🚀 neetcode.io/ - I created a FREE site to make interview prep a lot easier, hope it helps! ❤ This one turned out longer than expected, recommend 1.5x speed
@thelookofdisapproval8234
3 жыл бұрын
Lmao I've watched so many lectures on youtube, many require you see it at 1.5x or 2x this is the first time, I have seen the youtuber himself suggesting doing the same, thank you for the uploads also can you do a question on union find algo?
@brieflyfun
Жыл бұрын
I wonder if there's a problem in line 26~28. The "have" count is not balanced. Should we have the same == conditions as line 17? if s[l] in countT and windows[s[l]] == countT[s[l]]: have -= 1 windows[s[l]] -= 1 to make the count balanced. "have" is the number of unique characters that meets the required numbers.
@Axl124124
Жыл бұрын
to be honest if I meet you and I hear you talking at normal speed, I would think you have dementia or something. I watch all your videos at 2x speed.
@jackscott4829
2 жыл бұрын
I never thought I'd see the day I'd fully understand a Leetcode Hard problem in under a half hour. Bless you sir.
@harshitsingh2118
Жыл бұрын
same feeling 🥳
@srikrishnarohanmadiraju8688
3 жыл бұрын
Beautiful..! The blackboard explanation, clean code, the naming conventions, everything is just beautiful. Thank you.
@kiddem9247
2 жыл бұрын
For your solution, in line 17 comparing both map counts: if c in countT and window[c] == countT[c] I had to replace == with
@re-think7693
Жыл бұрын
Thanks! This helps
@yaboyjz
Жыл бұрын
same here!
@brieflyfun
Жыл бұрын
"need" is number of unique characters in t, not total number of characters. I think the problem is in line 26~28, which should have the same == conditions as line 17. if s[l] in countT and windows[s[l]] == countT[s[l]]: have -= 1 windows[s[l]] -= 1 to make the count balanced. "have" is the number of unique characters that meets the required numbers.
@felixdeng9824
9 ай бұрын
In his solution, he set need = len(countT) instead of len(t), that makes the while loop( have == need) will execute with no errors. Of course we can set window[c]
@govindrai93
9 ай бұрын
thanks @felixdeng9824 that was the reason. in python len(dictionary) == number of keys. So the case of "aaa", the dictionary would be { "a": 3 } and the len(a) would be 1. That's why == works for neetcode!
@airysm
3 жыл бұрын
Thank you for these videos! I'm switching from java to python for interviews and these have been really helpful
@NeetCode
3 жыл бұрын
Thanks! I'm really happy that they are helpful!
@ChanChan-pg4wu
2 жыл бұрын
Likely the most difficult question for sliding window. Again, watched it 3 times. Thank you, Neet!
@Notezl
Жыл бұрын
wait till you try Sliding window Maximum.
@Haseebkhan-yd9ud
Жыл бұрын
@@Notezl 😂
@Tyler-jd3ex
11 ай бұрын
I’m glad that somebody else had to watch it multiple times. Sometimes I just feel dumb, but I have to remind myself that these videos are prepared and Neetcode probably struggled with it himself at first.
@akagamishanks7991
8 ай бұрын
tbh honest Sliding window Maximum way easier haahahah@@Notezl
@hmanjun7260
Жыл бұрын
Nice. I figured out how we would need to slide the window, but was struggling with how to store it and check if it satisfies the t counts. This was the first leetcode hard I've tried and feel good that I got a majority of the idea down.
@SIAMEInekeidijdnen
Жыл бұрын
I agree. It was not as hard to actually figure out what was needed to solve the problem, but then when it came to implementing the solution, it got really hard and confusing. It was probably because we were trying to implement the brute force approach.
@srinadhp
2 жыл бұрын
as usual, your explanation makes a hard problem look simple! Thank you so much!
