Anytime I search for youtube solutions of any leetcode problem, if I find your solution to that problem then I always check out your video first because I always know for sure if you have explained it, then I'll surely understand your solution without the need to check out any other solution. You explain very well. Nice work. Keep it up !
This will replace the character from the board and will not be available to search in the following recursion calls. Or you may also save the element position it in a list and check it everytime to prevent it from reusing.
This will replace the character from the board and will not be available to search in the following recursion calls. Or you may also save the element position it in a list and check it everytime to prevent it from reusing.
Anytime I search for youtube solutions of any leetcode problem, if I find your solution to that problem then I always check out your video first because I always know for sure if you have explained it, then I'll surely understand your solution without the need to check out any other solution. You explain very well. Nice work. Keep it up !
ayyee thanks a lot I'm glad the videos help :)
Great explanation, thank you!
can you explain more how board[y][x] = '' " can help prevent re-use of the same character?
This will replace the character from the board and will not be available to search in the following recursion calls.
Or you may also save the element position it in a list and check it everytime to prevent it from reusing.
you made this problem so easy! thank you.
Glad it helped!
Great explanation ❤️
Glad it helped!
What is the time and space complexity of this solution?
Looks tough but u explained it so well
Why can't I apply BFS?
could not understand index and how it is used.
index is the pointer to the character in word for traversal
Cool
father ?
👍🤩
getting TLE from this soluion
can you explain more how board[y][x] = '' " can help prevent re-use of the same character?
This will replace the character from the board and will not be available to search in the following recursion calls.
Or you may also save the element position it in a list and check it everytime to prevent it from reusing.