Tushar, these videos are phenomenal. It's very apparent that you have spent a lot of time preparing this material - your explanations and examples are incredibly clear. Thanks!!!
I have never written comment on any video whether it is good or not, But your videos force me to write comment, By heart you are Genius man !!! Legend of Computer Science!!!!!
I can't put into words how much these videos help me and my friends understand the subject you cover! No way in a million years that my professor will go that deep into the explanation of the all the little (but very important) details you do explain. Please, please cover more topics in the future! :)
just my opinion: 95% of Indian engineering students waste average 6 hours per day in colleges learning shit. We can easily understand actual concepts on youtube in 2 hours and through books for extra. Shit and lazy ass professors and money making colleges. Don't want to hurt anyone's sentiments but IT's THE F TRUTH. Thanks Tushar bhai. I am grateful to you.
I give you thumbs up because the same can be said for any field of study. But no employer is going to hire you to an engineering position without a degree. So the paper degree has its value.
I agree with you at some point. But college its also important because there professors guide students threw the most important subjects. If many people didnt go to college, this videos wouldnt be watched by the same amount of people.
I have seen most of your videos and just wants to say a big thanks. you are a role model for many young people coming out and wants to achieve something big in their life....
For the diagonal condition there is an alternate. Given two cells with the coordinates (i1, j1) and (i2, j2) , you can say they are on the same diagonal if | i1 - i2 | = | j1 - j2 | . You can validate this by calculating with values, or you can deduce it from the two formulas you have stated. This way you only need a single formula to validate if the queens attack on diagonal.
You can deduce this with the slope of the line formula Y2 - Y1/X2-X1 = +1 or - 1 implies that the angle is either 45 or 135 degrees and hence is a diagonal
The tree visual was fantastic. It allowed me to visualize how the algorithm was eliminating possibilities and backtracking to the first level to start over.
Thank you Tushar! Your explanation is really clear and makes sense. It saves me tons of time and makes my day! I hope you can continue to upload more videos.
Thank you for making me understand Recursion.....What I have got from your explanations is something very valuable....I got to learn the concept of Recursion, Back Tracking and Dynamic programming. After understanding the concept, I am able to code it myself. Thank you
Thanks for the videos, they are great! Keep up the good work!! I went through this code which says it is of Time: O(n*n) but actually, it's of Time: O(n*n*n) because you fix rows (Time: O(n)), increment col for every row (Time: O(n)) and check for positions for every col which again is another Time: O(n). All of this adds up to Time: O(n*n*n). Please check and correct it.
Great job. Generally to find the attack in diagnal position abs(r1-r2)==abs(c1-c2) is used, now I got a new formula. No need of two dimensional array, single dimension is enough with index represents rows, values represents the columns.
Thanks a lot Tushar. Your Backtracking videos are really awesome. And I really appreciate you explaining things on a whiteboard rather that PPTs. Your code examples are also really clean and simple to understand which is very hard to find to algorithms videos. I hope you keep creating these videos and help us achieve our goals of working in bigger companies. Thanks a lot once again.
Thanks.. A Suggestion- You could simply check if mod(row1-row2)== mod(col1-col2) to check for the diagonal attacking position rather than using 2 separate formulas for 2 diagonals.
Thanks sir..Its very good explaination. One problem : when we are backtracking we should delete element from position array. if(solveutil) return true; else delete pos; ...... return false;
This is an awesome explanation. Helped me understand backtracking. And all the people telling you that your english is horrible do not deserve your attention. Fuck 'em
There are a lot of obviously unnecessary check that could be avoided: the four squares around each queen are for sure not safe so why even try it? I would create N square object with a queen property that holds the queen id if attacking it. After each queen position iteration I would jump to a square that is not already attacked by a queen and place the queen and register her id to all squares it attacks. So after each iteration the number of available unattacked squares will reduce and makes the the algorithm finish much faster.
hey i don't know about of number of views people have got in their video for explaining N queens problem. But i tell you people this is the best explanation. NOBODY by far can teach in such a simpler way as he has done. You have any tutorials for all this Tushar, i would love to join it.
