You're an excellent teacher, John. I find your videos not only extremely helpful, but impressively well-made - and it's not even so much about the format (which is ideal), but the way you teach and the examples you give. You explain things in a logical, progressive way, and you're excellent at providing the right level of exposure. I do wish there were more teachers like you in universities - but then, we might not have you on RUclips ;) Please keep going with these tutorials, you are contributing to the community in a huge way.
@@yahia1355 oh I didn't do anything smart. I literally brute forced a solution. Basically going over every single square and trying the first number that works, then if I got to one that had no possible correct number I would go back until the first square that I could have put a different number in and tried the different number. Do that until you can fill every square correctly. Just literally trying every combination until it works. There are much better ways to solve this. But the not so smart solution works.
I had recently discovered this channel, haha, and I started to watch your videos on TV Like I am watching some movies or some kind of entertainment, nice videos keep it up, subscribed after watching 2-3 videos
Sir before I watch your video I was a student who was struggling to solve recursion problems for my upcoming test. Now I get the idea about how to backtrack in recursion :) Huge thanks from South Korean student!
This is exactly what I was looking for to complement my java studies. Projects with step-by-step explained. Please, keep it going, unfortunately here on RUclips, everything seems to be only JavaScript or Python. It would be great more projects with OOP.
I wrote a Sudoku solver in 2008 in Javascript, but it didn't quite use full recursion (my mistake) so it wouldn't solve moderately difficult puzzles. Mine includes character recognition of uploaded websuduko puzzles. This week I added "paste from clipboard" so that I didn't have to save/load my puzzle images. Today, following your model I finally got mine solving even the "evil" puzzles. Thank You John for this great lesson. Although now I don't feel that "I wrote the code", my page works and I did right the optical character recognition parts. I don't believe that I can share a link here, but I can try.
Thank you! Amazing video, where I work I tried several times create something to solve our work schedule with some kind of algorithm but always fail, this was just exactly what I needed and hope to solve my needs. Thank you!
Hi John, i have learned a lot from your videos. I hope you won't stop making videos in upcoming days, it is really helpful for us. I can guarantee you will get a huge subscribers in coming days.
You teaching skills are really impressive, great work! I have one minor point to critique, you hardcoded GRID_SIZE 9 but used 3 as a magic number. You could have computed it once sqrt(GRID_SIZE) or just declared it as a constant.
@@Michael-se7ny Really? Doesn't a Sudoku box always have side length of sqrt(GRID_SIZE)? But I guess the really clean way would be to define the sudoku in terms of the BOX_SIZE, since GRID_SIZE has to be square number. so a normal sudoku is BOX_SIZE = 3. (GRID_SIZE = BOX_SIZE * BOX_SIZE)
It was just really awesome and in so nice way u described it's just really wow... M so surprised why just so less likes I have watched couple of videos before but u just explained really understanding with gotcha learn some new concept as well successfully found my final year project as well thanks for making this video... I will surely check out your more videos.. Great job 👍
I have reached the part where I didn't find a solution to erase the all the numbers of the board and to retry. I know it needed some recursion, but didn't figure out how. Your explanation was excellent and after reviewing your method, everything makes sense. Thanks and keep it up with these kind of videos!
Wow, great. So clear. I especially appreciate the backtrack! So clever, just a 0 and then the loop moves on at the point it stopped, so we get the next possible number that fits in the location. Continue with examples like this. It's excellent.
Thanks! Let me know if you have any ideas for programs like this. It's tough because they have to be complex enough to be interesting but small enough to fit in a reasonable video.
@@CodingWithJohn yes, thanks. I just bought a book about Ai and python where they also take small projects that can perfectly fit to your devs. There is the classic 8puzzle, The shortest way, Snake, Tetris, Spam search, And I don't know how you say in English : O|X|O _|X|O O|X|_
It's like O(k*10^(k^2 -n)) where k is the number of cells in a row and n is the number of placed numbers at the start. If you try with k greaters than 9 this algorithm would literally take years to find a solution. You can do better using sat solvers and translating the sudoko problem into cnf logical formulas.
@@robertoborelli3275 may I ask why do you have zero in the beginning of the expression 0(k * 10^(k^2 - n)) doesn't it just cancel everything to zero? Also, why do you have 10 to the power of cells to solve, how did you get 10? Did you mean k because with every added unsolved cell you need k times more traversing the loops (assuming that the number of rows = the number of digits you can fill a cell in) ? And why do you have k * in the beginning of the expression?
