Interesting! (j+k == (numInputs-1)) is cool.

me too same approch...

Same, though my second if statement was the mathematically equivalent (i == (numInputs - j -1)).

Yeah. You don't need to save the matrix into an array

I love this solution, neat and easy to understand... though for a beginner like me, array is our first choice. Thanks for sharing by the way.

surely your way of solution is quite good it saves a lot of space but if look 4 time it is not a cool solution if the number for matrix N is too high then for each entry there will be condition checking goes timely too much costly.

cool :) am noob coder! wasted 40 mins for this logic but finally happy.

same!

I performed similar logic on each row, but used a temporary array. Difference being one loop absent confusing for-loop nesting. Solution in C#.NET.

I used this as a basis for my c# submission as I was quite stumped.

int sum1=0,sum2=0;   

for(int i=0, k=size-1; i< size; i++){
    for(int j=0; j< size; j++){
        cin >> matrix[i][j];            
        if(i==j)
         sum1 += matrix[i][j];                        
    }
    sum2 += matrix[i][k];              
    k -= 1;
}

cout << abs(sum1-sum2);

Nice logic :)

there's no need for nested for loops:

matrix_size = int(input())

current_x_one = 0
current_x_two = matrix_size - 1

sum_one = 0
sum_two = 0

for line in range(matrix_size):
    input_string = input().split(" ")
    sum_one = sum_one + int(input_string[current_x_one])
    sum_two = sum_two + int(input_string[current_x_two])

    current_x_one += 1
    current_x_two -= 1

print(abs(sum_one - sum_two))

Great work rick.thanks for the idea of not using an array.....being a beginner all will think of using arrays which unnecessarilly will occupy some storage space.thanks once again for adding this code..

//I did it this way with one loop, but mine is less effective
for(i=0;i<n*n;i++)
    {
    cin>>t;
    if(i%n==(i/n)%n){Lsum+=t;}
    if(i%n==n-1-(i/n)%n){Rsum+=t;}
}
cout<<abs(Lsum-Rsum);

Could someone help me understand what's happening with the line "cin >> curInput;"?

awesome

Hi,

I have a question regarding this condition If our numInputs is an odd number

if(j == k) { leftD += curInput; }

at some point we will have a (j==k) in rightD as well and at this point the code will add to the leftD and rightD as well which would make the same to give you a wrong answer. Wont It?

Please forgive me. Noob Question if i havent seen something correctly.

it took me several minutes to understand your code im not that fast to analyse problems, what surprise me is how you decomposed the issue and found a solution within the loop, could you give me some tips about organize thoughts and everything to perform a better logic and not hardcoding things just like my solutions XD. thanks

whats about the number which is common in both digonal....

I don't even bother storing leftD and rightD.

if(j == k) { result += curInput; }
if(j+k+1 == numInputs) { result -= curInput; }
ans = abs(result);

#!/bin/bash

read n;

sum=0;

for i in n ); do

read line;

sum=sum + echoi -d " " - (echo((i )) -d " ") ))

done

echo $sum | sed -e 's/-//'`

Thx, dude! Nice approach! I did it before using array to store a literal matrix, but after reading it and the comments bellow, y'all convinced me to make it without storing data into arrays. Doing this way was a bit challenging using python, cause im not quite acquainted with the language, yet. But it worked. Thx!

And yes, it's better not to store data in arrays if you won't use it later.

Ah, your "(j+k == (numInputs-1)" condition is very nice. I was using "j == (numInputs-1)-k and k == (numInputs-1)-j". Your logical solution for this condition is much better than mine. Thanks for this learning too.

Can someone explain what this line does?:

cin >> curInput;

A very clean solution, thank's for share ur algorithm ;)

Cool!

I did something similar. :)

int main(){
int n,*p,i,j,sum,k;

scanf("%d",&n);
p = (int *)calloc(n,sizeof(int));

for(i=0;i <= n*n - 1;i++)
    scanf("%d",&p[i]);

j=0;k=n-1;sum=0;

for(i=1;i<=n;i++)
    {
    sum += p[j];
    sum -= p[k];
    j   += n+1;
    k   += n-1;
}

printf("%d",abs(sum));
return 0;

}

Super cool yar!!

