Question

Diagonal Difference HackerRank Pseudocode and C++: Given a square matrix, calculate the absolute difference between the sums of its diagonals.

Function Description

Complete the diagonalDifference function described below to calculate the absolute difference between diagonal sums.

diagonalDifference( integer: a_size_rows, integer: a_size_cols, integer array: arr)

Parameters: a_size_rows: number of rows in array a_size_cols: number of columns in array a: array of integers to process Returns: integer value that was calculated

Constraints

-100 < = elements of the matrix < = 100

Raw Input Format

The first line contains a single integer,  denoting the number of rows and columns in the matrix .
The next  lines denote the matrix 's rows, with each line containing  space-separated integers describing the columns.

Sample Input 0

3
11 2 4
4 5 6
10 8 -12

Sample Output 0

15

Explanation 0

The primary diagonal is:

11
   5
     -12

Sum across the primary diagonal: 11 + 5 - 12 = 4

The secondary diagonal is:

     4
   5
10

Sum across the secondary diagonal: 4 + 5 + 10 = 19
Difference: |4 - 19| = 15

Note: |x| is the absolute value of x

#include <bits/stdc++.h>

using namespace std;

/*
* Complete the diagonalDifference function below.
*/
int diagonalDifference(vector<vector<int>> a) {
/*
* Write your code here.
*/

}

Please write the pseudocode along with the C++ source and explain in full detail both the pseudocode and the source code. Thanks. Mathematically speaking, I know we are trying to sum both diagonal lines of a square matrix and so we would have to use matrix notation for the pseudocode to accompisl the diagonal difference. Is this right?

Answer 1

int diagonalDifference(vector<vector<int>> a) {

int sumFirst = 0, sumSecond = 0;

int N = a.size();

for (int i = 0; i < N; i++) {

sumFirst += a[i][i];

}

for (int i = 0; i < N; i++) {

sumSecond += a[i][N - i - 1];

}

  

return abs(sumFirst - sumSecond);

}

===========
It passed all the test

Test Case #0 Test Case #3 Test Case #6 Test Case #9 Test Case #1 Test Case #4 Test Case #7 Test Case #10 Test Case #2 Test Case #5 Test Case #8


Run and let me know if there is any concern


PSEUDOCODE

Algorithm DiagonalDifference(Vector<Vector>A)

START

N = A.size()
sumF = 0
while i<N
sumF = sumF + A[i][i]
i = i + 1

i = 0, sumS = 0
while i<N
sumS = sumS + [i][N - i - 1];
i = i + 1

RETURN abs(sumS - sumF)

END