@@smrtfasizmu6161 it's been a while since I've watched the video and now I don't remember exactly the code but here I give you some explanations: 1) It's not a 0, but is a big O. This thing is called "Big O notation". Simplifying, you can read it as "the complexity is less or equal than..." but there are some formal and accurate mathematical definitions behind this intuition I gave. 2) The intuition behind the formula is that this code is clearly exponential in the number of unsolved cells (k^2 - n). 3) each time you call the function to see if a number is valid in cell you spend O(k) time. 4) For each cell there are (exactly) k possible values (in the case of normal sudoku k is 9). So to sum the important thing here are: - exponential in the number of unsolved cells. - you take linear time in k, each time you check if a number is valid So as I wrote in the previous comment this is a pretty easy solution to write and implement but it's very inefficient since it uses the technique 'guess and verify'. It's much better (but also a bit more difficult) to use a sat solver for this kinds of problems.
@@juanmalpartida1333 no. just because you have 3 nested loops you can't always say that you have O(n^3). Inside this loop there is a recursion! This solution is exponential for sure, infact there are situations in which you guessed j correct values, than you are trying to fill the j+1-th value and you discover that there is no such a solution. In this case the algorithm does "backtracking" and there are situations in which all the j guessed valuesare wrong and the algorithm must re guess the entire solution. This leads to exponential cost!
I think that those nested for loops on the solve method are unecesary, you could add two additional params to send the current row and column and inside the function you just would need to either add the the rows or to the column depending on what you are doing... yes you will have to put some IF's, but the way you did it I think it will do unnecesary iterations, I mean, the last recursive call on a solvable board will put the needed number but there's nothing that would stop the initial call to the solve method from stopping it's iterations. Actually, on each recursive call it will have to traverse as far as possible on the board until sending back a false or a true if solved. You don't need the algorithm to traverse almost the whole board on each recursive call, you just need it to make a single step on the board instead.
cool project :) i'm fairly new to java, i programmed a small snake'ish game, so i understood the 2d array way better then i thought :D keep up the good work!
I think we can use dynamic programming too, to reduce some of the complexity. We can use a HashMap to store if a number at particular position was a valid placement or not.
The argument of the hashmap needs to be both the board and the placement of the number. How does backtracking in your algorithm work? I am not sure how you intended to do it with hashmaps and make the code less complex than what you saw in this video. Can you elaborate further? One idea popped up in my mind, but it is more or less a copy of what you saw in the video but without recursion. First you have an array which holds the position which are given in the beginning. Then, you proceed filling in the board and every time you find out that you can't continue filling in the board you backtrack the i and j variables of the loop until you reach the next number which you filled in and continue trying filling it after setting number To Try to be 1+the previous numberToTry.
How did you make it so simple, I always appreciate your content. your content is limited on youtube but whenever I get stuck in any problem and you have a video on that, which means, my last destiny is your video. Thank you
Thanks so much John. I spent a lot of time searching on youtube for a good tutorial on this. I wish yours was the first video so i could have saved time :(
Reduce sudoku to 7 colours as 1 fixed color as 2, comparative colour as 3. Then reduce the comparison to 2 sat compare with upto 8 colors to get main grid clues. For sudoku grid of any size is In O(n^5)
Pretty nice explanation. I made one sudoku solver but try to improve his performance using a priotityQueue. Basically save in the queue the empty spaces that can take the minimun posible values, so.. the queue delivery the spaces that have less probablity to change... and i reorder the probability of the other spaces that have been afected for add a new number in ther matrix. This consume so much memory but less time...
Interesting methodology! This definitely isn't the most efficient way to do it but the board is limited to 9x9, so even with a terrible time complexity any board is still solved pretty quickly
Great and concise explanation! When this video was suggested, I found it hard to believe that this was doable in a beginner friendly way in only 20 minutes, well done! I have one suggestion though: If you have the GRID_SIZE as a static final int, the box size (3, as used as a magic number in isNumberInBox(...) and print(...)) should be, too! Probably it should be initialized as the square root of GRID_SIZE, or vice versa.
In an intro to programming textbook I own, it mentions in the preface that one difference from prior printings is that it removes the sudoku solver example code from the multidimensional arrays chapter because apparently is it too complex at that point.
You don't need an else in "if (solveBoard(board)) " as you don't have to set board[row][column]to 0. It will just be reset to another number when it immediately after is assigned a new number in the for loop. It is only if the for loop fails that you need to set it to 0.
