HackerRank Diagonal Difference Solution Explained - Java

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[annakarenina]

every time he said "super-easy" my stressful tears just fell down

[yahia]

Im literally crying, this man solved it without even thinking, I spent hours trying to figure it out, I feel so dumb as usual...

[shyamutty]

With List<List<int>>
public static int arr)
{
int leftd = 0;
int rightd = 0;
int n = arr.Count;

for(int i =0; i < n; i++)
{
leftd+= arr[i][i];
rightd+= arr[i][n-i-1];
}
return Math.Abs(leftd - rightd);
}

[dcode]

the hardest part of solving for me is translating the solution in my head into actual code...

[krishnnasarrdah]

hey man, can you solve the "almost sorted" problem from hackerrank
been stuck on it for a while now

[ravichandrang]

Might be simplified as

static int diagnalDifference(int[][] arr) {
int leftToRightSum = 0;
int rightToLeftSum = 0;
int col = 0;
for(int row=0;row<arr.length;row++) {
col = arr.length - row - 1;
leftToRightSum += arr[row][row];
rightToLeftSum += arr[row][col];
}
return
}

[luigikart]

You don't actually need k in this case, you can just re-use i since it is just tracking the row iterations

[jackfrost]

in square array rows and columns are the same so why not cut the code.
btw I did this way in python. Can anyone suggest a more optimized solution?
code:
def diagonalDifference(arr):
# Write your code here
first=0
second=0
rows=len(arr)

for i in range(rows):
first += arr[i][i]
second+= arr[i][rows-1]
rows -= 1

return abs(first-second)

[MaximusPower]

By the way, java convention variable names are not snake case *left_to_right*, but camel case *leftToRight*.
And in your solution *j* is duplication of *i*, given that the input is always a square.

Here's a better solution, with only ONE variable instead of 4:
```
int leftRight = 0;
int rightLeft = 0;

for (int i = 0; i < array.length; i++) {
leftRight += array[i][i];
rightLeft += array[i][array[0].length - i - 1];
}

" + Math.abs(leftRight - rightLeft));
```

[anuragchaturvedi]

Need to be contacted for asking and sharing problem ..


Because you did very easy and smartly

[miltgee]

This was my solution:

public static int arr)
{
var aDiagonal = 0;
var bDiagonal = 0;

for (int i = 0; i < arr.Count; i++)
{
for (int k = 0; k < arr[i].Count; k++)
{
aDiagonal += i == k? arr[i][k] : 0;
bDiagonal += i+k == arr.Count-1? arr[i][k] : 0;
}
}

return Math.Abs(aDiagonal - bDiagonal);
}

Mine is O(n^2)... so Nick's is still more performant

[jaystockton]

You are my Hero....please continue with your wonderful videos...:)

[seydaozdemir]

this solution confusing too many variable
int a=0;
int b=0;
int j=arr.get(0).length-1;
for(int i=0;i<arr.get(0).length;i++){
a+=arr.get(i).get(i);
b+=arr.get(j).get(i);
j-=1;
}
return Math.abs(a-b);

[juanefrainpaquiyauri]

it took me 10 minutes to understand LOL

[avinashrana_blog]

Only solve the question on real parameter.
diffenceDiagonal() has List type parameter.Donot try to solve with your own modified way.

Best of luck bro

[adeleyeadedolapo]

is there an advantage to using the while loop and not a for loop?

[MidoValo]

pls back to these videos try to make more of it you help us solve question

[pifacetester]

Here is the updated problem solution:


int right_to_left = 0;
int left_to_right = 0;

int rows = arr.size();
int columns = arr.get(0).size();

int i = 0;
int j = 0;
int k = 0;
int l = arr.size()-1;


while(i<rows && j <columns && k<rows && l>=0)
{



i++;
j++;
k++;
l--;
}

return Math.abs(left_to_right - right_to_left);
}

[YouTubeisZionistTool]

My Solution :-

public static int arr) {
// Write your code here

int left = 0;
int right =0;
int indexLeft =0;
int indexRight=arr.size()-1;

for(int i = 0 ; i < arr.size() ; i ++){

left += arr.get(i).get(indexLeft);
right += arr.get(i).get(indexRight);
indexLeft ++;
indexRight --;

}

return Math.abs(left-right);

}

[AbcXyz-kz]

Damn guys every single time he says simple I cry