Thanks to Mailgun for sponsoring this video! Head to mailgun.com/john to try Mailgun free today. Looking forward to seeing how you guys make this solution even better!
Not able to understand how this algo will work for some string like "abcdcga" as we will keep our left pointer at first a and we will increment right pointer, at when right pointer at second c it will get repeated letter and will increment left pointer to b and in next iteration it will move to g and will register length of bcdcg but this substring is already having repeated character. Rather than incrementing left pointer by one we should increment it by lastCharOccuredIndex+1
You genuinely have one of the best programming explanation videos on this site, honestly. Funny, I was discussing learning about the Sliding Window algorithm to practice Leetcode questions with a friend yesterday, and lo and behold you've uploaded a great explanation literally 24hrs later, legendary! Could you please upload more Leetcode explanation questions? If not on TH-cam, perhaps a course?
I had just solved this problem and found this video. You explained each solution and your approach in the best way. Your way to explaining make things crystal clear John !!
This is probably one of the best tutorials out there and i think TH-cam is with me in this.❤️ I could you not, even though I have watched it already it is every time the first video in my recommended 😂 Guess I'll watch it twice. Keep up the good work!
The way you walked through the solution was like i have never seen anywhere. Most youtubers just jump to the optimal solution. please do more of these! I went to your channel hoping to find a video series of Blind 75. but wasn't there. Please do a playlist of Blind 75. I bet people would love to see those.
John I love your Java explanations, can you please add more leetcode to your channel or bootcamp? I would love to see more Java leetcode solution guides properly explained
Great job, please keep posting the leetcode solution videos in structural manner(for eg top 50 that includes most practices/datastructures) , this helps a lot in preparing for interviews.
John, you’ve helped me understand DS&A in Java better than before. The way you explained things are easy to understand and follow. I’m a visual learner and your explanations makes it easier for me visualize & understand. Thank you and please keep these leet code videos coming. Your other videos on Java has truly helped me understand Java so much more than my professors lol you are incredible in the work that you’re doing for me and many. We cannot thank you enough ❤ I appreciate you.
Definitely one of the best explanations or just the best, starting from the basic one (brute force) just to kick off and let viewers like me grasp the idea before jumping to more complex solutions. Thank you!
Hey John, I love all of your videos. Learning so many important skills. However, I have one suggestion, If you could make more of this type of problem-solving videos then it would be very helpful to the viewers as it will teach new programmers how to think of a solution to the given problem and how can we actually implement the solution using the programming language. By the way thanks again for all efforts that you put in to make this possible. Looking forward to seeing more problem-solving videos. Have a wonderful day.
Great and in-depth explanation for every approach. I really liked this video. I have found another better approach final int n = s.length(); int len = 0; int [] repeat = new int[128]; for (int c = 0, j = 0, i = 0; j < n; j++) { c = s.charAt(j); i = Math.max(repeat[c], i); len = Math.max(len, j - i +1); repeat[c] = j+1; } return len; I somewhat understood, but it would be better if you can explain. Thanks John and keep creating more videos for different problems.
Hi John. I previously implemented this solution (yours) using a while-loop (and using the map) instead of a for-loop and got similar runtime numbers to the last solution you found (6ms runtime). Not sure why that's the case. But yeah, the last solution, intuitively should be a worse runtime even though it seems like it performs better. It's also a clever solution too. Nice work!
For the brute force solution. I suggest checking if the max need to be updated in the "if contains part" that makes sure that we only do it right before we break. It is one time per substring. We fount a substring, now, let's check if it is better than the other we had so far. Second, I will suggest using HashSet instead of StringBuilder. HashSet has contains method and it is faster.
Hey John! First of all thanks for the amazing video! Short answer for why indexOf is faster in this case: It is simply because in this case it has an O(1) runtime, how? here is how: Before you slide your right pointer to the right you have already made sure that this substring has different characters, which in the worst case will be 24 characters long. Since you know it will never be longer than 24 characters, or 34 if numbers are included in the string, then it is constant time, because no matter how big your string will be, it will be in the worst case that maximum substring :D so it is related to the valid substring and not to the input! :D And since the map indeed has access and write time of O(1) it still has an overhead for hash calculation and storing and etc... which take more operations / time than indexOf in that case, but since none of them is related to n, the one with the shorter/faster operations wins :D I hope that was helpful!
