefix_sum: return_val += prefix_sum[prefix] if count in prefix_sum: prefix_sum[count] += 1 else: prefix_sum[count] = 1 return return_val

[@edwardteach2]

My python implementation: class Solution(object): def subarraySum(self, nums, k): """ :type nums: List[int] :type k: int :rtype: int """ occur = collections.defaultdict(int) occur[0] = 1 count = 0 cur_sum = 0 for n in nums: cur_sum += n if cur_sum - k in occur: count += occur[cur_sum-k] occur[cur_sum] += 1 return count

[@chilly2171]

9:04, shouldn't it be 3 instead of -3? since you need to add 3 to 0 to form K.

[@dipeshbhardwaj1747]

Hi @NeetCode. I am not able to understand the part where you say. we need to add the prefix sum to Hashmap simultaneously. even if are maintaining the Hashmap first. we can still get count of prefix - k and prefix = k.

[@anandmehrotra6170]

Take this example: [1]; k = 0. Without adding first your prefix sum array would be: 0->1 and 1->1. Now you will compute prefixSum-k for each key. 0 - 0 = 0, You say you have a match(which is just the empty prefix). Now you compute 1- 0 = 1, you check your map and the value exists, so you return a count of 2. That is why the update needs to happen simultaneously. So, that you can check the subarray between some starting index "i" and current index "k"(excluding k). Hope this makes sense.

[@thegod3500]

Finally some decent explanation of this magic

[@sgetti4dinner]

I see some confusion on why we initialize prefixSum = {0:1}, and that's because we can always produce 0 by taking no elements from our array (e.g. []). This works out nicely when we run into a subarray that has a sum == k, because sum - k = 0. Then the question we ask at this moment is "is there a prefix-subarray I can take away from my current subarray sum that also equals 0?" The answer is yes, because I can takeaway no elements from my current subarray and the current subarray sum will still equal k.

[@vishalanthony1529]

This is one of those rare videos of Neetcode where I could not really understand what he is trying to convey

[@Obligedcartoon]

This is a LC hard imo, gawd damn!

[@NeetCode]

Agree, it's definitely more difficult than some actual hard problems.

[@razisyed9481]

Amazing explanation--thank you!

[@ritwik_shivam]

9:59, we need to add 2:1 in hashmap or am I not getting this part, please help me out

[@one_step_sideways]

check pinned comment, you are correct

[@oscarchan5684]

Thank you for such a clear explanation!!!👍🙏

[@nemesis_rc]

Your channel is really underrated man 😢

[@TokyoXtreme]

Truth.

[@orellavie6233]

I am not sure, but take an example where the array is [-1,1,-1] and k>0(e.g., 2). The result would be 1, since after the 1 is covered, and then we went into the -1 again, the result will be 1, and that's not true (should be 0). Am I missing something? I think the results should be added only if the cursum is >=k

[@m04d10y1996]

Why did you initialised your dictionary before hand with (0:1)

[@dheepthaaanand3214]

7:24 i don't understand what he is trying to explain here. Why don't we want to remove prefixes which sum to 0?

[@nivethanyogarajah1493]

Very nice explanation!

[@emekaobasi4765]

why do we add to the hashMap after checking for diff? Can we add to the hashMap after checking for diff?

[@lonen3rd]

I watched it 3 times and finally I get it and can code it up. What I don't know is, how do I come up with such efficient solutions?

[@АлександрПетровский-м3з]

Very nice & clear. TY!

[@hwang1607]

great solution thank you

[@jameszhang9422]

Thank you for the great video!

[@ngndnd]

i think im just dumb af bc i still dont understand it

[@siriussem2]

yey first real english guide!

[@vatsalsharma5791]

Awesome explanation❤️ Can you plz solve Count submatrices with all ones?

[@divyanshmishra5121]

I have a doubt. If the array is 6,4,3 and k is 9. we find that sum till n is 13 and 13-k ie 13-9=4 is present in the array so that will get counted but in reality it is wrong as it is not a subarray is we dont pick contiguous element. please explain

[@slayerzerg]

get() is unnecessary can just add to hashmap regularly. this drawing analysis is goated though!!!

[@vinethasuresh3488]

Thanks for this video sir , it helped a lot.

[@NeetCode]

Happy to help!

[@fengliu975]

Did this problem but used an array to keep track of the running totals then reversed it... used a deque after but wow is deque slow and memory intensive

[@kalp2586]

Yeah it's complicated but don't forget to come up with the optimal solution in 10 minutes without any compiler errors and of course it needs to pass all test cases in the interview. smh. Tech interviewing has become a circus.

[@juliewiner5287]

but how do you know the difference will be contigous with the current sum ?

[@nguessaneric9830]

Because you substract a prefix, you don't select element in your window

[@mike0998unstopable]

Wayyyy less confusing than the explanation on leetcode

[@AmolGautam]

Thank you so much for this.

[@xmnemonic]

Added more comments, explanation, changed diff to prevSum which I think makes it easier to understand: class Solution: def subarraySum(self, nums: List[int], k: int) -> int: kCount = 0 curSum = 0 prefixSums = {0:1} for n in nums: curSum += n # We are at curSum, but we want to know how many sum = k # have occurred. k = curSum - prevSum, use the prefixSums # hashmap to determine how many prevSums occurred. Each # one represents a potential subarray that could be # subtracted to reach sum = k. prevSum = curSum - k kCount += prefixSums.get(prevSum, 0) # Update prefixSums AFTER we have gotten the kCount for # this curSum. We ensure that kCount considers only # prevSums prior to this iteration. Including this sum # potentially doublecounts the n of this iteration, # using the n in both curSum and prevSum. prefixSums[curSum] = 1 + prefixSums.get(curSum,0) return kCount

[@kirillzlobin7135]

Can it be considered as a dynamic programming solution?

[@nemesis_rc]

Explained well 👍

[@ygwg6145]

It appears easier to use hash table cum->deque[index]. It did pass all the tests.

[@odelyaholiday9519]

It's easier to use defaultdict instead of the .get

[@itachijonin]

really good explanation. my one nit is that you should make your code more pythonic (camel-casing?!)

[@anastasiiafilippova5212]

Brute force is n**3 not n**2 as stated. n**2 for all possible subarrays and n for the sum

[@TheK1KoS]

Just failed this one in an interview. I just don't get it even when with such detailed explanation

[@anujpatel2121]

This guy is Goated 🚀

[@jorgemejia1586]

thanks bro I needed this

[@gorilla4870]

Simple problem, but can be quite confusing to explain. You explained it very well Neet. Thanks a lot!

[@shaon6996]

Great explanation!

[@arifinrahman6]

amazing explanation :)

[@saul_no]

I was actually asked this question for a Facebook internship position in 2019.

[@dmwhite4787]

Did you find the optimal solution in an interview? Had you seen this problem before? If no, did you pass that round?

[@AbhishekJaiswal24]

Thanks neetcode !!!