Friday 10 July 2015

Diagonal read the matrix array - C++

 

Introduction

After a long time, here comes a post on parsing a NxN matrix. Our main intension here is to take the sum of the diagonals and take a absolute difference between them. At first, I simply forgot the C++ way to initialize an NxN array. Whoof!!

Let’s first understand what a double pointer is. It’s just a pointer to an array of pointers!!

That is,

What we do with a integer pointer *A? . Allocate an array of integers

Then what to do when u need a list of integer pointers **A ?  Allocate an array of integer pointers

How its done?

Now below is the code to allocate the double pointer

int N=10;
int **A = new int*[N]; // get array of N integer pointers
for(int i = 0; i < N; i++)
    A[i] = new int[N]  // now allocate to each pointer an array of N
 
// finally A[i][j] gets you into NxN!!

Next confusion comes with reading this array diagonally. Until we understand how coordinates of an diagonal behave, it’s tough to understand that clearly.

IMG_20150710_232251

 

 If you see the left diagram, you can guess the paper work done for it.

 

From left->right, you can notice that always you get two indices, (i == j) equal. So you got this diagonal for all NxN.

 

From right –> left, what is the pattern ? It has to be the combination of i and j. If you see closely, i increases and j decreases as we move. i always increases in our loop, we just need to take care of making j to decrease.. or equating like this as well  i+j = N+1 works!!

 

 

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
 
 
int main() {
    int N=1;
    cin>>N;
    int **A = new int*[N];
    for(int i = 0; i < N; ++i)
    {
        A[i] = new int[N];
        for(int j = 0; j < N; ++j)
        {
            cin>>A[i][j];
        }
    }
    
    int sumDiag1 = 0;
    int sumDiag2 = 0;
    for(int i = 0; i < N; ++i)
    {
        for(int j = 0; j < N; ++j)
        {
            if(i == j)
            {
               sumDiag1 += A[i][j];    
            }
            
            if(j + i == N-1)
            {
               sumDiag2 += A[i][j]; 
            }
        }
    }
    cout<<abs(sumDiag1 - sumDiag2);
    return 0;
}

DISCLAIMER: There are better ways to do this on internet without O(N^2) loop. You can do this in a single loop as well.