Hey Tushar, Thank you for all your videos. They have helped me a lot to understand the problem and the solution. The code you provide just makes it the best. Regarding this video, I am a little confused with the time complexity. Is it exponential as you say in the video or is it O(n*n) as mentioned in the code? Any clarification is appreciated. Thank you once again! :)
I noticed that in the code example you have the space complexity as O(n*n), however, isn't it O(n) where n represents the total number of queens we have?
Awesome video, thank you so much. If I may suggest, it might be better if you can also use a mic for the white board explanation since the sound quality will be much better.
Hey, which part of you code does backtrack? I think somewhere you have to remove the elements from position when its not safe. I can't get that part in your code. Can you explain
I got the same doubt and tried to understand that piece of logic. Looks like, the elements won't get removed as such from Positions, they rather get replaced/updated by a new element in the next iteration.
Awesome video, very well explained. I think that I would now be able to solve this exercise even a little later in the future, as I have understood the principle and the rules to determine a valid placement for a queen. One question regarding the backtracking step though: If a callee function return false to the caller, that's to say a deeper level of recursion can't find a valid place for a queen, the calling function continues to look for a valid place one column ahead where it stopped before. However, the previously placed queen in the array is never cleared and remains there. Doesn't that cause problems? On the right hand side you always remove the notes but you don't do it in the code. WOuldn't you have to add an else statement after the if(foundSafe) block to remove an falsly placed queen?
Very clear cut explanation of n Queen problem. Thanx a lot Sir. Please upload the videos on bellman ford with directed graph with algorithm. Sir it is very hard to understand. Please Sir help me In this.
Great video! How do you derive the formula upwardDiagonal = i+j and downwardDiagonal = i-j ... is it just by observation or is there a way to find this from using line equations?
You can use the equation of a straight line to find the relation. Treating the row and column value as x y coordinates , and taking the slope as 1 and -1 for the 2 diagonals. :)
Tushar I know you must be familiar with the following approach too but just in case you are not then - Let the current queen to be placed = cur_q Let all the other queens that have been placed earlier = placed_q If( abs(cur_q.row - placed_q.row) - abs(cur_q.col - placed_row.col) == 0) then the queens are on the same diagonal
If they gave me a problem in real time I can solve it with the basic serial and according to the family that belongs to the board. This is only for cases of an empty board. For the case of a board with a queen on the board, I can solve all the prime boards in the position that the queen is indicated, be of the size that is from n = 5, 7,11,13,17,19,23. ........ 61, ..... etc
Hi Tushar, Thanking you for nice article. I see a loop hole here. I may be wrong. Please correct me if so. When we are checking if this row and column is under attack from any previous queen. When it comes to check diagonally, we are checking only next row from previous queen. There can be possibility that a queen placed in 1st row, attacking a position in 4th row, placed diagonally. It looks like we have missed this check. Yours response awaited.
Can you please make a video on "How to make recursion tree" and thinking of solving it recursively, i often confuses when i see two or more functions calling to same functions again and again.
Tushar, these videos are phenomenal. It's very apparent that you have spent a lot of time preparing this material - your explanations and examples are incredibly clear. Thanks!!!
Awesome explanation. You inspired me to make my own version of this video for explanation.
And actually you made it!
I have never written comment on any video whether it is good or not, But your videos force me to write comment, By heart you are Genius man !!!
Legend of Computer Science!!!!!
thanks for writing the comment for the first time.Now i can die peacefully.
I can't put into words how much these videos help me and my friends understand the subject you cover!
No way in a million years that my professor will go that deep into the explanation of the all the little (but very important) details you do explain.
