Surrounded Regions | Boundary DFS ✅ Graph Explained Hindi Leetcode | Complete Graph Videos

[@oqant0424]

i always try the problem for 1-1.5hrs before watching your solutions
.................... i respect that you always keep reminding people to try the problem on their own.......Thanks a lot bhaiya for such awsome videos...

[@HelloWorldbyprince]

Thanks bhai

[@shridharsarraf2188]

Bhaiya i respect that you always keep reminding people to try the problem on their own, and i always try the problem for 1-1.5hrs before watching your solutions

[@HelloWorldbyprince]

Thanks a ton for this positive words
Please, share this channel in your college, groups, or LinkedIn bcoz it's cost nothing to you 😀

[@RohitKumar-cv7qb]

Thanks, done without seeing the solution code, Explanation was enough, 🌹

[@deep1998-j1v]

Wah kya code likha hai, ❤ Modular, clean, just gr8, perfectly understood.

[@HelloWorldbyprince]

😍 Thanks a lot

[@manisharya2138]

very easy explanation (1 bar me samajh me aa gya👌👌)

[@Ybash2979]

pyaara smjhaya thanku bhai

[@anshul6516]

Waaah bhaiyaaa ULTIMATE, mazaaa aa gyaaa...

[@HelloWorldbyprince]

Thanks a ton 😀

[@apsey2162]

can you explain the time and space complexity of each solution after giving the solution

[@pratyushbhatt1712]

Very nicely explained, Thanks a lot!

[@adarshrajput8590]

u are so self-less, always thinking of the benefit of students

[@HelloWorldbyprince]

Thanks Adarsh
Please, share this channel in your college, groups, or LinkedIn bcoz it's cost nothing to you 😀

[@pritishpattnaik4674]

Bahut hi OP approach , mazza agaya

[@realAB459]

Best Explanation For Boundary Traversal... Thanks

[@HelloWorldbyprince]

Thanks yaar

[@ajoydev8876]

Easy solution!Btw thanks for concept!

[@HelloWorldbyprince]

Most welcome 😊

[@sudhadevi6692]

I tried different approach but this approach is EXCELLENT 😂🎉

[@HelloWorldbyprince]

Glad you liked it!

[@sceptile_editz]

I always follow your tips.😇😇

[@HelloWorldbyprince]

Thanks 🙏

[@kushalshukla444]

time complexity jyaada hai bro but iski like boundary aur fir andar ka . btw nice explanation .

[@anjigolla4853]

superb explanation brother thank you

[@snehalmdave]

Good one, boundary DFS, catchy name as well to remember.

[@HelloWorldbyprince]

yesss

[@LearnWithSWAG]

Amazing explanation sir.

[@HelloWorldbyprince]

Thanks Anjali

[@VikashSingh-sf2xp]

Good explanation

[@rahulgautam6387]

very great explanation....

[@HelloWorldbyprince]

Glad it was helpful!

[@vishalmishra1937]

clear crisp

[@divya-ei8du]

Great vid!😀

[@HelloWorldbyprince]

Thanks! 😁

[@Rohitraj-70]

thanks

[@chandnibhatia1211]

If there ie a O at boundary-1 ..how are we dealing with that ...

[@vishalmishra1937]

dfs

[@abbasali9730]

Bhai I m confuse I have abdul bari dsa videos and coding ninja c++ dsa videos and sometimes watch love babbar videos as well as ur videos also but not ale to choose where i have to learn

[@HelloWorldbyprince]

😅😅
You have lots of resources buddy
Aap aise socho ki ...at the end aapse questions hone chahiye
Wo kaha se padhkar ho raha hai ....
Ye dekho aap

[@Sp_Global-oc4tf]

shaktiman hi gangadhar hai!!! graoh br like.

[@ishajain7879]

best trick.....

[@HelloWorldbyprince]

Yeah Thanks

[@pritishpattnaik4674]

I attempted this on my own but i think there is slight error in my code ,
This is the code i wrote
bool isValid(int i, int j, int n, int m, vector& board){
if (i>=0 and i=0 and j

[@VishalKumar-ce9pe]

class Solution {
public:
bool isValid(int i, int j, vector& board)
{
int m = board.size();
int n = board[0].size();

if(i>=0 && i=0 && j

[@nish0798]

class Solution {
public void solve(char[][] board) {
boolean[][]vis=new boolean[board.length][board[0].length];
for(int i=0;i

[@hanumanvaktsankha]

bhaiya bfs wala
class Solution {
public:
void solve(vector& board) {
int n = board.size();
int m = board[0].size();
vector image(n, vector(m, 0));
queue q;

for (int i = 0; i < m; i++) {
if (board[0][i] == 'O') {
q.push({0, i});
image[0][i] = 1;
}
}

for (int i = 0; i < n; i++) {
if (board[i][0] == 'O') {
q.push({i, 0});
image[i][0] = 1;
}
}

for (int i = 0; i < m; i++) {
if (board[n-1][i] == 'O') {
q.push({n-1, i});
image[n-1][i] = 1;
}
}

for (int i = 0; i < n; i++) {
if (board[i][m-1] == 'O') {
q.push({i, m-1});
image[i][m-1] = 1;
}
}

int delrow[] = {-1, 0, 1, 0};
int delcol[] = {0, -1, 0, 1};

while (!q.empty()) {
int row = q.front().first;
int col = q.front().second;
q.pop();

for (int i = 0; i < 4; i++) {
int nrow = row + delrow[i];
int ncol = col + delcol[i];

if (nrow >= 0 && ncol >= 0 && nrow < n && ncol < m && image[nrow][ncol] != 1 && board[nrow][ncol] == 'O') {
q.push({nrow, ncol});
image[nrow][ncol] = 1;
}
}
}

for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (board[i][j] == 'O' && image[i][j] != 1) {
board[i][j] = 'X';
}
}
}
}

};

[@HelloWorldbyprince]

good job

[@parasadhwaryu7532]

My solution :->
class Solution {
public:
void dfs(int i,int j,int n,int m,vector& board,vector&vis){
if(i=m || vis[i][j]==1 || board[i][j]=='X'){
return;
}
vis[i][j]=1;
dfs(i+1,j,n,m,board,vis);
dfs(i,j+1,n,m,board,vis);
dfs(i-1,j,n,m,board,vis);
dfs(i,j-1,n,m,board,vis);
}
void solve(vector& board) {
//convert all those zeroes to x which are not a part of boundary region
int n=board.size(),m=board[0].size();
vectorvis(n+1,vector(m+1,0));
for(int i=0;i

[@sudhadevi6692]

Able to solelve it
But in 10 attempts.😂

[@HelloWorldbyprince]

😁 nice

[@oqant0424]

22/32 done (19.8.22)

[@HelloWorldbyprince]

nice

[@VishalKumar-ce9pe]

----------------------------------------------------------------CODE-----------------------------------------------------------------------------
class Solution {
public:
bool isValid(int i, int j, vector& board)
{
int m = board.size();
int n = board[0].size();

if(i>=0 && i=0 && j

[@abhisheksamboddu6580]

ye th0 bawasir hai......c0de |ikha hai ya textb00k ka chapter

[@rachnamalik7605]

this code is giving runtime error

[@VishalKumar-ce9pe]

@@rachnamalik7605 try to submit on Leetcode, yes it can be optimize....