Hi John I like how you present your leetcode solutions, so easy to follow and understand. Hope to see more of this or if you have anothe platform teaching leetcode problems , I would be happy to know.. Thank you
Hi, first time watching your videos, and I love the clarity of your explanation. I think I figured out the reason of Map been more slowest than last example, when you use the hash map you as writing and updating the data, to have the current letter position. On the other side the s.index only search if exist this value.
You have to be careful when using a MAP to lookup something because, even though its O(1) in time complexity, there is a lot of overhead. Let's say that the constant time for a lookup in a MAP is always 500 nanoseconds, if the indexOf (that has a time complexity of O(n)) takes only 300 nanoseconds because it finds the answer in the first few characters, then the indexOf will perform better. This obviously can be measured and we can come up with some threshold that tells us what algorithm to use in each situation. My guess is that in the English language indexOf will always perform better because we are dealing with words that are small in size (there is always the repeating space character).
You're videos are so helpful, one 5 minute video is worth 6 hours reading a chapter in a book. Please can you do some videos or a playlist on design patterns. Thanks!
I'm currently just at the brute-force method in the video, but for the lookups I'd probably use a hashset. I would therefore prevent an O(n²) lookup. But the brute-force method was the first idea for me as well.
Interesting! I am very honest with myself, I am not seeing the full picture when it occurs to recursion; However, I did grasp the idea of the "quicksort" method a bit better, not 100% but we are learning. I can see this is almost similar to an array.
hi John, The way of explanation which you are giving is excellent. Since, I am mostly working JAVA. You videos are very mich helpful for me to achieve greater heights in my life. I mostly use ECLIPSE for my Java Projects. If possible please guide me in using Intellij with each and every shortcuts. Thanks in Advance, Logesh
Hi there!! I really like your videos so much you make them easy to understand!! I want to ask you something can you recommend a book or books to improve our problem solving skill and to learn data structures
I know this is a bit late, but I believe Map is quite slow in runtime from what I heard and so it doesnt give off the best speed, Map is a great to use if you dont care much about speed as much as you want cleaner and simple code (tho run time is and should be priority)
Awesome video!!! You are by far the best coding tecaher i could find in internet. Coud you do a video explaining how to read a csv file in Java?? It would be great. Really very good content, keep doing it
The reason for the final solution being faster is because while hash tables have "constant" lookup time it still takes time. Depending on the hashing function and the initial size of the hashmap the lookup, add, and resize functions may be more time consuming than just linearly looking through an array of characters. Especially if the hashing function has division, because of how bad computers are at it. So rather than your hashmap having O(1) time complexity for lookup its probably more like O(100) which is potentially smaller than the size of n. Sorry for the long rambling answer, but yeah hashmaps are great, if you know how the data is being put into them. They also take up a *bunch* of space, so finding an algorithm that is similar in functionality, but doesn't use them is always the way to go.
True, just to add your answer, hashmap has constant time complexity in average case, but O(n) in worst case. So it's not 100% perfect n effective in all cases
Thanks John for the amazing video!! As for maps - what is the low level implementation of maps in JVM code? I mean, does it calculate the hash for the key when put or get methods are invoked? If it does, then adding or reading keys will be quite expensive , and explains why indexOf works faster.
Thanks John for awesome video and great explanation. For the Map solution wouldn't it also help to add an exit condition to break the loop if (maxLength > input.length() - right + 1) - so we need not iterate the remaining characters in the input.
Maybe HashMap takes longer because I the calculation of the hash? To get to a value HashMap first calculate hash value, and only then go to the location.
I would never thought of HashMap is it possible to do the same thing with a list just if list.contains(a) list.remove(a) list.append(a). Could you pls make a video about how fast all these datastructures in comparison are. you can explain very good :)
given that there's a finite and constant number of letters, you could probably replace the HashMap with a regular array for a negligible decrease in time and memory
Yeah it could be the case that despite the technical time complexity being lower, because n will never be more than 50,000, simple sequential array searches might just be more performant
@@cipherxen2 and a character can be found in this array in O(1) time by - int lastEncountered = charIndexes[currentChar]; Initialize charIndexes[255] with all -1;
Thanks for the clear explanation. I'm struggling with tech interviews. How am I supposed to complete a challenge in a short time? I just got stuck in the last one. I'm trying so hard to understand. I've been practicing LeetCode exercises, but there's always something tricky, and I've never seen anything similar before. It's so frustrating
@31:42 -> is there a better way to find length of the substring(right - left + 1)? Can we do something like s.subString(left, right) and pass this as an argument to s.length() somehow and return the length of that substring? thoughts?
as for the better time, my best guess is a combination of the sample data and the relatively massive overhead of the default HashMap implementation, using arrays to implement your own map SHOULD in theory produce better results, I'll test it later. Or you could just pass 127 as the initial capacity of the map?