Please, please cover more topics in the future! :)
just my opinion: 95% of Indian engineering students waste average 6 hours per day in colleges learning shit. We can easily understand actual concepts on youtube in 2 hours and through books for extra. Shit and lazy ass professors and money making colleges. Don't want to hurt anyone's sentiments but IT's THE F TRUTH. Thanks Tushar bhai. I am grateful to you.
I give you thumbs up because the same can be said for any field of study. But no employer is going to hire you to an engineering position without a degree. So the paper degree has its value.
I agree with you at some point. But college its also important because there professors guide students threw the most important subjects. If many people didnt go to college, this videos wouldnt be watched by the same amount of people.
The legend of Computer Science. Yesterday I had lots of problems regarding coding a solution for this problem, now I don't have any. Brilliant video!
I've watched 3 videos prior to yours' on this topic and so far you've done the best job explaining it. Thanks.
I have seen most of your videos and just wants to say a big thanks. you are a role model for many young people coming out and wants to achieve something big in their life....
For the diagonal condition there is an alternate. Given two cells with the coordinates (i1, j1) and (i2, j2) , you can say they are on the same diagonal if | i1 - i2 | = | j1 - j2 | . You can validate this by calculating with values, or you can deduce it from the two formulas you have stated. This way you only need a single formula to validate if the queens attack on diagonal.
Totally agreed. Taking absolute of " |X2 - X1| == |Y2 - Y1| " is better conditional statement for checking if the queen is safe or not.
You can deduce this with the slope of the line formula
Y2 - Y1/X2-X1 = +1 or - 1 implies that the angle is either 45 or 135 degrees and hence is a diagonal
this is a great video. Always helps me a ton to see stuff drawn out like this.
The tree visual was fantastic. It allowed me to visualize how the algorithm was eliminating possibilities and backtracking to the first level to start over.
Thank you Tushar! Your explanation is really clear and makes sense. It saves me tons of time and makes my day! I hope you can continue to upload more videos.
every time a problem bothers me, I find video of you explaining it :) you are awesome .
You are a hero at explaining things.WELL DONE SIR.THANK YOU SOOOOOO MUCH.
Your explaination is always clear and easy to understand! Thanks for your videos!
Thank you for making me understand Recursion.....What I have got from your explanations is something very valuable....I got to learn the concept of Recursion, Back Tracking and Dynamic programming. After understanding the concept, I am able to code it myself. Thank you
Thoroughly explained! I love this! Thank you for your video!
Best explanation I have seen yet. Thank you!
I admire your visual presentation and clean code matching your flow of explaining the algorithm.
Thanks for the videos, they are great! Keep up the good work!!
I went through this code which says it is of Time: O(n*n) but actually, it's of Time: O(n*n*n) because you fix rows (Time: O(n)), increment col for every row (Time: O(n)) and check for positions for every col which again is another Time: O(n). All of this adds up to Time: O(n*n*n). Please check and correct it.
Great job. Generally to find the attack in diagnal position abs(r1-r2)==abs(c1-c2) is used, now I got a new formula. No need of two dimensional array, single dimension is enough with index represents rows, values represents the columns.
Ya. It is a single array. My mistake.
best video available for the N queen problem using backtracking .
Great Explaination Tushar. Kindly consider doing the Graph Coloring problem using backtracking as well!
Thanks a lot Tushar. Your Backtracking videos are really awesome. And I really appreciate you explaining things on a whiteboard rather that PPTs. Your code examples are also really clean and simple to understand which is very hard to find to algorithms videos. I hope you keep creating these videos and help us achieve our goals of working in bigger companies.
Thanks a lot once again.
These videos are incredibly helpful! Thank you for the clear step by step explanation of the algorithms!
very good by far the best explanation for this problem the explanation of index of attacks was very very helpful
I watched many backtracking videos, but this is so simple and understandable. thank you man
concise explanation. understood the use of row + i, row - i .
Thanks.. A Suggestion- You could simply check if mod(row1-row2)== mod(col1-col2) to check for the diagonal attacking position rather than using 2 separate formulas for 2 diagonals.