Gr8 vid. Spent the last 2days working on this really breaking it down. I got a lot outta following and really understanding what’s going on here. Keep them coming…🙌🏿🙌🏿
1. You return only one solution. 2. It's the brute force method instead of a logical solution. But it finally works !! I try to make a smart solution with the possible value exclusion method. It's more interesting in my opinion ))
I like that you also play binding of issac 🙂. As for the program I loved it. It was well made video for beginners and one can't not understand the concept. Keep it up! 👍
I‘d love to see this exact algorithm but more efficient. Knuths Algorithm X for sudoku sounds very interesting and shows a pretty good method for backtracking. I‘m too dumb to understand how to implement it in java but it could be a cool challenge for you.
Very nice. I do this stuff for fun. Recently did one of those word search generators which you see kids books all the time. Way more interesting to do than you may initially think.
Its clever, I understood right away how it worked but the get to this simplicity is quite amazing. If it's you who found the solution then you're quite a clever man! Amazing work!
I made once a Sudoku solver en Excel using Visual Basic using a more human like strategy... placing various plausible numbers in each empty space, and recursevly eliminating invalid ones
I did the same in Python in AI class in college. I think that's a cooler solution, but I liked the simplicity that this one offered for someone newer to coding, for purposes of this video
Please make a video on the Hand Of Poker (Poker Hands) Game, it seems to be used alot on interview assessments this days. I am sure alot of people need it.
I didn't use the tutorial but after seeing some of it I decided to try it myself without any help and this strategy worked, only issue was that my loop variable went from 0-8, so if a number was 9 then it failed
Hey, John I loved the way you went forward with explaining the video , you made it quite easy to learn the algorithm and fun too, i was curious if we can attach some image recognition tools like OpenCV etc for Java , so that we can scan a sudoku at run time and give an image back? I would love if you could show us how to do this if you like the idea.
You're an excellent teacher, John. I find your videos not only extremely helpful, but impressively well-made - and it's not even so much about the format (which is ideal), but the way you teach and the examples you give. You explain things in a logical, progressive way, and you're excellent at providing the right level of exposure. I do wish there were more teachers like you in universities - but then, we might not have you on RUclips ;) Please keep going with these tutorials, you are contributing to the community in a huge way.
This was, legitimately, the most interesting java tutorial I've seen in years! EXCELLENT!!!!!
Making a sudoku solver was actually the very first project I programmed once I'd learned the basics. This takes me back.
Wow ! I have been coding for arround 5 years and could'nt come up with a solution!
@@yahia1355 oh I didn't do anything smart. I literally brute forced a solution. Basically going over every single square and trying the first number that works, then if I got to one that had no possible correct number I would go back until the first square that I could have put a different number in and tried the different number. Do that until you can fill every square correctly. Just literally trying every combination until it works. There are much better ways to solve this. But the not so smart solution works.
@@19joni69 🤣🤣😂
I'm so glad channels like this one exist. It was a part of my project and I would've never figured that out myself...
I had recently discovered this channel, haha, and I started to watch your videos on TV Like I am watching some movies or some kind of entertainment, nice videos keep it up, subscribed after watching 2-3 videos
Same)
Sir before I watch your video I was a student who was struggling to solve recursion problems for my upcoming test. Now I get the idea about how to backtrack in recursion :) Huge thanks from South Korean student!
It’s beautiful when something that’s educational is also entertaining
This program works perfectly. What a flawless algorithm you have written .
I didn't think this solution would have an adequate time complexity. Thanks!
Great video John. Congrats on 500 subs!
Thanks!!
Well at least I've certainly gone well past 500
@@CodingWithJohn by far 😎
This is exactly what I was looking for to complement my java studies. Projects with step-by-step explained.
Please, keep it going, unfortunately here on RUclips, everything seems to be only JavaScript or Python. It would be great more projects with OOP.
No wonder you’re a lead! Very clear sir, thank you
I wrote a Sudoku solver in 2008 in Javascript, but it didn't quite use full recursion (my mistake) so it wouldn't solve moderately difficult puzzles. Mine includes character recognition of uploaded websuduko puzzles. This week I added "paste from clipboard" so that I didn't have to save/load my puzzle images. Today, following your model I finally got mine solving even the "evil" puzzles. Thank You John for this great lesson. Although now I don't feel that "I wrote the code", my page works and I did right the optical character recognition parts. I don't believe that I can share a link here, but I can try.
This is the best video I found online for this problem. You are a great teacher, thank you.
