Rotting Oranges | Directional Arrays & BFS | LeetCode

Rotting Oranges is a popular LeetCode problem asked by Amazon and Microsoft. For this problem, we are given an integer matrix containing only 0's, 1's, and 2's representing an empty cell, a fresh orange, and a rotten orange respectively. The intuition to solve this problem with BFS is that it mentions a grid as input, finding a minimum, and having very small constraints. Solving this problem with a breadth first search leads is the most optimal approach using a queue and directional arrays.
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[@iameenr]

Micheal, you are literally the most underrated creator I know now. Keep up this amazing work.
And your code quality is Kick Ass. 🔥

[@AlgosWithMichael]

Haha thanks so much, I appreciate it!

[@micosair]

Nice trick to store i,j in a single variable.Thanks for sharing.

[@AlgosWithMichael]

No problem!

[@vivek.tiwary]

As always, best explanation. Best part, how to approach this kind of problem, no one talks about intuition except you. Thank you

[@yunyixu5994]

So underrated!!!💯 Thank you for your work!

[@TheElementFive]

Couldn’t have been explained or visualized more effectively!

[@odrare]

Man, that's amazing, thank you

[@odrare]

also we can write this
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(grid[i][j] == 1) {
++freshOranges;
} else if(grid[i][j] == 2) {
queue.add(i * n + j);
}
}
}

[@yaboyspence]

Could the if-check (lines 19-21) be included in the top for loops? Just curious why we need to iterate the grid twice?

[@kavyak.p8878]

Could you please make a video on Leetcode 753. Cracking the safe. It’s a hard problem

[@dvfrost6227]

Unable to pass all the testcases on Leetcode

[@prithvirajrathore4256]

Congrats for Google bro🎉🎉 . I also want to switch my current job to google . Can you help by covering the topics which google developers ask most.