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Edvins Antonovs

Diagonal Difference code challenge

My approach to solving HackerRank’s Diagonal Difference code challenge.


Problem

Given a square matrix, calculate the absolute difference between the sums of its diagonals.

For example, the square matrix is shown below:

11 2 3
24 5 6
39 8 9

The left-to-right diagonal = 1 + 5 + 9 = 15. The right to left diagonal = 3 + 5 + 9 = 17. Their absolute difference is |15-17| = 2.

Input format

The first line contains a single integer, n, the number of rows and columns in the square matrix arr.

Each of the next n lines describes a row, arr[i], and consists of n space-separated integers arr[i][j].

Output format

Return the absolute difference between the sums of the matrix's two diagonals as a single integer.

Returns

Returns int value of the absolute diagonal difference.


Solution

This was a reasonably exciting challenge to solve. I've created empty arrays to store primary (pri) and secondary (sec) diagonal values.

Then I created an arrSum function, which we will use later, to sum up, all the array values using the Array.reduce() method.

Finally, here we have a for loop where all the fun starts to kick in. Let's return to our original example and review our two-dimensional array.

11 2 3
24 5 6
39 8 9

Following the example above, our primary diagonal will have 1 + 5 + 9, while the secondary diagonal will be 9 + 5 + 3. Okay, what now? We must access a 2D array - arr[i][j]. Where [i] represents the row's index, while [j] is the element's index within the selected row.

Below is the two-dimensional visualisation of our initial example.

1[0][0] [0][1] [0][2]
2[1][0] [1][1] [1][2]
3[2][0] [2][1] [2][2]

Primary diagonal consists of 1 ([0][0]) + 5 ([1][1]) + 9 ([2][2]). Secondary diagonal consists of 9 ([2][0]) + 5 ([1][1]) + 3 ([0][2]).

The idea is to understand the pattern. For the primary diagonal, we can see that we start at [0][0], and we increment both numbers by 1 with each step.

Below is an example of our for loop for the primary diagonal.

1for (let i = 0; i < arr.length; i++) {
2 pri.push(arr[i][i]);
3}

For the secondary diagonal, things are a bit different. We need to descend, so we start at the last index (arr.length - 1). In our case, it will be 2. Then with each iteration, we also need to subtract - i as we descend.

Below is an example of our for loop at this stage.

1for (let i = 0; i < arr.length; i++) {
2 pri.push(arr[i][i]);
3 sec.push(arr[arr.length - 1 - i][i]);
4}

Now it's time to calculate sums of primary and secondary diagonal values computed using the earlier created arrSum function and subtract the secondary from the primary. Finally, we can return the result of the Math.abs() where we passed the substracted value of both arrays.

Solution complexity: O(n).

1function diagonalDifference(arr) {
2 let pri = [],
3 sec = [];
4 const arrSum = (arr) => arr.reduce((a, b) => a + b);
5
6 for (let i = 0; i < arr.length; i++) {
7 pri.push(arr[i][i]);
8 sec.push(arr[arr.length - 1 - i][i]);
9 }
10
11 return Math.abs(arrSum(pri) - arrSum(sec));
12}

Appendix

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