I'm not entirely sure, but the answer to the difference in time complexity might be in how indexOf is implemented in the first place. I'm not an expert at all, but what if the indexOf method itself already utilizes the exact same Map implementation? Meaning, your Map implementation could be a duplicate of indexOf(), except indexOf() doesn't have the extra steps within the implementations of contains() and get() because it's already encapsulated in core Java. I'm just guessing here, of course. Pretty sure I could be wrong.
The max possible substring is bounded by the number of possible characters in the string, m, which given the problem constraints seems to be at most ~100ish. And, I would imagine on average the actual distance between the left and right pointers will be much less than this. This then means you are *effectively* comparing two linear solutions (even though the indexOf solution is technically nm - m is small). I would imagine that for small values of m the simplicity of linearly looping through a few characters of a string would be much faster than map lookups and manipulation. It would be interesting to see if the map-based solution would surpass the indexOf solution by increasing m, by, for example, using all UTF-8 chars.
Well the problem states that the input string may be up to 50,000 characters long. But I do think it's a good insight that the max length of the longest substring is bounded by the number of unique characters in the character set being used. In this case it says "English letters, digits, symbols and spaces." I'd have to add all those up to be sure, but let's say it's maybe 100, which may be a big reason the map solution is overkill at this level
I'm assuming this last approach is faster because the window defined by 'left' and 'right' never ends up being long enough to operate less efficiently than when using a map. I mean, Imagine the 5*10^4 (50000) String where ALL its characters are different (this means, maxLength would be 50000). Everytime you call 'indexOf()', Java is basically running a O(N) algorithm, where N is the length of the window we use. Said window starts with length 0, it's increased each loop and its length ends up being 50000. We have a complexity O(50000!), don't we? For this kind of scenario, unless I am mistaken, using a map is a way better choice.
for the version that uses the HashMap, I would suggest to have an else. when we are in the "if contains" we are always shortening the substring and hence no need to update the max.
Thanks to Mailgun for sponsoring this video! Head to mailgun.com/john to try Mailgun free today.
Looking forward to seeing how you guys make this solution even better!
Can you let us know the compiler you use for the videos?
Not able to understand how this algo will work for some string like "abcdcga" as we will keep our left pointer at first a and we will increment right pointer, at when right pointer at second c it will get repeated letter and will increment left pointer to b and in next iteration it will move to g and will register length of bcdcg but this substring is already having repeated character. Rather than incrementing left pointer by one we should increment it by lastCharOccuredIndex+1
The complexity of your first algorithm is n cubed, not n squared! There is one more loop hidden in String.indexOf, looking for a duplicate character.
Best java chanel on TH-cam
Excellent!!
You genuinely have one of the best programming explanation videos on this site, honestly. Funny, I was discussing learning about the Sliding Window algorithm to practice Leetcode questions with a friend yesterday, and lo and behold you've uploaded a great explanation literally 24hrs later, legendary! Could you please upload more Leetcode explanation questions? If not on TH-cam, perhaps a course?
i already signed up in his course but it does not contain any exercises i wonder if there is another course where he solves problems
Same feelings here, we need more leet code problems...
please keep doing these, your explanations are so much more in depth than other youtube channels
I had just solved this problem and found this video. You explained each solution and your approach in the best way. Your way to explaining make things crystal clear John !!
John, you are a wizard. Everythings looks so logical and simple. Please make more videos like this.
This is probably one of the best tutorials out there and i think TH-cam is with me in this.❤️ I could you not, even though I have watched it already it is every time the first video in my recommended 😂
Guess I'll watch it twice. Keep up the good work!
The way you walked through the solution was like i have never seen anywhere. Most youtubers just jump to the optimal solution. please do more of these! I went to your channel hoping to find a video series of Blind 75. but wasn't there. Please do a playlist of Blind 75. I bet people would love to see those.
I appreciate that you go into the brute force solution and also the more clever solution. Thanks for these!!
I love the final thoughts on the differences of performance between the 2 last algorithms, answered all the questions I had to myself. Thnak you!
Would love to watch a complete java dsa course from you
DSA would be more easy and interesting with John 😀
An excellent in-depth explanation of two approaches to solving this problem. Thank you very much, John.
You are really a coding geek. Even the concepts known, still visit them and keep something. Your explanations are far reaching. Thank you.