@trueworth638
3 жыл бұрын
great video, you're helping out a ton for interviews! Your channel is too underrated
@arupdas2210
2 жыл бұрын
C++ Implementation of the above explanation: string minWindow(string s, string t) { if(t==""){ return ""; } unordered_map um; for(int i=0;i
@riteshsrivastava3447
9 ай бұрын
Thanks , this helps .
@mahithamadduluri5233
7 сағат бұрын
Helped a lot
@roman_mf
Жыл бұрын
Thanks to you and NeetCode 150 list, I was able to solve this hard problem on my own without peeking at your solution! The runtime sucked really bad, but it was nonetheless accepted. Hooray!
@DavidInga7
Жыл бұрын
@NeetCode Thanks so much for the video! I enjoyed watching how you solved this problem. I wanted to point out that the Leetcode problem states that we are dealing with only the 26 lowercase characters in the English alphabet. As such, our hashmap would be a constant size. Therefore our hashmap would never grow to a size of 100 as stated. All your videos are awesome. I hope my comment is constructive and helpful!
@user-SerhijA
9 ай бұрын
That's not the case. Even at the beginning of the video it is seen that letters are capital.
@SanketBhat7
6 ай бұрын
This is one of a very few hard leetcode problems that I solved without any help. But indeed this video dives to the depth of the concept, watching to understand what other ways I could have solved it. Thanks!!
@Andrew-dw2cj
13 күн бұрын
do you usually draw out a diagram and plan everything out before solving or do you just dive straight into the problem?
@SanketBhat7
13 күн бұрын
@@Andrew-dw2cj Depends on problem. If I get an idea of what would be the flow, then I just jump to code. Otherwise I will think, write a pseudo code and then proceed to coding.
@Andrew-dw2cj
10 күн бұрын
@@SanketBhat7 how long have you been doing leetcode for you to be able to just jump straight into the code?
@SanketBhat7
10 күн бұрын
Had been in leetcode platform for more than 2 years but never had been this consistent. Recently started a consistent streak of 2+ months
@mohamadilhamramadhan6354
Жыл бұрын
It tooks me 12 hours to solve hard problem for the first time lol. I even use a linked list haha. And the runtime is bad but at least pass the test. I always look at your solution after I try by myself and... I often learn something new. Thanks 👍
@lomoyang3034
2 жыл бұрын
Another issue. In your code, the window map also count the character which does not exist in T, which is different from what you explained in slides. AFAK, your implementation is different than what you explained, though both of them are sliding window.
@bas5rocker311
Жыл бұрын
no, I think the implementation matches the code. Try and pop from left. If the character popped is not in 't', it throws a key error
@amrutaj28
Жыл бұрын
I was going to point out the same. In both push (for right char) and pop (for left char) places, we need to just push the chars that exist in t-map. And this code modification works perfectly according to his explanation. PS. Thank you so much Neetcode for this great video!
@hwang1607
10 ай бұрын
this is my modified solution using the explanation, the curr hashmap is a map of the chars in T all set to 0, and chars not in T are not added from S class Solution: def minWindow(self, s: str, t: str) -> str: tcount = collections.Counter(t) curr = {k : 0 for k in t} l = 0 have = 0 need = len(tcount) minlen = float('inf') res = [0,0] for r in range(len(s)): if s[r] in curr: curr[s[r]] += 1 if curr[s[r]] == tcount[s[r]]: have += 1 while have == need: if (r - l + 1) < minlen: minlen = min(minlen, r - l + 1) res = [l,r] if s[l] in curr: curr[s[l]] -= 1 if curr[s[l]] < tcount[s[l]]: have -= 1 l += 1 if minlen == float('inf'): return "" l,r = res return s[l:r+1]
@mruduladdipalli5417
Жыл бұрын
With new test case of s = "bbaa" , t = "aba", it's not going inside while loop because "need" end up to 3 and "have" end up with 2, Since we check for map values of currentChar, which is updating have to 3, but need is at 3, because of length of t, it has 3 as value so we need to update "need" = uniqueCharactersOf(t); which we can achieve using set
@davidmar8612
Жыл бұрын
I made this change and it worked thank you!
