Matrix Diagonal Sum - Leetcode 1572 - 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.

[@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, 😄😄

[@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

[@RobinHistoryMystery]

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

[@deekshitht786]

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

[@juanmacias5922]

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

[@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

[@MP-ny3ep]

Thank you!

[@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;
}
}

[@aahiljunaidnext]

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

[@krateskim4169]

Thank you so much

[@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

[@suneosama939]

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

[@felixboachieyiadom4457]

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