Why complexity of nested loop? This is how I solved it:

long sum1 = 0; long sum2 = 0;

        for (int i = 0; i < n ; i++)
        {
            sum1 += a[i][i];
            sum2 += a[i][n - 1 - i];
        }
        Console.WriteLine(Math.Abs(sum1 - sum2));

I like this idea. I see that you don't need O(n^2) storage space but does it need O(n^2) time complexity? If so, can it be reduced?

Nice!

really cool approach

It means you are just considering the input values when the if conditions are true, otherwise all other values are discarded. Right ?

Thats Perfect !!

superb sir hats off.......

nicely done

Same here. Except since we are calculating the difference, I merely declared an int of 0 to hold the difference. In the first if conditional analogous to yours, I added to this int. In the second conditional, I subtracted from this int. My logic was, these early subtractions help keep the numbers small preventing early overflows.

This one is so cool and clear, thanks for sharing.

public static int arraySum(int[][] array){ int total = 0;

for (int row = 0; row < array.length; row++)
{

    total += array[row][row];
}

return total;

}

Nice approach

You are using 2 variables "left and rigth" when you can use only one. wich one should be incremented and decremented.

Second loop (k) is not needed, index can be caluculated from array dimension and j

this logic (j+k==n-1)is short and precise

nice one!

Oh nice, I did this too in Javascript! I was afraid it wouldn't work because the pattern "seemed" to be that the x/y coordinates added up to be 1 less than N.

var backSlash = 0;
var forwardSlash = 0;
for (var i = 0; i < n; i++) {
    for (var j = 0; j < n; j++) {
        if (i == j) {
            backSlash += a[i][j];
        }
        if (i + j == n-1) {
            forwardSlash += a[i][j];
        }
    }
}

console.log(Math.abs(backSlash - forwardSlash));

What is the curInput used for?

wow, awesome man you showed me a whole new way to look at this problem.

now it's look like a algorithm domain...

cool..bro

it is the best solution of the question given and it is good even for large value of N;

I did about the same, but using a single for loop and a single sum variable:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
	int sz; scanf("%d", &sz);
	int sum = 0;
	for (int i = 0; i < sz*sz; i++) {
		int x; scanf("%d", &x);
		if (!(i % (sz+1))) sum += x;
		if (i > sz-2 && i < sz*sz-1 && !(i % (sz-1))) sum -= x;
	}
	printf("%d\n", abs(sum));
}

nice approch , solution is quite good

Cool !!!

it was my condition for show an different example

if(a_i==(n-a_j-1))

We can reduce the complexity by both loop at same time execute and no if condition like this:

    i think its better approach instead of using two if condition.

    for(int i = 0,j = n; i < n && j > 0; i++,j--){
        diag1 += arr[i][i];
        diag2 += arr[i][j];
    }

this won't work for an odd square matrix like 3x3, 5x5, 7x7

Could you maybe explain the: "cin >> curInput;"? I know that >> is used for bitshifting, but that is about it. Eveen after research I still have no coue how your solution works.

Thanks in advance.

Very intersting solution.

i tried with this method but it is showing terminated due to timeout

Great Dude! if(j+k == (numInputs-1))

if(j+k == (numInputs-1))

really instresting logic. how you find this?

How can I use like this code in Java

Sorry I'm a beginner haha

Great !!

great thinking :D

Interesting !!

sorry but i did not understand the "cin >> curInput" line :(

awesome bro !!

Haha without using array approach is very cool.

Brilliant!

what would be the numInputs here, example plz.

please explain it.

Cool answer! I couldn't figure out until i draw the total of x and y coordinate in paper. Total of right diagonal is always same!

Two for loops expensive.. function solve(n, arr){ var primarySum = 0; var secondarySum = 0; arr.forEach(function(v, row){ primarySum+=v[row]; secondarySum+=v[n-row-1]; });

return Math.abs(primarySum - secondarySum); }

What does this do? If this is a bit shift, what is the purpose?

    cin >> curInput;

if(j+k == (numInputs-1)) You are awesome :)

include

include

include

include

include

include

include

include

include

int main() { int a[100][100],n,c=0,d=0,i,j,sum=0; scanf("%d",&n); for(i=0;i for(i=0;i

you are good in math,certainly!

second=0; n-1; for(i<i++){ a[i]; a[j]; $j--; }

wow.. surely a thing to learn

nice approach!

that is more nicer way than the way i did it k == (size-1)-j

awesome logic dude

Awful solution, it takes O(n^2) which is quadratic time, this does not scale. No idea why people are cheering this code.

