Spiral Matrix | Leetcode-54 | Amazon, Microsoft, Paytm | Explanation ➕ Live Coding 🧑🏻‍💻👩🏻‍💻

Hi everyone, this is the 10th of our Array Playlist.
In this video we will try to solve a very famous and frequently asked 2-D Array Qn “SPIRAL MATRIX”.
We will do live coding after explanation and see if we are able to pass all the test cases.

Problem Name : SPIRAL MATRIX
Company Tags : Paytm, Zoho, Morgan Stanley, Accolite, Amazon, Microsoft,
Snapdeal, D-E-Shaw, MakeMyTrip, Oracle, MAQ Software,
Nearbuy, Nagarro, BrowserStack

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#coding #helpajobseeker #easyrecipes #leetcode #leetcodequestionandanswers #leetcodesolution #leetcodedailychallenge #leetcodequestions #leetcodechallenge #hindi #india #hindiexplanation #hindiexplained #easyexplaination #interview #interviewtips
#interviewpreparation #interview_ds_algo #hinglish

[sarathkumardunga]

Hi Bhaiya! I am following your your videos for past few weeks and the way you are putting all the thoughts and the way you are going about a solution is jus awesome.Thank you for the brilliant content. I have also taken a look of the github repository.Such a good collection of questions. Eagerly waiting for the new videos and graph playlist😊

[nitkarshchourasia]

*FOR THOSE WHO ARE WONDERING FOR THE FULL SOLUTION WITH NO MISSING PART, HERE IT IS*

class Solution {
public:
vector<int> matrix) {
int m = matrix.size(); // row
if (matrix.empty() || matrix[0].empty()) return {};
int n = matrix[0].size(); // column

int dir = 0;

// dir = 0: left to right;
// dir = 1; top to bottom;
// dir = 2; right to left;
// dir = 3; bottom to top;

int left = 0;
int right = n - 1;
int top = 0;
int bottom = m - 1;
vector<int> result;

while (result.size() < (m * n)) {
if (dir == 0) {
// left to right
// constant: row(top)
for (int i = left; i <= right; i++) {

}
top++;
} else if (dir == 1) {
// top to bottom
// constant: column(right)

for (int i = top; i <= bottom; i++) {

}
right--;
}

else if (dir == 2) {
// right to left
// constant: row(bottom)

for (int i = right; i >= left; i--) {

}
bottom--;
}
else if (dir == 3) {
// bottom to top
// constant : left;

for (int i = bottom; i >= top; i--) {

}
left++;
}
dir = (dir + 1) % 4;
}
return result;
}
};

[sanjaykatta]

One of those problems where naming the variables meaningfully is very very important, otherwise it would lead to a lot of confusion.

[Prakash-xnn]

Your explanation is better than striver 🎉

[souravjoshi]

Ah I found this video in the morning . Masterpiece bro ❣

[coldcoke]

Nice one...
also friendly advice to all the student: after learning the algo try to code yourself without watching sir code.

[AnandKumar-kzls]

i still remember when i watched that video today on a streak of 403 never able to do this without you thanks bro

[darshandm]

Bhai aaj tak itne DSA ke video dkh liye particular javascript me code krne k liye wahi video dkha pdta tha
Tumhari video dkh kr smjh kr khud javascript me code likh deta hu
Well explained 🎉

[yashpaunikar]

the very very easiest solution i have ever seen for this question. thank you bhaiya, i was stucked in this question for last 3 very helpful

[subhamjha]

very clear and easy explanation. Thanks.

[insidious_]

what is the condition under while loop ..aapne baad me btane ko bola video me pr apne submit kr dia

[malavipande]

Where you defined left, right, top down!
Btw

[vishal-sret]

Such a wonderful explanation, thanks sir

[anonymous-qyvw]

again saying i saw the same question from striver stuck in this from the last 1 hour or sooo got it now mannnn

[PakIslam]

Great job :)

BTW For dir you can also simply do a % operator

[lakshsinghania]

bhaiya, spiral matrix 3 ka solution daal do pls samjh nai aa rha kaise solve kare

[vineetkumar]

Hey I think you miss to tell about the breaking condition in while loop? Btw I got it from from your explanation part So no problem at all. Thanks.

[namanjain]

you have not defined variables top, down, left and right ?

[aayushranjan]

pls sara vdo sb me sc nd tc de do har sol ka

[shreyash__]

started on 8th feb completed till lec 10 @ 3:10 AM .

thanks bhaiya and once again 🤎 if possible plzz accept my meeting request .