No bro
You will have to use two conditions in that case too
yeah
Great Explanation. the best one on n-queen problem.
Agree with others. Your explanation made so much sense. Great job.
Good Explanation. Helped me to understand the problem easily. Thank You.
You are the best Tushar sir, can't wait to see your next video!
thanks Tushar Roy, your videos are really helpful
Thanks sir..Its very good explaination. One problem : when we are backtracking we should delete element from position array.
if(solveutil)
return true;
else
delete pos;
......
return false;
You are amazing. I code in python but the way you explain your code is so good. Thanks : )
Greatest explanation (for me). Thank you so much
This is the best explanation I got yet.
Grear job man! So clear and vivid! Thanks!
Appreciate your hard work. Very well explained.
This is an awesome explanation. Helped me understand backtracking. And all the people telling you that your english is horrible do not deserve your attention. Fuck 'em
Neat Explanation. Love your Dynamic Programming Series!
I have learned a lot from your videos. thanks
very very well explain. now i can do my college task. thank you.
Very good and clear explanation. Thank you for the video.
Thank you very much... I was searching for a algorithm like this to make a Sudoku solver. Thanks...
Finally an explanation that makes sense! THANK YOU!!!
There are a lot of obviously unnecessary check that could be avoided: the four squares around each queen are for sure not safe so why even try it? I would create N square object with a queen property that holds the queen id if attacking it. After each queen position iteration I would jump to a square that is not already attacked by a queen and place the queen and register her id to all squares it attacks. So after each iteration the number of available unattacked squares will reduce and makes the the algorithm finish much faster.
You can keep a Map for the finding how many queens are attacking a row, column or diagonal and check if the current cell is under attack or not
hey i don't know about of number of views people have got in their video for explaining N queens problem. But i tell you people this is the best explanation. NOBODY by far can teach in such a simpler way as he has done. You have any tutorials for all this Tushar, i would love to join it.
thanks a lot for your clear and brief tutor and it insight me about this topic thanks again keep go on
The best explanation for backtracking..........!!!!! thanks tushar... :D
i saw your lots of videos which helps me a lot do as many as possibles plz...
Simply superb explanation!!!
Thanks man! You are an excellent teacher.
Awesome video, especially the code explanation.
nice work on explaining the recursion tree!!!
Thanks
Hey Tushar, Thank you for all your videos. They have helped me a lot to understand the problem and the solution. The code you provide just makes it the best.
Regarding this video, I am a little confused with the time complexity. Is it exponential as you say in the video or is it O(n*n) as mentioned in the code? Any clarification is appreciated. Thank you once again! :)
Fabulous lecture Bro! Thank's a lot
Best tutorial. Thanks for the clear explanation!!
If anyone's struggling with the indices for the diagnols think of straight line equations but this array is in the 4th quadrant
Thanks Tushar for explaination.
I noticed that in the code example you have the space complexity as O(n*n), however, isn't it O(n) where n represents the total number of queens we have?
You are awesome Tushar.
Awesome video, thank you so much.
If I may suggest, it might be better if you can also use a mic for the white board explanation since the sound quality will be much better.
really clear, easy to understand, thanks a lot
you alright bro? you got the thousand algorithm stare goin
When we backtrack shouldnt we clear the previously set foundSafe position in the positions array?
Hey, which part of you code does backtrack? I think somewhere you have to remove the elements from position when its not safe. I can't get that part in your code. Can you explain
I got the same doubt and tried to understand that piece of logic. Looks like, the elements won't get removed as such from Positions, they rather get replaced/updated by a new element in the next iteration.
can you link to the chessboard visualizer?
was wondering if someone could elaborate on time complexity of algorithm
What is the time complexity of this problem?
kaash aaise teacher hume padhane aaye hote
Hi Tushar, In your N-queen video, you said the time complexity is exponential but your github comment has time complexity as n2.