John I love your Java explanations, can you please add more leetcode to your channel or bootcamp? I would love to see more Java leetcode solution guides properly explained
More leetcode videos plzzz! love this series.
💯
Yes please, solve all the challenges :D
I love this series. Thank you...
Great job, please keep posting the leetcode solution videos in structural manner(for eg top 50 that includes most practices/datastructures) , this helps a lot in preparing for interviews.
A unique way of explanation, clear concise.
John, you’ve helped me understand DS&A in Java better than before. The way you explained things are easy to understand and follow. I’m a visual learner and your explanations makes it easier for me visualize & understand. Thank you and please keep these leet code videos coming. Your other videos on Java has truly helped me understand Java so much more than my professors lol you are incredible in the work that you’re doing for me and many. We cannot thank you enough ❤ I appreciate you.
Thanks so much for the very kind words!
when we get full playlist of leetcode in java you are amazing i am from india @@CodingWithJohn
This is the most detailed and accurate explanation of this problem I've seen. 👍
Definitely one of the best explanations or just the best, starting from the basic one (brute force) just to kick off and let viewers like me grasp the idea before jumping to more complex solutions. Thank you!
Hey John, I love all of your videos. Learning so many important skills. However, I have one suggestion, If you could make more of this type of problem-solving videos then it would be very helpful to the viewers as it will teach new programmers how to think of a solution to the given problem and how can we actually implement the solution using the programming language.
By the way thanks again for all efforts that you put in to make this possible.
Looking forward to seeing more problem-solving videos. Have a wonderful day.
At 35:00 we should use indexOfFirstAppearenceInsubstring
Best channel ever ! , please keep going with the leatcode series
I liked your videos, you made them all easy to grasp! please do more leetcodes tutorials and data structures and algorithms content
I watched till 25:32 and was able to come up with a solution. Thanks for this amazing explanation!
Great and in-depth explanation for every approach. I really liked this video. I have found another better approach
final int n = s.length();
int len = 0;
int [] repeat = new int[128];
for (int c = 0, j = 0, i = 0; j < n; j++) {
c = s.charAt(j);
i = Math.max(repeat[c], i);
len = Math.max(len, j - i +1);
repeat[c] = j+1;
}
return len;
I somewhat understood, but it would be better if you can explain. Thanks John and keep creating more videos for different problems.
Excelent explication of a leetcode coding challenge, thanks i hope to see more videos of coding challenges medium and hard
this series is actually so good.
Hi John. I previously implemented this solution (yours) using a while-loop (and using the map) instead of a for-loop and got similar runtime numbers to the last solution you found (6ms runtime). Not sure why that's the case. But yeah, the last solution, intuitively should be a worse runtime even though it seems like it performs better. It's also a clever solution too. Nice work!
It's great! Please, do more Leetcode Exercise explanations!🙏
this channel has wonderful content, and easy to understand explaination.
Nice to learn from you John, can you add more solutions from LeetCode?
For the brute force solution. I suggest checking if the max need to be updated in the "if contains part" that makes sure that we only do it right before we break. It is one time per substring. We fount a substring, now, let's check if it is better than the other we had so far. Second, I will suggest using HashSet instead of StringBuilder. HashSet has contains method and it is faster.
Excellent explanation and solution. I clearly understood the problem and solution. thank you
Hey John! First of all thanks for the amazing video!
Short answer for why indexOf is faster in this case:
It is simply because in this case it has an O(1) runtime, how? here is how:
Before you slide your right pointer to the right you have already made sure that this substring has different characters, which in the worst case will be 24 characters long. Since you know it will never be longer than 24 characters, or 34 if numbers are included in the string, then it is constant time, because no matter how big your string will be, it will be in the worst case that maximum substring :D so it is related to the valid substring and not to the input! :D
And since the map indeed has access and write time of O(1) it still has an overhead for hash calculation and storing and etc... which take more operations / time than indexOf in that case, but since none of them is related to n, the one with the shorter/faster operations wins :D
I hope that was helpful!
Love the series and quite a nice explanation on the different approaches, although the last one is confusing for me.
This is awesome way of teaching! Thanks John!
Best Explanation. Please make videos on All LeetCode's 145 top interview questions. You will be immortal for the Computer Programmers Community.
Can you please make more of these leetcode java solutions videos. They are of great help and you make us understand wonderfully.
I will when I can!
Hi John I like how you present your leetcode solutions, so easy to follow and understand. Hope to see more of this or if you have anothe platform teaching leetcode problems , I would be happy to know.. Thank you
You make programming easy and simple
Hi, first time watching your videos, and I love the clarity of your explanation.
