Continuous Subarray Sum of PAIN

We add {0 : -1} in dict not to handle case where 1st element is divisible by k. We add it to handle case where subarray starting from 0 is divisible by k. For example when k = 6 and nums = [1,2,3,4] and we don't initialize 0, we will get False.

@xinl9259

2 жыл бұрын

Thanks! I watched the video and could not understand why {0:-1} and you answered my question. Very appreciate!

@shin-wu

2 жыл бұрын

Great addition to the video! Thanks for the clarification.

@halahmilksheikh

2 жыл бұрын

Yep. Also a way to get rid of that is to have `if (remainder == 0 && i > 0) return true` in the for loop

@a_maxed_out_handle_of_30_chars

7 ай бұрын

yeah, this makes sense

@user-mh4lq5ce6u

6 ай бұрын

you Can Just Add this Condition if sum%k==0 and i>=1: return True

Again, how is someone able to come up with a solution like this in a time limit interview? This is the question that needs answering next video or something

@easynow6599

2 жыл бұрын

i am thinking the same..in most of these problems..I guess there are three ways: a) you should know the problem beforehand and solve it..so it would be relatively easy to repeat during interview..b) you get plenty of help from the interviewer and you kinda solve it with bugs or/and pseudocode..c) you have spend your life with LC and you solve it in 15' seeing first time.. I dont see myself even close to option c and i guess i have to practice 10 times more..but i would be curious if someone who has passed the interview can tell

@halahmilksheikh

2 жыл бұрын

Step 1: Solve it before the interview Step 2: Solve it during the interview and pretend you've never seen it

@rams2478

2 жыл бұрын

I second that... looks interview is pure luck...

@josecabrera7947

Жыл бұрын

@@easynow6599 I had an interview with a big tech company like last week. I was able to solve roughly 2 problems that I had never seen before that I would categorize as leetcode easy's. They had edge cases that took time and killed like 35 min. The other two however one was hard and the other was either hard or medium. I had no idea how to solve them. I was able to get some work in but couldnt pass all the cases. I later looked them up for hours and was able to find 1 similar but not entirely the same. I spent like 2 hours trying to understand and solve the problem I found. There is no way someone can solve these in 15 minutes.

@easynow6599

Жыл бұрын

@@josecabrera7947 i dont know man..this leetcode looks like a BS system.. but discussing about that i think it's the optimal way to filter people..with a lot of false negatives (a lot of good people stay out because a problem was too hard to solve) and a some false positives (people that grind LC as hell, but in reality they dont know much stuff).. Anyway..i wish you good luck!

yeah, fuck this problem.

This is not a medium question. It is hard, borderline extreme. If you see the solution, it becomes easy since it's memorable, but you cannot average easy and hard to get a medium question 😃

@shanemarchan658

Жыл бұрын

ikr. for days ive been trying LOL... really out of the box thinking to come up with this

@shivamshah6579

Жыл бұрын

100% true!

@CS_n00b

8 ай бұрын

I was asked this question for a new grad job at a small company fml

Is it possible to think of hashing the remainder in an actual interview? What 'intuition' could anyone even come up with it in the first place?

I wrestled with this one for a couple hours... I was trying to store the prefix sum, not the remainder, and I was having a tough time making that work. Storing the remainders is a great idea, and made the code much more concise! Thanks!

wouldn't space complexity be O(k) since you are storing at most k entries in the hashmap? i.e. if k = 6 you are storing at most 6 entries since the remainder would never be more than 5 (+1 more for the 0 edge case)

Nice solution and explanation. I was struggling to find a faster solution, but this duplicate remainder approach is very creative!

Thank you for the clear explanation, I saw the solution before, but I couldn't get behind why it was neccessary to have a duplicated modulo result value in the array. You were able to explain it in a simple manner, appreciate it a lot! Keep up the good work! :)

awesome vid! i'd like to add that the optimized solution is like two-sum in the sense that we're caching complementary values. once you see it this way it becomes intuitive, however, there's no way in hell the average person is going to figure that out without tons of practice doing leetcode questions specifically 🙃 i'd ALSO like to add a one line intuition for the key point of this video: if we see two numbers that when % by k have the same remainder, then those numbers are either k away from each other (or multiple of k away from each other) - and this is exactly what the problem is asking us to find

great vid! No idea how u come out with these solutions but they make perfect sense

I hate math questions. To solve this problem all we need to know is that (a-b)%c = ((a%c) - (b%c)) % c. so to get a-b mod c = 0, just need to have this: a mod c = b mod c

Awesome explanation.... for an insane problem. Great work !!!!!