@miserablepile
4 ай бұрын
I'm at a point where I draft up my approach, then watch Neet explain the approach to see how I did. Then I draft up the code on my whiteboard, then watch how Neet writes the code. Feelin good!
@Yuipser
2 жыл бұрын
Thanks for another greater video ! after watching your video on 567. Permutation in String, I could figure out solution for this by myself. ur videos are always very inspiring and helpful ! also, just want to share how I handle the left position differently: 1. keep the left pointing at a char in t, 2. if countS[nums[l] ] > countT[nums[l] ], increase the left because we can shorten the substring by making left pointing at the next num in t these condition will make sure that redundant elements on the left are dropped before we calculate the len of current substring
@sar3388
Ай бұрын
Were you able to code it yourself?
@SeifGneedy2
Жыл бұрын
Great Explanation, Thanks for everything! a small note: if we checked the entire hashmap, it will also be in O(1) because we have just lowercase and uppercase characters (Just 52 entries in the hashmap).
@algopenne
Ай бұрын
agreed, I think something like this works just as well (at least it passes on LC) class Solution: def minWindow(self, s: str, t: str) -> str: t_counts = {} w_counts = {} #window_counts for c in t: t_counts[c] = t_counts.get(c, 0) + 1 w_counts[c] = 0 l = 0 r = 0 def is_good_window(): for c in t_counts: if w_counts[c] < t_counts[c]: return False return True curr_len = len(s) + 1 res = "" while r < len(s): w_counts[s[r]] = w_counts.get(s[r], 0) + 1 while is_good_window(): if r - l + 1 < curr_len: curr_len = r - l + 1 res = s[l:r+1] w_counts[s[l]] -= 1 l += 1 r += 1 return res
@tanaykamath1415
Жыл бұрын
I love how he explains and codes stuff elegantly !
@sravanikatasani6502
3 жыл бұрын
Hello, Bob Ross!. Amazing explanation as always :)
@tomonkysinatree
4 ай бұрын
Glad you made this solution video. I felt like i was close but wasn't able to test all the test cases. My approach was slightly off by trying to perform the solution with a single hashmap. Made some of the conditionals/updating super confusing and complicated and would have not finished in a real interview.
@infinity-clips9414
Жыл бұрын
The legend comes and make it a piece of cake as usual. Your explanations are way more on another level. thanks man.
@shashwatkumar6965
2 жыл бұрын
If String t has repeated characters, then instead of have+= 1 we can do have += countT[c] and similarly instead of have -= 1, we do have -= countT[s[l]]
@JulianBoilen
2 жыл бұрын
7:56 Comparing the maps It wouldn't be bounded by t, it would be bounded by the number of possible characters, which is 52, or O(1).
@milapshah1075
2 жыл бұрын
I agree code needs one modification where you count need
@sudhanshusingh-cy9wp
Жыл бұрын
this might be submitted, but in an interview, this solution will considered much much better, i think
@MrTulkonas
Ай бұрын
I suggest an alternative solution, which I think it is far easier to understand and code. 1. Start defining a goal dictionary (with counts of "t" string) and worst dictionary (with counts of "s" string) -> current solution. 2. Check if solution valid (counts in goal smaller or equal to current solution) => solution exists, else return "". 3. Initiate a while loop with pointer l, r = 0, len(s) -1 3a If utmost left character not in goal, increase l. 3b Elseif utmost left character in goal and its count in current solution bigger than in goal, increase l and reduce count in current solution by 1 3c,3d elseif ... (analog for utmost right character) 3e return s[l:r+1] 4 return s[l:r+1]
@danomov
Ай бұрын
To optimize our window shrinking process, I think we can store the index of the new valid value. This way, next time we can move our start point directly to the next valid value and begin counting from there.
