Matrix Diagonal Sum - Leetcode 1572 - Python

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Solving Leetcode 1572, Matrix Diagonal Sum, today's daily leetcode problem on May 7.
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Problem Link: leetcode.com/problems/matrix-diagonal-sum/
0:00 - Read the problem
0:30 - Drawing Explanation
5:40 - Coding Explanation
leetcode 1572
#neetcode #leetcode #python

[@uptwist2260]

Thanks again for the daily. I used a 2 pointers approach: var diagonalSum = function(mat) { let left = 0; let right = mat[0].length - 1; let sum = 0; for (let row = 0; row < mat.length; row += 1) { sum += mat[row][left]; if (left != right) { sum += mat[row][right]; } left += 1; right -= 1; } return sum; };

[@jonnyskybrockett8844]

Doing the if l == r on every iteration would slow it down by a factor of n instead of doing modular arithmetic after or before.

[@jonnyskybrockett8844]

Oh, I did two separate for loops that only go across the diag. I guess it's O(n) either way. edit: shorted it to one loop and it beats a lot more now: class Solution { public: int diagonalSum(vector& mat) { int n = mat.size(); int sum = 0; if(n % 2 == 1) sum -= mat[int(n/2)][int(n/2)]; int j = 0; for(int i = 0; i < n; i++){ sum += mat[i][j]; sum += mat[n-i-1][j++]; } return sum; } };

[@MeanSoybean]

this looks like C++. if that's the case then you don't need to typecast it to int because n/2 is already int

[@Muslim.mode.]

I actually picked up the trick as my first EVER leetcode question, I Made 2 variables though instead of using i as the increments and had the if/else in the loop, 😄😄

[@krishnachoudhary4924]

Amazing. Could you please update your 350 questions list with these new questions. Thanks

[@NeetCodeIO]

Yeah, I'll prob find a way to just automate it. But i'm nervous about making a list that keeps growing indefinitely.. will need to add pagination at least...

[@shubhamraj25]

@@NeetCodeIO do you handle your website on your own?

[@utkarshdewan8736]

@@shubhamraj25 He has a team

[@def__init]

@@NeetCodeIO hahahahaha

[@juanmacias5922]

I ended up using a set, and checking if the tuple([row,col]) was in it lol

[@deekshitht786]

Please upload more videos on leet code problems. No one can explain like u❤

[@RobinHistoryMystery]

"did you notice something? probably did if you have eyes"... sassy aren't we 4:50

[@sondidu]

Did you notice something?? You probably did if you have eyes.

[@NeetCodeIO]

👀

[@acceleratorlevel645]

Bro just roasted and made us laugh at the same time lmao

[@MP-ny3ep]

Thank you!

[@aahiljunaidnext]

just make whole primary digonal 0 after adding it into the sum

[@Karan-java-99]

class Solution { public int diagonalSum(int[][] mat) { int row = mat.length; int col = mat[0].length; int sum = 0; for(int i = 0; i < row; i++){ // Check if the current index is not the center (to avoid double counting in odd-sized matrices) if(i != col - i - 1){ int primary = mat[i][i]; int secondary = mat[i][col - i - 1]; sum = sum + primary + secondary; continue; } // If the current index is the center, to avoid double counting sum += mat[i][i]; } return sum; } }

[@krateskim4169]

Thank you so much

[@suneosama939]

Can anybody explain line if n% 2 else 0, i don't understand 🤓

[@shagggyyx]

class Solution: def diagonalSum(self, mat: List[List[int]]) -> int: n = len(mat) res=0 for i in range(n): res+=mat[i][i]+mat[i][~i] if n%2==1: mid=n//2 res-=mat[mid][mid] return res This will also do sir.we can remove the mid by using negative indexing.

[@felixboachieyiadom4457]

Awesome solution I’ve watched so many videos from you that my code for this problem looks just like yours 😂😂🫶🏾