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Bros explaination 🚀🚀🚀🚀

Super explanation sir 🎉

bhai keep it going 👊 subscribed your channel

Someone recommended me this channel for GFG POTD and it is worth it 💯🔥🔥

Thank You sir

Best Explanation Bhaiya 🚀🚀🚀🚀🚀🚀🚀🚀

Thank You So Much Bhaiya, Yaar Itna Easy Question Than 🥳🥳🥳🥳🥳🥳

Nice Explanation 😊😊😊

Nice Explanation 😊😊😊

Friendly and good explanation

explained so well sir 😊😊 You should definitely make videos for LEETCODE POTD

achha pdate ho bhai, java me code diya kro if possible . good going thanks

Keep going sir..

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Bhai bahut hard tha ye question tumne ekdum easy kridya 😌

22

nice explanation

Don't go without subscribing

Thank you bhaiya ❤

Face reveal bhai

best explained .......your intuition is best bhaiya

Not executing..

Bro keep going .

You’re working so hard, may all your wishes come true.

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Behtriiiiinnn explanation 🙃🙃

What a understanding 😮

Thanks!

your solution is of O(n^3) right? expected time complexity is O(n^2).

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Thank you sir

bhaiya vo column nikalna samajh nhi aya

Bhai mai 10 email se subscribe kar dunga tu bass video dalana band mat kar

Great question

Did they removed 100 coin t-shirt

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

int covverage(vector<vector<int>> &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

Nice 👍

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

class Solution { public: int findCoverage(vector<vector<int>>& 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.

Thanks for sharing explanation bro 💜

why are you using row.rend() and not row.end() and row.begin()

Thanks again sir ❤

sir itna detail me explain kiya hai koi bhi smjh jayega .... amazing sirr🙇