@theresabarton858
4 ай бұрын
Much simpler solution: keep incrementing the left pointer as long as the substr cond is satisfied.. Also don't worry about comparing the 2 dicts, its still O(1). def minWindow(self, s: str, t: str) -> str: cond = lambda c1, c2: all([c1.get(k, 0) >= c2[k] for k in c2]) sCounts = defaultdict(int) tCounts = dict(Counter(t)) minWindowSz = float('inf') minWindowIdx = 0, 0 L = 0 for R in range(len(s)): sCounts[s[R]] += 1 while cond(sCounts, tCounts) and L 0: sCounts[s[L]] -= 1 L += 1 return s[minWindowIdx[0]:minWindowIdx[1]+1] if minWindowSz < float('inf') else ""
@sumitsharma6738
Жыл бұрын
398 ms done. Took me 40 minutes to do so. but now lets see optimised version too :)
@joshstevens2779
Жыл бұрын
I didn't do this in Python but another language. There seems to be an issue with line 16. The condition should be checking if window[c]
@najwadafir217
11 ай бұрын
yes, you're right! else, the test s = "aa" t = "aa" won't pass
@jey_n_code
11 ай бұрын
thats true, thank youuu!
@straightlegmusic5269
10 ай бұрын
Was watching this on the neetcode site, literally opened it up to scan the comments for this issue cause I was doing this with JS lol.
@cinimodmil
9 ай бұрын
thanks for this! so in this case, have represents the number of characters needed to build substring t? i.e., we take into account duplicates when counting too.
@amitgupta1202
Ай бұрын
you can check dictionary sizes if we treat absence as zero. Complexity remains same but less no variables
@ishanpadalkar9072
6 ай бұрын
I had this solved intuitively, in quadratic time i'm guessing which is why it was failing the last test case on Time. This is a great explanation 👍
@harshavardhanranger
3 жыл бұрын
whenever I find your video for a question I'm looking for, happiness = float("inf") !!
@rishabh05
2 ай бұрын
Instead of maintaining a set for visited nodes, we can just mark a visited node as 0 and we can then skip it the next time. This spaces us the space complexity involved with the set.
@uvarajupdates4464
6 ай бұрын
Keep posting videos regularly brother❤
@lal_1404
22 күн бұрын
thanks for explaining in this simple words, otherwise people had made this question so hard and complex
@wangfred
Жыл бұрын
When it runs, it runs like Rolex, so intriguing and accurate! There are two windows actually: the l r and have need.
@Goodm26
2 ай бұрын
The time complexity should be O(n + k) instead of O(n), where k is the length of t. This is because in the beginning of the code we iterate over t to count its letter frequency.
@MalushJ
Ай бұрын
since we're dealing with capital letters (assuming thats all we have to check) the maximum number of checks is 26, its not dependent on the size of t.
@jegadheeswarank6290
5 ай бұрын
Awesome! Now I am going to code this in Java 🙂
@alejandrodardon7091
2 жыл бұрын
Amazing video with a great efficient solution, despite the code being a little bit all over the place at first glance I could perfectly follow your explanation great job!
@bostonlights2749
11 ай бұрын
I was asked this in an interview yesterday
@RajiurRahman1
Жыл бұрын
Hey! Big fan here. Just wanted to point out, this code will fail for s= "bbaa" and t="aba" (Leetcode testcase). In line#17, you have the following (C# version) if (countT.find(c) != countT.end() && window[c] == countT[c]) For the aforementioned test scenario, countT['a'] is 2, while looping through s, for the first time when it sees 'a', window[c] == countT[c] this part of the statement will become false, hence have will not be increased. After finishing the first for loop, count will be 3 and have will be 2. If there are 10 'a's in s and 10 'a's in t, the have will have 1 after the loop. Therefore it'll return an empty string "". Changing it to following will pass the case. window[c]
@xmnemonic
Жыл бұрын
It's a good solution, but I think it's very unlikely anyone could come up with the have, need variables in an interviewing first time seeing this problem. That is a very clever optimization.