I think I figured out the reason of Map been more slowest than last example, when you use the hash map you as writing and updating the data, to have the current letter position. On the other side the s.index only search if exist this value.
The best explaining ever, hoping to make more videos plz or DS and Algo videos. Thank you.
Can you please do more videos like this as you explain it in a great way.
Wow so easy to learn and understand , you have a new student sir.
More videos like this with solutions for diferent letcode, hackerank problems this is very much usefull to improve skills, you explain as a master
You have to be careful when using a MAP to lookup something because, even though its O(1) in time complexity, there is a lot of overhead. Let's say that the constant time for a lookup in a MAP is always 500 nanoseconds, if the indexOf (that has a time complexity of O(n)) takes only 300 nanoseconds because it finds the answer in the first few characters, then the indexOf will perform better. This obviously can be measured and we can come up with some threshold that tells us what algorithm to use in each situation. My guess is that in the English language indexOf will always perform better because we are dealing with words that are small in size (there is always the repeating space character).
My thoughts exactly
Your explanations are very good. More leetcode!!!!!!!!!
You're videos are so helpful, one 5 minute video is worth 6 hours reading a chapter in a book. Please can you do some videos or a playlist on design patterns. Thanks!
I'm currently just at the brute-force method in the video, but for the lookups I'd probably use a hashset.
I would therefore prevent an O(n²) lookup.
But the brute-force method was the first idea for me as well.
Thanks for great explanations!
Continue this series in java please.Thanks john.
Yes, please!
Most LeetCode videos out there are either Python or C++
Definitely want to see more LeetCode walkthroughs in Java (or Kotlin)
i'm really impressed 👍👍. keep on going 👏👏👏
buddy make an in depth video for regex expressions please
Yeah, good idea. Regex really messes with the brain.
Спасибо Джон. Продолжайте и дальше радовать нас своими видео. Thanks John. Continue to delight us with your videos
Thanks John. Always putting these awesome videos.
if possible more videos like this, especially interested in thought process behind solutioon
Interesting!
I am very honest with myself, I am not seeing the full picture when it occurs to recursion;
However, I did grasp the idea of the "quicksort" method a bit better, not 100% but we are learning.
I can see this is almost similar to an array.
hi John,
The way of explanation which you are giving is excellent.
Since, I am mostly working JAVA. You videos are very mich helpful for me to achieve greater heights in my life.
I mostly use ECLIPSE for my Java Projects. If possible please guide me in using Intellij with each and every shortcuts.
Thanks in Advance,
Logesh
Hi there!! I really like your videos so much you make them easy to understand!! I want to ask you something can you recommend a book or books to improve our problem solving skill and to learn data structures
I know this is a bit late, but I believe Map is quite slow in runtime from what I heard and so it doesnt give off the best speed, Map is a great to use if you dont care much about speed as much as you want cleaner and simple code (tho run time is and should be priority)
need more tutorials please :) thank you!
Awesome video!!!
You are by far the best coding tecaher i could find in internet.
Coud you do a video explaining how to read a csv file in Java?? It would be great.
Really very good content, keep doing it
Thank you for the amazing and clear explaination.
Can you explain this next? 992. Subarrays with K Different Integers.
The reason for the final solution being faster is because while hash tables have "constant" lookup time it still takes time. Depending on the hashing function and the initial size of the hashmap the lookup, add, and resize functions may be more time consuming than just linearly looking through an array of characters. Especially if the hashing function has division, because of how bad computers are at it. So rather than your hashmap having O(1) time complexity for lookup its probably more like O(100) which is potentially smaller than the size of n.
Sorry for the long rambling answer, but yeah hashmaps are great, if you know how the data is being put into them. They also take up a *bunch* of space, so finding an algorithm that is similar in functionality, but doesn't use them is always the way to go.
True, just to add your answer, hashmap has constant time complexity in average case, but O(n) in worst case. So it's not 100% perfect n effective in all cases
great video. Keep them coming!
wow pls, do more Leetcode exercise explanations!
Thanks Please do more
Thank you, John.
love from india❤️ Pls upload a problem on daily basis
Great video, also I think the last part might be bc of best case scenario
perfect. you have more question answer videos, like a playlist of them? Thanks
This playlist is amazing, Why don't you continue this series of videos?
Thanks John for the amazing video!! As for maps - what is the low level implementation of maps in JVM code? I mean, does it calculate the hash for the key when put or get methods are invoked? If it does, then adding or reading keys will be quite expensive , and explains why indexOf works faster.
