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.
@xinl92592 жыл бұрын
Thanks! I watched the video and could not understand why {0:-1} and you answered my question. Very appreciate!
@shin-wu2 жыл бұрын
Great addition to the video! Thanks for the clarification.
@halahmilksheikh2 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_chars10 ай бұрын
yeah, this makes sense
@AbhishekMakwana-p1v10 ай бұрын
you Can Just Add this Condition if sum%k==0 and i>=1: return True
@erikshure360 Жыл бұрын
yeah, fuck this problem.
@dantesmith78597 ай бұрын
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!
@hanklin46332 жыл бұрын
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?
@AlexN202220 күн бұрын
the easier-to-come-up-with solution can still work: 1. calculate prefix array and store in a map (sum->leftmost index where found) 2. r = 0; r < size; r++ 3. calculate the current rmdr = sum%k 4. if we could find a sub-array on the left with sum=rmdr, we could remove that subarray from [0:r] and sum%k would be 0 5. look for rmdr in the map. If found && index < r - we have found the sub-array to remove
@mirceskiandrej2 жыл бұрын
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_n00b11 ай бұрын
I was asked this question for a new grad job at a small company fml
@ObtecularPk2 жыл бұрын
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
@easynow65992 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
@halahmilksheikh2 жыл бұрын
Step 1: Solve it before the interview Step 2: Solve it during the interview and pretend you've never seen it
@rams24782 жыл бұрын
I second that... looks interview is pure luck...
@josecabrera79472 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.
@easynow65992 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!
@redblacktech4 ай бұрын
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
@ivankok83992 жыл бұрын
Nice solution and explanation. I was struggling to find a faster solution, but this duplicate remainder approach is very creative!
@adam-zy5tb2 жыл бұрын
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)
@Dumbastic2 жыл бұрын
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! :)
@CreeperFace752 жыл бұрын
great vid! No idea how u come out with these solutions but they make perfect sense
@finchshi43422 жыл бұрын
bro you are my favorite leetcode channel
@NeetCode2 жыл бұрын
Appreciate that :)
@algosavage7057 Жыл бұрын
just best. I'm watching you for about a year and just wanna say that u'r the best! Thanks for ur job!
@priyankachoudhary36942 жыл бұрын
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!!!!
@rongrongmiao30182 жыл бұрын
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
@Ryan-g7h26 күн бұрын
Hey guys also don't forget this video is missing a key detail: look into why we use elif i - remainder[r] > 1 instead of >=. It ties to the {0:-1} and a edge case such as [6,1,2], k = 6 -> should false but if we do >= return True
@danielcontreras93432 жыл бұрын
I am a continuous sub array
@vdyb7452 жыл бұрын
Awesome explanation.... for an insane problem. Great work !!!!!
@vadimkokielov21732 жыл бұрын
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.
@hershjoshi35494 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.
@numberonep54042 жыл бұрын
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 :)
@SomneelMaity Жыл бұрын
Amazing Explanation Bro. Your explanation videos made my DSA skills far better.
@tusharnain6652 Жыл бұрын
Only explanation that my brain understood. Thank you Neetcode.
@sitam-meur43172 жыл бұрын
This solution is just awesome !!! Thanks @NeetCode .
@jitpatel76929 күн бұрын
Nice Explanation,Thank You
@aadityakiran_s4 ай бұрын
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.
@andrewpagan Жыл бұрын
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.
@shadowthehedgehog27272 жыл бұрын
Damn this solution is crazy.. i don’t think I would even think of this
@avanishgvyas1992Ай бұрын
Nice explanation. Impossible to come up with this solution in 25 minutes while under stress in an interview.
@shelbys42522 жыл бұрын
only leetcoder who makes sense tbh. Well done.
@samridhanand49264 ай бұрын
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.
@beinghappy92234 ай бұрын
Amazing intuition
@anirbannandy9104 Жыл бұрын
Definitely Like Your Work Bro......Your Explanation Rocks!!!!
@mohammadkareem11872 жыл бұрын
Very elegant explanation. Thanks a lot. Can you please do Leetcode 2060 as well?
@henryrussell73922 жыл бұрын
Amazing explanation as always
@vinuk.vijayakumar8023 Жыл бұрын
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
@ayushidalal5488 Жыл бұрын
Hey, I'm new to this channel. I loved how you explained it so easily! Thanks :)
@vedbhanushali608 Жыл бұрын
thanks, great intution.
@anand.prasad502 Жыл бұрын
Thanks !
@shikharathaur5179 Жыл бұрын
Very well explained !
@chandamwenechanya86142 жыл бұрын
Hey neet, this is neat!