@RandomShowerThoughts
9 күн бұрын
I hate how clear you make things lol, like I had an idea, but then once I heard you talk about it, I figured it out
@nightmarauder2909
2 жыл бұрын
you can use Counter(t) to avoid the first loop
@gouravsingh9509
3 жыл бұрын
Thanks a lot for the amazing explanation with each video. It has been so useful to understand things.
@thisisnotok2100
9 ай бұрын
I improved upon this by using a single hash map containing the amount of characters in the t string, and decrementing the value as it was encountered in s. As we closed the window, we would re increment back up. If all the values in the map were 0 or less, we knew we had a valid solution.
@sumeshbharathiramasamy8559
8 ай бұрын
checking the whole map for 0 or less --> will take 0(m). In the video, the approach has 0(1) time complexity
@harishankarkarthik3570
6 ай бұрын
I solved this by myself, before seeing this solution. And I checked here to know that it was the most optimal solution :)
@mingjuhe1514
2 жыл бұрын
Thanks bro ,I am keeping watching your video these days. I feel I am improving in a fast speed.
@yufeiqiu4519
3 күн бұрын
Took me two hours to solve this before watching your video gosh
@Alexis-ym9ph
10 ай бұрын
Difficult to image that someone would genuinely come up with the solution on the fly. Only if one have solved exactly this problem before. So, what companies like Google are looking for --- diligent applicants who solve 200-300-400-500+ problems on leetcode and memorize approaches? I got my job in the startup this way. I solved about 80 the most popular problems on leetcode. And on the interview I got 2 out of 3 which I solved perfectly, and the 3-rd one(which I didn't see before) I solved in a brute-force way. That was enough to get in. So for Google, Facebook, etc. you just need to solve more, especially hard ones. That's it!
@aditi1786
Ай бұрын
Hi Alexis Which questions did you solve? I am going to interview for google next month. Your input would be great. Even if you see this message later than a month, do reply, it might help someone else. Thanks
@pavankumarraja6848
3 жыл бұрын
Shouldn't line 17 be " if c in countT and window[c]
@Kai-iq2ps
3 жыл бұрын
Same question had. Watch from 13:48 . Should help you.
@tengma1020
3 жыл бұрын
need = len(countT) but not the len of String t, that takes care of duplicated char
@KANKSHMusic
3 жыл бұрын
Thank you pavan
@brieflyfun
Жыл бұрын
"need" is number of unique characters in t, not total number of characters. I think the problem is in line 26~28, which should have the same == conditions as line 17. if s[l] in countT and windows[s[l]] == countT[s[l]]: have -= 1 windows[s[l]] -= 1 to make the count balanced. "have" is the number of unique characters that meets the required numbers.
@kareni7572
2 ай бұрын
Thank you for showing why we need that extra variable on top of hashmap & also that it will only be updated if it is ==
@JanneSauvala
Жыл бұрын
On line 10, it should be "have, need = 0, sum(countT.values())" because if a character is present more than once in t, then it is not enough to check the length of the dictionary (number of keys) but you would need to sum up the dict's values.
@anonymoussloth6687
3 жыл бұрын
At 19:17 why is need set to size of countT? If t is "aab" then we would need 2 a and 1b right? But ur code will set need at 2
@RV-kl2wl
8 күн бұрын
Why does the need variable at 19:22 is set to count the unique characters only and not the count/num of occurrence of chars? It's because the windowCount map[s[l]] has >= count[s[l]] i.e during the comparison we will take care of the count. If t has "aaa" then step 1 is to check if string s has 'a' or not, later by imposing >= we make sure that sliding window has equal or greater occurrence of chars.
@sastecoder3421
Жыл бұрын
Lord NeetCode SUPREME Teacher !!
@MP-ny3ep
6 ай бұрын
Phenomenal explanation !!!! Thank you sooo very much !!!