Thanks John for awesome video and great explanation. For the Map solution wouldn't it also help to add an exit condition to break the loop
if (maxLength > input.length() - right + 1) - so we need not iterate the remaining characters in the input.
Maybe HashMap takes longer because I the calculation of the hash? To get to a value HashMap first calculate hash value, and only then go to the location.
Yum just a subscriber with the initial sponsor ad
Awesome best explanation 👏
Great Video! But what book would you recommend for learning java?
I would never thought of HashMap is it possible to do the same thing with a list just if list.contains(a) list.remove(a) list.append(a). Could you pls make a video about how fast all these datastructures in comparison are.
you can explain very good :)
wow thanks 🙏
Thanks for this explaination🥰
given that there's a finite and constant number of letters, you could probably replace the HashMap with a regular array for a negligible decrease in time and memory
Yeah it could be the case that despite the technical time complexity being lower, because n will never be more than 50,000, simple sequential array searches might just be more performant
@@CodingWithJohn you don't need array of 50000 elements, just 255 for ASCII characters
@@cipherxen2 and a character can be found in this array in O(1) time by -
int lastEncountered = charIndexes[currentChar]; Initialize charIndexes[255] with all -1;
Lots of love from India...
Thanks for the clear explanation. I'm struggling with tech interviews. How am I supposed to complete a challenge in a short time? I just got stuck in the last one. I'm trying so hard to understand. I've been practicing LeetCode exercises, but there's always something tricky, and I've never seen anything similar before. It's so frustrating
More leetcode problems!!! I like this style of video 🧌
@31:42 -> is there a better way to find length of the substring(right - left + 1)? Can we do something like s.subString(left, right) and pass this as an argument to s.length() somehow and return the length of that substring? thoughts?
Also at 30:00 why not just do a left++?
as for the better time, my best guess is a combination of the sample data and the relatively massive overhead of the default HashMap implementation, using arrays to implement your own map SHOULD in theory produce better results, I'll test it later. Or you could just pass 127 as the initial capacity of the map?
thanks for the video !!
I'm not entirely sure, but the answer to the difference in time complexity might be in how indexOf is implemented in the first place. I'm not an expert at all, but what if the indexOf method itself already utilizes the exact same Map implementation? Meaning, your Map implementation could be a duplicate of indexOf(), except indexOf() doesn't have the extra steps within the implementations of contains() and get() because it's already encapsulated in core Java. I'm just guessing here, of course. Pretty sure I could be wrong.
I tried a better idea to use a fixed size array for character mapping it got more optimised thanks for the video need more leetcode content
The max possible substring is bounded by the number of possible characters in the string, m, which given the problem constraints seems to be at most ~100ish. And, I would imagine on average the actual distance between the left and right pointers will be much less than this. This then means you are *effectively* comparing two linear solutions (even though the indexOf solution is technically nm - m is small).
I would imagine that for small values of m the simplicity of linearly looping through a few characters of a string would be much faster than map lookups and manipulation. It would be interesting to see if the map-based solution would surpass the indexOf solution by increasing m, by, for example, using all UTF-8 chars.
Well the problem states that the input string may be up to 50,000 characters long. But I do think it's a good insight that the max length of the longest substring is bounded by the number of unique characters in the character set being used. In this case it says "English letters, digits, symbols and spaces." I'd have to add all those up to be sure, but let's say it's maybe 100, which may be a big reason the map solution is overkill at this level
29:53 What if we just did Left++ , I think still works the same, correct me if I am wrong.
I'm assuming this last approach is faster because the window defined by 'left' and 'right' never ends up being long enough to operate less efficiently than when using a map. I mean, Imagine the 5*10^4 (50000) String where ALL its characters are different (this means, maxLength would be 50000). Everytime you call 'indexOf()', Java is basically running a O(N) algorithm, where N is the length of the window we use. Said window starts with length 0, it's increased each loop and its length ends up being 50000. We have a complexity O(50000!), don't we? For this kind of scenario, unless I am mistaken, using a map is a way better choice.
If we're limited to the ASCII character set, then the maximum length will be < 128.
@@toby9999 fair enough
for the version that uses the HashMap, I would suggest to have an else. when we are in the "if contains" we are always shortening the substring and hence no need to update the max.
Really you are awesome 🎉
Hi Johnny sin, good to see in as a software developer role as well
Thank you
Could you please do one about Polymorphic load/save methods for Java Io file handling