Omg the the solution is so damn cool ! I tried out the same lines you talked about in the beginning but it led me nowhere so i was feeling pretty dumdum :p Thank you for your clear explanation as always, it really helps :)

Amazing Explanation Bro. Your explanation videos made my DSA skills far better.

bro you are my favorite leetcode channel

@NeetCode

Жыл бұрын

Appreciate that :)

I am a continuous sub array

This solution is just awesome !!! Thanks @NeetCode .

Got asked a harder variation of this in a Google interview. No real hint was also provided by the interviewer. I suppose luck is also a factor. If you've seen the problem before or if you're able to fit whatever problem, you get into the pattern that you've seen before.

just best. I'm watching you for about a year and just wanna say that u'r the best! Thanks for ur job!

Amazing explanation! I don't think I would have been ever able to understand this question without your elegant and detailed explanation about what works and what doesn't work and why it doesn't work.

@varunshrivastava2706

Жыл бұрын

I was asked the same question in my Accolite Digital interview, glad I had already solved it before otherwise I would have been doomed!!!!

Only explanation that my brain understood. Thank you Neetcode.

Thank you so so so so soooo much for this. I was on the right track with modulo and thinking of a sliding window, but was missing the last piece with prefix sum.

Hey, I'm new to this channel. I loved how you explained it so easily! Thanks :)

Very elegant explanation. Thanks a lot. Can you please do Leetcode 2060 as well?

Mind blown. Who would've come up with that trick from the get go haha?

Amazing explanation as always

Damn this solution is crazy.. i don’t think I would even think of this

Definitely Like Your Work Bro......Your Explanation Rocks!!!!

only leetcoder who makes sense tbh. Well done.

I feel you need to be good at math to crack this one on your own.

You made it very easy for me 🔥🔥💯

first i want to make sure you, but also anyone else knows. I am not writing suggestions because I don't need your videos and the help they give. I need them a lot. I am intensively preparing for an interview. Whatever I write is anything I discover on top of your solutions. That said...You can spare the map and use a hash set. Just keep it trailing behind by one index, so you don't get any invalid values. I am sure you know this solution, and if you decided not to explain it because the mind xxxx of the problem itself is pain enough, then I absolutely agree with you. It's worth pointing out, however.

@hershjoshi3549

Ай бұрын

yeah you can def do it with a set and a prev value where prev = sum % k. Personally I find the hashmap solution a little easier to understand and remember.

The thought process that helped me get to the solution was by thinking of the problem Largest Subarray with sum 0, as we use hashmap to store the sum with index.

Very well explained !

I have been watching your videos and May be it helped I don’t know but I had the same idea to solve the problem before seeing the solution 😃

What if we wanted to know where or wich list entries give us this multiple?

You got a new subscriber!!

Hey neet, this is neat!

great explanation bro💗

thanks, great intution.

Amazing intuition

used this to solve 974. made a few changes. def checksumarray_k(nums, k): remainder={0:[-1]} total=0 count=0 for idx, element in enumerate(nums): total+=element r=total%k if r not in remainder: remainder[r]=[idx] else: count+=len(remainder[r]) remainder[r].append(idx) return count

The time complexity of the brute force solution (by summing up the sub-arrays in sliding windows with cumulative sum) seems to be O(n!), which is worse than O(n^2).

Thanks!

I think because of this approach this should have been marked as hard problem.

2 sum or sliding windows. Your false ideas were exactly what I was trying to do.

Well done on this video

Could you please make videos on systems designs as well, if possible?

Yep, it's hard to come up with the part 'i-remainder[r] > 1' by my own. It's great that we leave in the era of streaming to gain best practices

I managed to do this on my own with prefix sum approach. My approach is different though and is more than O(n). Following is the code. bool checkSubarraySum(vector& nums, int k) { int sum = 0; unordered_map hash; for (int i = 0; i sum += nums[i]; if (!i) { hash[sum] = i; continue; } if (sum % k == 0) return true; int j = 0; while (k * j int rem = sum - k * j; if (hash.count(rem) && i - hash[rem] >= 2) { return true; } j++; } if (!hash.count(sum)) hash[sum] = i; } return false; }

Does any one know why when we encounter a remainder that already exists in the hash map, there exists a sum that is %k ?