Is it easy or difficult to find one pattern (one solution) of 2500 or 5000 queens with Backtracking Algorithms?
you are doing an awesome job bro! keep up the good work
Great Job Tushar !
+Tushar Roy Keep up your good work ! You are doing a stupendous job and helping countless people ! :)
Can anyone tell me why Tushar has taken 1 box for row 0, 2 for row1, 3 for row 3 and again two for the last row?
What did you use to trace your program in the matrix? Looks like something that might come in handy for tracing programs.
Thanks for the great videos.
interview.io
Awesome video, very well explained. I think that I would now be able to solve this exercise even a little later in the future, as I have understood the principle and the rules to determine a valid placement for a queen.
One question regarding the backtracking step though: If a callee function return false to the caller, that's to say a deeper level of recursion can't find a valid place for a queen, the calling function continues to look for a valid place one column ahead where it stopped before. However, the previously placed queen in the array is never cleared and remains there. Doesn't that cause problems? On the right hand side you always remove the notes but you don't do it in the code.
WOuldn't you have to add an else statement after the if(foundSafe) block to remove an falsly placed queen?
Very clear cut explanation of n Queen problem. Thanx a lot Sir. Please upload the videos on bellman ford with directed graph with algorithm. Sir it is very hard to understand. Please Sir help me In this.
Great video! How do you derive the formula upwardDiagonal = i+j and downwardDiagonal = i-j ... is it just by observation or is there a way to find this from using line equations?
You can use the equation of a straight line to find the relation.
Treating the row and column value as x y coordinates , and taking the slope as 1 and -1 for the 2 diagonals.
:)
Tushar I know you must be familiar with the following approach too but just in case you are not then -
Let the current queen to be placed = cur_q
Let all the other queens that have been placed earlier = placed_q
If( abs(cur_q.row - placed_q.row) - abs(cur_q.col - placed_row.col) == 0)
then the queens are on the same diagonal
Isn't the time complexity O(n!)??
Loved this solution!! Your solution is amazing.. would be better if you had explained time complexity, why it is exponential.
Very helpful! Thanks for sharing
If they gave me a problem in real time I can solve it with the basic serial and according to the family that belongs to the board.
This is only for cases of an empty board. For the case of a board with a queen on the board, I can solve all the prime boards in the position that the queen is indicated, be of the size that is from n = 5, 7,11,13,17,19,23. ........ 61, ..... etc
What if don’t know how many queens can fit on the board? Or is it assumed as part of the problem
What about the 8 queen problem? Can your code be scaled to larger problems?
can you help me ; i have an exercice ; last letter first letter problem ; i dont understand how i use backtracking thanks
Amazing explanation 🙏
Hey, Nice explaination. Just wanted to know about the codevisualization. How you did it??
+Tushar Roy thanks man. Is there any option for the following or you have coded it?
it is very helpful sir...can please explain global parallel genetic algorithm for solving n queen problem n also naive algorithm for n queen
what if i fixed 2 queens in 8 queen problem , then how many solutions exists in 8 queen problem?
Hi Tushar,
Thanking you for nice article.
I see a loop hole here. I may be wrong. Please correct me if so.
When we are checking if this row and column is under attack from any previous queen. When it comes to check diagonally, we are checking only next row from previous queen. There can be possibility that a queen placed in 1st row, attacking a position in 4th row, placed diagonally. It looks like we have missed this check.
Yours response awaited.
We are checking whether the queen at particular level is under attack from all the previous levels
for(int queen=0; queen
hey! looking for the python code of this algorithm ... Any help please ?
Any trick to make this code to get all possible solutions?
is there any solution
excellent !! thank you for making this video .
Can you please make a video on "How to make recursion tree" and thinking of solving it recursively, i often confuses when i see two or more functions calling to same functions again and again.
Excellent explaination!
Could anyone tell me what is the visualization tool he is using?