Awesome!

It's Simply so brilliant!!!

Aren't nested for loops a performance bottleneck? They perscribe to O(n^2) right? Isn't that unscalable?

why is there (numinputs-1) and not (numinputs+1)?

Explain j+k == (numInputs-1) ?? For every matrix element a_jk. When j+k would be equal to numInputs+1, then and only then it is right Diagonal. For example, in matrix of three, every element with j+k=4 (where 4 is numInput 3 + 1). Kindly explain this to me.

a₁₁ b₁₂ c₂₁ c₂₂

Here 1+2 and 2+1 = 3.

Did this in python

n = len(a)
ans = 0
for i in xrange(n):
    ans+=a[i][i] - a[i][n-i-1]
print abs(ans)

Mine followed a similar logic, but you don't actually have to use two for loops. You already know the relationship between j and k, so you can use just one for loop to increment j.

int rightD = 0;
        int leftD = 0;
        for(int i = 0; i < n; i++){
            leftD += a[i][i];
            rightD += a[i][n-i-1];
        }
        System.out.println(Math.abs(leftD - rightD));

Came to the same result, but ruby way. Just one loop

diag1=diag2=0
for i in (0...n)
    diag1 += a[i][i] 
    diag2 += a[i][n-i-1] 
end
puts (diag1-diag2).abs

O(N)Solution -

l=0 p=0 for i in range(len(a)): l+= a[i][i] p+= a[i][len(a)-1-i]

print(abs(l-p))

dang.... what.. is curInput, and cin and leftD, as I finished by myself I was very proud but realized, i dont know anything lol

Any Kotlin guys here? I'm trying to use the same logic here, but seems like always getting Runtime Error all the time. Could somebody tell me what is wrong here?

fun main(args: Array<String>) {

    val numInputs = readLine()!!.toInt()
    var leftD = 0
    var rightD = 0

    for (j in 0..numInputs) {
        for (k in 0..numInputs) {
            val curInput = readLine()!!.toInt()
            if (j == k) {
                leftD += curInput
            }
            if (j+k == (numInputs - 1)) {
                rightD += curInput
            }
        }
    }

    println(Math.abs(leftD - rightD))
}

Nice, this is a rigth algorithm.

static int diagonalDifference(int[][] a) { int l=a.length; int d1=0; int d2=0; for(int i=0;i

Really innovative!

int x = 0; int y = 0; int index = a.length-1; for(int i=0; i } return Math.abs(x-y);

that's pretty cool that you calculate both in one go of for loops

Very Creative....Great Work dude

Testing

Yeah me too. Mine was the same.

Thats clever. I like this solution +1

nice..but it should be raw_input().split() in line4 and not input().split()

cool, now the complexity is O(n);

brilliant!

This is what I was referring to in earlier comments. No if logic is required. This is similar to my solution.

My version:

r1, r2 = 0, 0

for i in range(int(input())):

T = list(map(int,input().split()))
r1 += T[i]
r2 += T[-i-1]

print(abs(r1-r2))

I like this solution!

Hi Gamanoid! Can you please explain me how row[-(i+1)] is automatically evaluated to row[N - (i + 1)]? New to Python!

This solution doesn't work. first, row = input().split() is inside the for loop. input has to be entered in each iteration. second, int(row[i])-int(row[-(i+1)]) is not properly incremented. you are not skipping to the next row.

here is a solution that works (inspired by yours):

N = int(input())
matrix = input().split( )
a = 0
b = -1
total = 0
for i in range(N):
    try:
        b -= N - 1
        total += int(matrix[a]) - int(matrix[b])
        a += N + 1
    except IndexError:
        break
print(abs(total))

you need the exception or you will will get indexError list index out of range.

sadly, my solution still needs to be optimized. it runs in 6.943235158920288 seconds

Same solution I used :)

This one is very nice

result=[] a=int(input("Enter the no of rows")) b=int(input("Enter the no of cols")) c=[[int(input("enter the elements:"))for j in range(a)]for i in range(b)]

for i in range(a): for j in range(b): s=0 if i==j: print(sum(c[i][j])) s=(c[i][j]) #s += c[i][j] result.append(sum(s)) print(sum(s))

what a fool and fucking guy you r!

Your solution assumes that the array is already available for you to reference. Other people have nested for loops as they need to create or read in the matrix from the input in order to access it. Other than that difference, your solution is similar to the others.