@dragonoid296

2 жыл бұрын

because if the remainder wraps around, we know that it changed by k

Can you do 410. Split Array Largest Sum? Thank you!!

great solution

There is a step after calculating r you can also check if r == 0 return True

thank you neet

Your original trains of thought about the 2sum problem as well as a prefix sum approach were also doable, if you had dived deeper into them and made minor modifications. Here is a 2sum-like approach to solving this problem in one pass: class Solution: def checkSubarraySum(self, nums: List[int], k: int) -> bool: seen = set() prv = cur = 0 for x in nums: cur, prv = (cur + x) % k, cur if cur in seen: return True seen.add(prv) return False

So key point here is if prefixsum%k=r and (prefixsum+x1+x2+•••••+xn)%k=r then undoubtedly x1+x2+•••••+xn is multiple of k

AMAZING IDEA 😍🌹

i did not try it but it looks like this could be solved with segment trees

Another way to handle the edge case when prefix sum will itself be a multiple of k, arr = [24,0,2] k = 6, [24,0] subarray sum is a multiple of 6. if(prefSum % k == 0 && idx > 0) return true;

thanks

Could it be solved using Brute force method? If solved, then how?

Dope

OMG, so tricky!!!!!

You are so smart!! I would like to elect you president.

I was recently asked this question in an interview and I was solving it through a two-pointer approach but I failed to optimize and got rejected 😥

@sureshsingam7291

Жыл бұрын

which company??

this should have been a hard problem.

hi @neetcode i Try to solve it Using Sliding Window It passes Most of the Test Csaes but if array consist of 0 then its failing 93/99 test cases are passed but this one if failing nums = [1,3,6,0,9,6,9]. , k = 7 output : True if len(nums) return False currSum = 0 totalSum =nums[0] l= 0 for r in range(1, len(nums)): currSum = nums[l]+ nums[r] totalSum +=nums[r] if currSum % k == 0 or totalSum % k == 0 : return True l+=1 if totalSum % k == 0: return True return False could you please See It Please

This problem is mainly optimization.

me as a day of life an unemployed dude :(

Ninja level

Still unable to understand how it has checked all subarrays' sum

i laughed when i saw the solution its sooooooo clever

oh god

Java Solution: class Solution { public boolean checkSubarraySum(int[] nums, int k) { HashMap map = new HashMap(); // To store int prefixSum = 0; map.put(0, -1); for (int i = 0; i prefixSum += nums[i]; int remainder = k == 0 ? prefixSum : prefixSum % k; if (map.containsKey(remainder)) { int length = i - map.get(remainder); if (length >= 2) { return true; } } else { map.put(remainder, i); } } return false; } }

Is there a real life use case for an algorithm for this?

@sabukuna

26 күн бұрын

doubt it

Java solution for the same public boolean checkSubarraySum(int[] nums, int k) { HashMap map = new HashMap (); int sum = 0; map.put(0, -1); boolean found = false; int n = nums.length; for(int i=0; i sum = sum + nums[i]; int hash = sum % k; if(map.containsKey(hash)) { int diff = i - map.get(hash); if(diff>=2) { found = true; break; } } else { map.put(hash, i); } } return found; }

Great concept but really this approach is difficult to come up with.

Python is so easy for OA. Sadly, I'm more comfortable writing in Java. Imagine "remainder = { 0, -1 }". Java would be first import Map and hashMap, then Map remainder = new HashMap(); map.put(0, -1);. So sad. So so sad!

Great yaar i was missing some cases ... thanks

How much IQ is required here to come up with the solution on the first look?

[5,0,0,0] 3 Expected output : True How can it possible ??

@jsalaat

Жыл бұрын

0 and 0 are a continuous subarray hence making 0+0 = 0%k expects to fulfil the condition. hence returns true

why you need to add 0, -1 TC; arr= [2,4,1] k = 2 if you not add 0,-1 in map it return false at index 1 ==> 6 % 2 == 0 index - map.get(0) >=2 ==> return true

Thanks !

Thanks!

@NeetCode

2 жыл бұрын

Thank you so much!!