@bluesteel15 ай бұрын
I feel you need to be good at math to crack this one on your own.
@mariotheguy1237 ай бұрын
thank you neet
@suyashpurwar631 Жыл бұрын
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 < nums.size(); i++) { sum += nums[i]; if (!i) { hash[sum] = i; continue; } if (sum % k == 0) return true; int j = 0; while (k * j < sum) { 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; }
@Will-dr9cf8 ай бұрын
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).
@vinayakhaunsnur1327 Жыл бұрын
great solution
@mrunalajitjoshi15122 жыл бұрын
You got a new subscriber!!
@rishabhraj8233 Жыл бұрын
great explanation bro💗
@tomtran693611 ай бұрын
There is a step after calculating r you can also check if r == 0 return True
@lesterdelacruz50886 ай бұрын
Mind blown. Who would've come up with that trick from the get go haha?
@clumsyroad40264 ай бұрын
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
@venkatasundararaman2 жыл бұрын
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 😃
@sidazhong201910 ай бұрын
2 sum or sliding windows. Your false ideas were exactly what I was trying to do.
@meetsoni19382 жыл бұрын
You made it very easy for me 🔥🔥💯
@mojedsamad7184 Жыл бұрын
What if we wanted to know where or wich list entries give us this multiple?
@ShivangiSingh-wc3gkАй бұрын
Yeah, after coming to prefix sums, I couldnt think about this so tricky either you see the math or you dont
@kaushik.aryan04 Жыл бұрын
I think because of this approach this should have been marked as hard problem.
@IK-xk7ex Жыл бұрын
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
@ujjaldas91794 ай бұрын
thanks
@srinivasareddychalla-i2o3 ай бұрын
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
@mattjm0072 жыл бұрын
Well done on this video
@mohamed_ellithy Жыл бұрын
AMAZING IDEA 😍🌹
@abhishekgururani6993 Жыл бұрын
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;
@ariaXP12 жыл бұрын
Can you do 410. Split Array Largest Sum? Thank you!!
@yatri6329 Жыл бұрын
Great yaar i was missing some cases ... thanks
@yashpathak92852 жыл бұрын
You are so smart!! I would like to elect you president.
@DJSTEVE422 жыл бұрын
Does any one know why when we encounter a remainder that already exists in the hash map, there exists a sum that is %k ?
@dragonoid2962 жыл бұрын
because if the remainder wraps around, we know that it changed by k
@SaahasBurichaАй бұрын
this is insane
@bhushanmahajan21502 жыл бұрын
Could you please make videos on systems designs as well, if possible?
@codeguru44 ай бұрын
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) < 2: 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
@madhuj6912Ай бұрын
Can someone tell me whats this , what is x and n are that values of array " An integer x is a multiple of k if there exists an integer n such that x = n * k. 0 is always a multiple of k "
@avanishraj386 Жыл бұрын
Could it be solved using Brute force method? If solved, then how?
@zz-yy-xx4 ай бұрын
OMG, so tricky!!!!!
@masternobody18962 жыл бұрын
me as a day of life an unemployed dude :(
@fazilshafi80837 ай бұрын
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 < nums.length; 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; } }
@kaioken1999 Жыл бұрын
Dope
@rajeevkri123 Жыл бұрын
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< n; 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; }
@santoshr4212 Жыл бұрын
this should have been a hard problem.
@vbcarnage2 жыл бұрын
i did not try it but it looks like this could be solved with segment trees
@Dannnneh4 ай бұрын
Is there a real life use case for an algorithm for this?
@sabukuna4 ай бұрын
doubt it
@youssefel-shabasy8334 ай бұрын
oh god
@charanpasupula37632 жыл бұрын
Ninja level
@rahulbhardwaj4380 Жыл бұрын
Great concept but really this approach is difficult to come up with.
@maheshjamdade14 ай бұрын
This problem is mainly optimization.
@heetshah72922 жыл бұрын
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 😥
@sureshsingam72912 жыл бұрын
which company??
@MDEMANURRAHAMAN- Жыл бұрын
[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
@PREETIGUPTA-l8j4 ай бұрын
Still unable to understand how it has checked all subarrays' sum
@experiment000311 ай бұрын
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!
@imjusttryingtobesafelol5909 Жыл бұрын
i laughed when i saw the solution its sooooooo clever
@eddiej204 Жыл бұрын
How much IQ is required here to come up with the solution on the first look?
@Ginger_Hrn9 күн бұрын
idk, last time I solved a matrix problem to which the algorithm was made by a russian computer scientist to solve THAT particular type of problem.
@АльбусДамблодр2 ай бұрын
yo, neetcode, bro, your solution doesn't work
@vijayanks17144 ай бұрын
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