@AlessioPepe
4 ай бұрын
I wan't able to figure out how to reduce complexity from O(n*k) to O(n). Thank you for explaining the trick so easily!
@aimie2837
2 жыл бұрын
def was given this at google
@PippyPappyPatterson
2 жыл бұрын
did u slay ?
@bouzie8000
7 ай бұрын
This was so so fun lol. You're doing the lord's work
@kchemutai3483
Ай бұрын
Best explanation ever. Thank you neetcode
@sucraloss
11 ай бұрын
I was so close on this one! I didn't think of that have vs need thing you did so I only solved half the test cases because I moved the left pointer properly but only changed the min_window_size when the count in t matched the count in s, which failed cases where we had duplicates. I couldn't think of a quick way to compare that we have enough of each specific value in the "s" hashmap such that we could check to see if our window was smaller. I was thinking maybe you would just have to make a helper function to go through the "s" hashmap's keys and make sure you have at least enough of each letter to match the count in t, but making a single integer increment when you have enough for each key is smart. I didn't even bother implementing my hashmap count helper function because I figured I was missing some way of tracking the size of the s and t hashmap and didn't want to spend 20 minutes just to get to a TLE. *Edit: I actually did check to see how slow my helper function would be and it was 80% slower than NC's solution. So it could work to check through the hashmaps key by key each time, but it's super slow. Good to know I was really very close to solving this.
@cathyhuang8557
3 жыл бұрын
Thank you very much~ It is really helpful~
@yuxuanc
2 жыл бұрын
Best explanation ever for this question, thank you for the great work!
@garitina987
Жыл бұрын
Wow that was incredibly explained. Hats off once again.
@RanjuRao
Жыл бұрын
Thank you for your detailed explanation on the approach!
@scottishfoldmocha5875
2 жыл бұрын
19:46 - why are we updating window hash map before checking if that character is in countT hash map? Aren't we only storing and counting characters we need and not ALL characters?
@linli7049
3 жыл бұрын
You are such a good trainer
@aaronkranzler4830
20 күн бұрын
beautiful explanation as always
@amanand727
2 жыл бұрын
one more optimization that we can do is to skip sliding the window once the size becomes equal to the current minimum window size in the result.
@jawaharSelvaraj
Жыл бұрын
What if this time we can shrink to even minimum window size - 1
@yangjiawenxu2319
Жыл бұрын
It is very clear for a hard leetcode problem.
@sports_op
Жыл бұрын
Whoa, it's something I'd never have come up with by myself 🔥
@naehalmulazim
10 ай бұрын
Skipping the coding solution part for now. I watched the explanation but it still seems worth it to try to code it on my own first, because it was a bit abstract.
@cricophobiya4178
7 ай бұрын
from collections import Counter def minimum_window(str1,str2): window="a"*(len(str1)+1) left=0 count=0 hashmap=Counter(str2) for i in range(len(str1)): if str1[i] in hashmap: hashmap [str1[i]]-=1 if hashmap [str1[i]]>=0: count+=1 while count==len(str2): if len(str1[left:i+1]) < len(window): window=str1[left:i+1] if str1[left] in hashmap: hashmap [str1[left]]+=1 if hashmap [str1[left]]>0: count-=1 left+=1 return window str1='ADOBECODEBANC' str2='ABC' print(minimum_window(str1,str2)) str1 = "PRWSOERIUSFK" str2 = "OSU" print(minimum_window(str1,str2))
@challengeyourmind3937
Жыл бұрын
24:55 is not an off by one "error"... substring is just non-inclusive for end of range
@feesabilillah101
10 ай бұрын
The best explanation hands down
@TheSSB007
3 жыл бұрын
Loved the explanation!!
@jonathandoe69
Жыл бұрын
I think this solution is failing test cases with duplicate letters in t. For example, s="aa" and t="aa". When you reach 2 a's, your 'have' variable becomes 1, but with 'need' set to length of t, that variable is 2 and they will never be equal.
