Coverage of all Zeros in a Binary Matrix | gfg potd | 26-06-24 | GFG Problem of the day

[@shah4304]

Nice 👍

[@CodeGenius316]

Thanks 😊

[@mayankmishra7520]

Very helpful video! The explanation is clear and easy to understand. Great job!

[@CodeGenius316]

Thanks 😊

[@sweabhishek]

Bro teri level of understanding bohot acchi hai how you achieve this level of knowledge are you iitian of what

[@CodeGenius316]

Thanks bro...no I'm not iitian
All this can be achieved by regular practice 😊

[@sapthasagar7840]

Thanks for sharing explanation bro 💜

[@CodeGenius316]

Welcome bro 😊

[@printfiamd5254]

class Solution {
public:
int findCoverage(vector& matrix) {
// Code here
int ans = 0;
int n = matrix.size();
int m = matrix[0].size();
for(int i = 0; i < n; i++){
for(int j = 0; j < m-1; j++){
ans+=matrix[i][j]^matrix[i][j+1];
}
}
for(int i = 0; i < m; i++){
for(int j = 0; j < n-1; j++){
ans+=matrix[j][i]^matrix[j+1][i];
}
}
return ans;
}
};
I have tried this approach and works correctly. Just basic XOR operation.

[@CodeGenius316]

Awsm👏🏻

[@printfiamd5254]

@@CodeGenius316 Thanks😃

[@souravsingh7635]

int covverage(vector &matrix)
{
int r = matrix.size();
int c = matrix[0].size();
int sum = 0;
for (int i = 0; i < r; i++)
{
for (int j = 0; j < c; j++)
{
if (matrix[i][j] == 0)
{
if (j > 0 && matrix[i][j - 1] == 1) // left element
{
sum += 1;
}
if (i < r - 1 && matrix[i][j + 1] == 1) //right element
{
sum += 1;
}
if (i > 0 && matrix[i - 1][j] == 1) // upside element
{
sum += 1;
}
if (i < r - 1 && matrix[i + 1][j] == 1) // downside element
{
sum += 1;
}
}
}
return sum;
}
} bhai ye code chal to gaya but muje right side element and downside element mai i < r - 1 , j < c - 1 ye boundary cases ka logic samaj nahi aya can you help in this

[@CodeGenius316]

Bhai ye code same mere Wale jaisa hai...video dekho yehi cheez smjha rkhi hai maine