Thanks John, I was really confused when I tried to solve questions like this based on the backtracking algo. You explain really well. Cheers :)
This was such a good explanation!!! I feel a lot more confident about implementing this. Also thanks for not acting like this is easy/obvious👏
Thank you! Amazing video, where I work I tried several times create something to solve our work schedule with some kind of algorithm but always fail, this was just exactly what I needed and hope to solve my needs. Thank you!
Fun way to learn about recursion 👍
Hi John, i have learned a lot from your videos. I hope you won't stop making videos in upcoming days, it is really helpful for us.
I can guarantee you will get a huge subscribers in coming days.
Thanks! Glad you're getting something out of them. I'll keep making them if people keep watching them!
This is excellent .. i simply love the naming conventions of the vars..which makes its easy to read and understand the logic
This tutorial is excellent John.
Your explanaition is soo cool! You even make feel this stuff is easy.
This was so nice and easy to follow! Thank you so much
it is amazing to see various algorithms that be used to solve sudoku, i made one but using while loop
You teaching skills are really impressive, great work! I have one minor point to critique, you hardcoded GRID_SIZE 9 but used 3 as a magic number. You could have computed it once sqrt(GRID_SIZE) or just declared it as a constant.
sqrt(GRID_SIZE) would make no sense at all
@@Michael-se7ny Really? Doesn't a Sudoku box always have side length of sqrt(GRID_SIZE)? But I guess the really clean way would be to define the sudoku in terms of the BOX_SIZE, since GRID_SIZE has to be square number.
so a normal sudoku is BOX_SIZE = 3. (GRID_SIZE = BOX_SIZE * BOX_SIZE)
@@JannisAdmek right
It was just really awesome and in so nice way u described it's just really wow... M so surprised why just so less likes I have watched couple of videos before but u just explained really understanding with gotcha learn some new concept as well successfully found my final year project as well thanks for making this video... I will surely check out your more videos.. Great job 👍
Simply Nailed It, AWESOME EXPLANATION
Awesome, exactly the kind of RUclips channel I wanted:
Regularly uploaded coding videos.
Great explanation and great mini project
Thank you so much. Looking for more project like this
I have reached the part where I didn't find a solution to erase the all the numbers of the board and to retry. I know it needed some recursion, but didn't figure out how. Your explanation was excellent and after reviewing your method, everything makes sense. Thanks and keep it up with these kind of videos!
Wow, great. So clear.
I especially appreciate the backtrack!
So clever, just a 0 and then the loop moves on at the point it stopped, so we get the next possible number that fits in the location.
Continue with examples like this. It's excellent.
Thanks! Let me know if you have any ideas for programs like this. It's tough because they have to be complex enough to be interesting but small enough to fit in a reasonable video.
@@CodingWithJohn yes, thanks.
I just bought a book about Ai and python where they also take small projects that can perfectly fit to your devs.
There is the classic 8puzzle,
The shortest way,
Snake,
Tetris,
Spam search,
And I don't know how you say in English :
O|X|O
_|X|O
O|X|_
@@CodingWithJohn pack man is more complex but can maybe make a video for the moves?
Great explanation.
Will be great if we can also talk about the time complexity of the solutions
It's like O(k*10^(k^2 -n)) where k is the number of cells in a row and n is the number of placed numbers at the start.
If you try with k greaters than 9 this algorithm would literally take years to find a solution.
You can do better using sat solvers and translating the sudoko problem into cnf logical formulas.
@@robertoborelli3275 may I ask why do you have zero in the beginning of the expression 0(k * 10^(k^2 - n))
doesn't it just cancel everything to zero? Also, why do you have 10 to the power of cells to solve, how did you get 10? Did you mean k because with every added unsolved cell you need k times more traversing the loops (assuming that the number of rows = the number of digits you can fill a cell in) ? And why do you have k * in the beginning of the expression?
@@smrtfasizmu6161 it's been a while since I've watched the video and now I don't remember exactly the code but here I give you some explanations:
1) It's not a 0, but is a big O. This thing is called "Big O notation". Simplifying, you can read it as "the complexity is less or equal than..." but there are some formal and accurate mathematical definitions behind this intuition I gave.
2) The intuition behind the formula is that this code is clearly exponential in the number of unsolved cells (k^2 - n).
3) each time you call the function to see if a number is valid in cell you spend O(k) time.
4) For each cell there are (exactly) k possible values (in the case of normal sudoku k is 9).
