01 Matrix Leetcode July Daily Challenge || Intuition + Code + Explanation

Given an m x n binary matrix mat, return the distance of the nearest 0 for each cell.
The distance between two adjacent cells is 1.
Link to challenge: leetcode.com/e...

[@shibamdebnath683]

Everytime I see you video in recommendation page, I just don't even think of other explanation videos ❤ Love from NIT Silchar ✌

[@RohitKumar-qh5bq]

Also send some water from NIT Silchar, so we can go to campus again 😂

[@venkatakalyan4953]

Very great explanation. The way you explains, it will make easy to understand the logic. ❤ Love from IIIT Dharwad

[@brilltwoant27]

The IIT Bombay girl explained it the best! Thank you. You're a huge inspiration :)

[@geekpods2824]

You don't need an additional answer array, you can directly modify the original array by adding '-1' and use it, so that the space complexity is just for the additional queue you are using.

[@alapatisrikanth3412]

I really liked it. one arrow 2 birds, learnt Graphs min distance and this problem.
i mean i really haven't understood other channels videos about this problem, honestly at first 2 mins i had same disappointment as with other channels but gradually at 6 and 7th min i got your way into mind.
completely without a doubt easily understandable, you got new subscriber.
Thank god i can sleep peacefully today.

[@probabilitycodingisfunis1]

Thank you so much! Glad it was helpful

[@preetikodwani9842]

gurl, not dying to watch ur vdeos

[@ivanez4456]

Thanks a lot, I was able to understand and solve the leetcode problem because of your great explanation!

[@ashupratap9960]

Hey Alisha can you pleases tell me why dfs with memoization isn't working here.....although for the same problem recursion is working fine,
See this is my code please help if I am conceptually wrong somewhere.
class Solution {
public:
int R;
int C;
vector dp;
vector updateMatrix(vector& mat) {
R = mat.size();
C = mat[0].size();
vector ans(R, vector(C, 999999));
dp.resize(R, vector(C, 999999));
for (int i = 0; i < R; i++) {
for (int j = 0; j < C; j++) {
vector visited(R, vector(C, -1));
dp.resize(R, vector(C, 999999));
ans[i][j] = dfs(i, j, mat, visited);
}
}

return ans;
}
int dfs(int i, int j, vector mat, vector& visited) {
if (i < 0 || i >= R || j < 0 || j >= C) {
return 999999;
}
if (visited[i][j] == 1)
return 999999;
if (mat[i][j] == 0)
return 0;
if (dp[i][j] != 999999)
return dp[i][j];
visited[i][j] = 1;
int top = 1 + dfs(i - 1, j, mat, visited);
int bot = 1 + dfs(i + 1, j, mat, visited);
int left = 1 + dfs(i, j - 1, mat, visited);
int right = 1 + dfs(i, j + 1, mat, visited);
visited[i][j] = -1;
return dp[i][j] = min(top, min(bot, min(left, right)));
}
};
For this input its wrong
[[1,0,1,1,0,0,1,0,0,1],[0,1,1,0,1,0,1,0,1,1],[0,0,1,0,1,0,0,1,0,0],[1,0,1,0,1,1,1,1,1,1],[0,1,0,1,1,0,0,0,0,1],[0,0,1,0,1,1,1,0,1,0],[0,1,0,1,0,1,0,0,1,1],[1,0,0,0,1,1,1,1,0,1],[1,1,1,1,1,1,1,0,1,0],[1,1,1,1,0,1,0,0,1,1]]

[@smitakandari911]

Same doubt

[@Neko-Konekko]

Same doubt (2)

[@prasadk289]

Same doubt, I initially solved using that.

[@gayathri6400]

The code doesn't work, thoughthe logic seems fine. It shows TLE.

[@Vijay-fi8bm]

at 9:21 you forgot to push the coordinate (1,1) into the queue. but the explanation of concept is good.

[@ravinsher]

Also in queue example you did not add (1,1)

[@sumitsinha995]

good explanation ..

[@ravinsher]

Hi can you post the solution link aswell it will be helpful.

[@gowrisankar98jammu49]

no words ur explanation simply superb✌✌✌

[@manisha9001]

Great video ❤ thanks a lot 🙏

[@rachitbadoni5717]

Nice explanation Di ❣️
I am having à doubt since for shortest distance We Can use Floyd warshall also so Why Here Bfs is preffered only ?

[@b28venkatasivakalyankaliki81]

But Now the question has changed ig in the leetcode platform

[@vaibhavgupta973]

thanks alisha

[@Chandraprakash-kx4ic]

Thanks bro..❤

[@shikharsrivastava7312]

Amazing explanation

[@reallanonym2015]

what's the time complexity of the code?

[@sathya1657]

Great explanation

[@shivamsharma-zv7zl]

history still can't explain how can one open so many youtube tabs in browser😁

[@sushant8686]

why do we need to check visited or not ?

[@nikhilflautist]

Thankyou so much for this solution its very easy to understand and grasp, U are from IIT with metallurgical engineering. A big respect for a big level of hardwork to reach at this position. Thanks again.

[@MP-ny3ep]

Great explanation as always. Thank you !!!

[@TarunKumar-cn6in]

Very nice explanation thanks 🙏

[@mxc7222]

very helpful thank you very much

[@Arya20012]

thank you mam

[@aryanyadav3926]

Wonderful explanation!

[@chrisdunphy7085]

I finally understand BFS lol

[@sumitsinha995]

class Solution {
public:
bool checkValid(int i,int j,int m,int n){
if(i=n){
return false;
}
return true;
}
vector updateMatrix(vector& mat) {
queueq;
int m=mat.size();
int n=mat[0].size();
for(int i=0;i

[@bcsshrutiJaiswal]

Good explanation carry on

[@abhishek04204]

good one.

[@ashvinkumhar5819]

great explanation🙌

[@codingarena4297]

Didi. Can you please explain why can't we use a DFS approach here?

[@respicio2990]

You can but it will complicate the logic but BFS any way gives you nearest distance always so BFS is preferred.

[@systemforge]

Woah🔥

[@prashantgupta7631]

very nice explanation ,thx

[@SuperWhatusername]

Thank you

[@dailydestress6189]

use openboard

[@AyushSingh-lt2rq]

Amazing explanation.

[@sakshi-ok8zu]

great explanation!

[@mdkashif7808]

Very good

[@karthikk7136]

Clear explanation👏

[@master_persi]

code link ??

[@preetikodwani9842]

U don't even use language which I use. I feel, u don't know coding

[@preetikodwani9842]

Kaam kr apna, mere against krne k liye log ko calls na kr

[@mrprime557]

lol u tlk funny

[@preetikodwani9842]

mind ur business

[@gowrisankar98jammu49]

awesome explanation

[@bhawna1997]

Amazing explanation