@atulkumar-bb7vi
Жыл бұрын
Very nicely explained the hard problem. Really liking your videos. Thanks for this...
@wuzi9657
Жыл бұрын
Thank you for the video! It helps a lot! I think they might have updated the constraints. It is guaranteed that the length of s and t is >= 1 as of 11/3/22, so it is not necessary to check the edge case where t is empty now.
@danny65769
2 жыл бұрын
Should line 27 be: if s[l] in countT and window[s[l]] + 1 == countT[s[l]]: ? Otherwise, if we try to shrink the window twice with the same left character, the "have" would be decremented twice for the same character?
@danny65769
2 жыл бұрын
I realized once "have" is decremented, the while condition of "have == need" would fail. So "have" would not be decremented more than once for the same character during any one window shrink operation.
@anmoljhamb9024
Жыл бұрын
Also, isn't your solution O(n+m) because you're creating a hashmap for the smaller string which is M long. I had a doubt, isn't the bruteforce solution also just O(56*N+M) and it also reduces ultimately to O(N+M)? but yeah, ofcourse it isnt' at all optimised, but doesn't that satisfy the paramters of the follow up asked in the description?
@chillsjiujitsu
6 ай бұрын
Neetcode is the goat!!!
@shriharihallur633
3 жыл бұрын
Very well explained, Thanks!!
@_dion_
6 ай бұрын
excellent explanation as always. i wonder how can someone come up with this solution in an actual interview of 45 min if it's a totally new problem?!!!
thank you for the crystal clear explanation and code
@aryanyadav3926
2 жыл бұрын
Wonderful explanation!
@dorondavid4698
2 жыл бұрын
During your explanation you missed listing CODEBA as another substring of length 6
@cortex-technologies
4 ай бұрын
Does this particular technique has any name? I mean how the two hashmaps are being compared without going through all the elements? This optimization technique seems so cool.
@thinhnguyenvan7003
2 жыл бұрын
Hi NeetCode. Your video is amazing and helped so much in some problems that i did not have any idea to solve. By the way, when you apply in GG, are there any other questions related about System Design or something like that?
@kafychannel
Жыл бұрын
Thank you NeedCode I got the entire idea of your solution!
@genericstospecifics1856
8 ай бұрын
*****RANT***** I started solving this problem 4 hours ago, watched multiple other solutions before this video. But in all the 3 solutions I could write in these 4 hours, none worked. This is so mentally draining; I feel like quitting the 3rd or 4th time in last 3 years. I have so much aggression in me, it feels like my head is bloating. I feel stupid and worthless.
@sudhanshusingh-cy9wp
Жыл бұрын
Instead of left, and right I was saving the substring in a string variable, that solution gave me memory error, but when using left and right It gives the correct answer, I am curious of how much extra overhead that might be adding that a new variable which cant be greater then the length s, is giving for a memory error
@hoyinli7462
3 жыл бұрын
finally you uploaded your code! thx
@eknathyadav8744
2 жыл бұрын
Hi neetcode, One question Suppose s = "ACDFCA and t = FCA. if we have found one window "ACDF" which contains "FCA". Can't we just start our window from "F" instead of going through "CDF..." Since we know our shortest window will always start from there only. btw thanks for the clear explaination as usual.
@dahomosapien
Жыл бұрын
Yeah exactly!! The code can be optimised more. You’ve probably use a first-in-first-out queue data structure
@case6339
Жыл бұрын
I think we cannot. Suppose s = "ADCAFCEA" and t = "FCA". If we have found one window "ADCAF", the next minimum window found would be "CAF". If we just start our window from "F" we miss that window.
@Spham99
2 жыл бұрын
doing java in interviews always messes me up, I get so caught up in trying to figure out how to do the little things :/ I tried switching to Python for that, but that messed me up too trying to learn a different language
@tokphys
Жыл бұрын
you don't need to store all characters in window dict, just those that exist in countT
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