So to sum the important thing here are:
- exponential in the number of unsolved cells.
- you take linear time in k, each time you check if a number is valid
So as I wrote in the previous comment this is a pretty easy solution to write and implement but it's very inefficient since it uses the technique 'guess and verify'.
It's much better (but also a bit more difficult) to use a sat solver for this kinds of problems.
Time complexity looks to be O(n^3) worst case. Average, eyeballing it, I would say is n^2*log(n)
@@juanmalpartida1333 no. just because you have 3 nested loops you can't always say that you have O(n^3).
Inside this loop there is a recursion!
This solution is exponential for sure, infact there are situations in which you guessed j correct values, than you are trying to fill the j+1-th value and you discover that there is no such a solution. In this case the algorithm does "backtracking" and there are situations in which all the j guessed valuesare wrong and the algorithm must re guess the entire solution.
This leads to exponential cost!
Just one word for John 🙏🏼
Big ThankYou ❣️
it is 100% working teacher john and thank you for the source code
I think that those nested for loops on the solve method are unecesary, you could add two additional params to send the current row and column and inside the function you just would need to either add the the rows or to the column depending on what you are doing... yes you will have to put some IF's, but the way you did it I think it will do unnecesary iterations, I mean, the last recursive call on a solvable board will put the needed number but there's nothing that would stop the initial call to the solve method from stopping it's iterations. Actually, on each recursive call it will have to traverse as far as possible on the board until sending back a false or a true if solved. You don't need the algorithm to traverse almost the whole board on each recursive call, you just need it to make a single step on the board instead.
WOW.. you are amazing....clean.. detailed.. explanation.
Please do Leetcode problems also.. It will help us.
cool project :) i'm fairly new to java, i programmed a small snake'ish game, so i understood the 2d array way better then i thought :D keep up the good work!
very nicely explained John, I saw few other videos as well for Sudoku Solver, but this is amazing!
As always, thank you John.
I think we can use dynamic programming too, to reduce some of the complexity. We can use a HashMap to store if a number at particular position was a valid placement or not.
i guess this one is for beginners and that's why he chose this way of doing it
But that is dependent on the previous placements. So I do not immediately see how that improves this algorithm.
The argument of the hashmap needs to be both the board and the placement of the number. How does backtracking in your algorithm work? I am not sure how you intended to do it with hashmaps and make the code less complex than what you saw in this video. Can you elaborate further?
One idea popped up in my mind, but it is more or less a copy of what you saw in the video but without recursion. First you have an array which holds the position which are given in the beginning. Then, you proceed filling in the board and every time you find out that you can't continue filling in the board you backtrack the i and j variables of the loop until you reach the next number which you filled in and continue trying filling it after setting number To Try to be 1+the previous numberToTry.
iirc the whole problem is a system of linear equation , and u can solve it in O(n) if u use multigrid method
How did you make it so simple, I always appreciate your content. your content is limited on youtube but whenever I get stuck in any problem and you have a video on that, which means, my last destiny is your video. Thank you
Watching couple of videos from this channel every day in any order. But still makes sense
Thanks so much John. I spent a lot of time searching on youtube for a good tutorial on this. I wish yours was the first video so i could have saved time :(
Beautifully well explained, thank you sir.
You just nailed it. Mind blowing tutorial, no doubt.
Great video! Thanks for this kind of tutorials John! :D
OMG, I understand how to solve this complex problem now. Thank you so much!
Reduce sudoku to 7 colours as 1 fixed color as 2, comparative colour as 3. Then reduce the comparison to 2 sat compare with upto 8 colors to get main grid clues. For sudoku grid of any size is In O(n^5)
Great coding John! I got a "Hard Sudoku" solved in a few ms !
Pretty nice explanation. I made one sudoku solver but try to improve his performance using a priotityQueue. Basically save in the queue the empty spaces that can take the minimun posible values, so.. the queue delivery the spaces that have less probablity to change... and i reorder the probability of the other spaces that have been afected for add a new number in ther matrix. This consume so much memory but less time...
Interesting methodology! This definitely isn't the most efficient way to do it but the board is limited to 9x9, so even with a terrible time complexity any board is still solved pretty quickly
Awesome Tutorial
Now I can solve it by myself, thanks a lot.
Excellent explanation 👏👏👏
Amazing! Thanks for this, this helps me stay engaged in java
Well explained!
Thanks John
Bro, thank you for teaching. You are the go to guy!
Loved this!
This was really nice, fun and informative!
I think, i just found my best youtuber. Keep up the good work John.
Same
Awesome algorithm. It would be nice, if you could also explain the time complexity of the algorithm in your video.
Great and concise explanation! When this video was suggested, I found it hard to believe that this was doable in a beginner friendly way in only 20 minutes, well done!
I have one suggestion though: If you have the GRID_SIZE as a static final int, the box size (3, as used as a magic number in isNumberInBox(...) and print(...)) should be, too! Probably it should be initialized as the square root of GRID_SIZE, or vice versa.
Great video, your explanation was pretty clear!
Great video, I really enjoyed your explanation. Super thanks, this was very helpful
I like what you doing, Hope you get the chance to do Dijkstra’s algorithm, Time complexity and a few more interview related.
I like your explanations
In an intro to programming textbook I own, it mentions in the preface that one difference from prior printings is that it removes the sudoku solver example code from the multidimensional arrays chapter because apparently is it too complex at that point.
You don't need an else in "if (solveBoard(board)) " as you don't have to set board[row][column]to 0. It will just be reset to another number when it immediately after is assigned a new number in the for loop. It is only if the for loop fails that you need to set it to 0.
Oh, yes. Right. It just makes it clear for us maybe.
So, “board[row][column] = 0” can be moved to the place before return false
For more informations, It must be one of these places, right?
wouldn't having it there cause elimination of 1 possibility while going for previously placed index?
@@Leo_banana That is correct.
@@fandusmercius723 It wouldn't.
wow,next level explanation
Very interesting tutorial!
love this!
Gr8 vid. Spent the last 2days working on this really breaking it down. I got a lot outta following and really understanding what’s going on here. Keep them coming…🙌🏿🙌🏿
Thanks for the great explanation! Keep it up with the channel!
Really helpful!
Wow! That’s so cool! This was an awesome video! Very interesting :)
Wow sir you can explain things very well, let aside being multitasker.
1. You return only one solution.
2. It's the brute force method instead of a logical solution.
But it finally works !! I try to make a smart solution with the possible value exclusion method. It's more interesting in my opinion ))
You make truly great videos
amazing professor!
this is a quality tutorial
I like that you also play binding of issac 🙂.
As for the program I loved it. It was well made video for beginners and one can't not understand the concept. Keep it up! 👍
I‘d love to see this exact algorithm but more efficient. Knuths Algorithm X for sudoku sounds very interesting and shows a pretty good method for backtracking. I‘m too dumb to understand how to implement it in java but it could be a cool challenge for you.
Fantastic stuff!
such a enjoyable algorithm!!!
That was fantastic. Thanks 🙏
if you could make a short tutorial about binary search trees and how to understand recursion there..
Very nice. I do this stuff for fun. Recently did one of those word search generators which you see kids books all the time. Way more interesting to do than you may initially think.
Wow! Sooo nice and free! Instant sub and hooked. Binge time.
Its clever, I understood right away how it worked but the get to this simplicity is quite amazing. If it's you who found the solution then you're quite a clever man!
Amazing work!
I learn more in 1 video than in 2 months worth of other tutorials
Awesome johnny...
Leaving a comment because I feel this channel deserves gold. 🤗
Loved it! 🔥
Great explanation, easy to understand, even for beginners :-) Thanks!
I made once a Sudoku solver en Excel using Visual Basic using a more human like strategy... placing various plausible numbers in each empty space, and recursevly eliminating invalid ones
I did the same in Python in AI class in college. I think that's a cooler solution, but I liked the simplicity that this one offered for someone newer to coding, for purposes of this video
Please make a video on the Hand Of Poker (Poker Hands) Game, it seems to be used alot on interview assessments this days. I am sure alot of people need it.
Fantastic video and fantastic channel, your videos are great thanks .
I didn't use the tutorial but after seeing some of it I decided to try it myself without any help and this strategy worked, only issue was that my loop variable went from 0-8, so if a number was 9 then it failed
I was thinking about that as I was watching it, wondering if it was going to capture the 9, being that it is 0 indexed.
Loved your explanation..keep it up buddy 😊
Shit took me a while but I think that's why it was worth it. This video made recursion seem so much more intuitive. Thank you!
Thank you very much!
Hey, John I loved the way you went forward with explaining the video , you made it quite easy to learn the algorithm and fun too, i was curious if we can attach some image recognition tools like OpenCV etc for Java , so that we can scan a sudoku at run time and give an image back? I would love if you could show us how to do this if you like the idea.