Square matrix exercise

[Kernul]

The exercise is:
Manage a square matrix of order N by:
- Determining and storing the maximum of each line;
- Determining and storing the minimum of each column;
- Calculate the difference between the relative maximum and minimum of the same location;
- Print a table showing the maximum, minimum, and their differences.

I was able to do the first two point and I'd like you to check if it's alright.
This is my code:

#include <iostream>

using namespace std;

int main () {
	//Initialization:
	const int N = 10;
	const int M = 20;
	int mat[N][N];
	int vecmax[M] = {0};
	int vecmin[M] = {0};
	int n = 3;
	mat [0][0] = 1; mat [0][1] = 4; mat [0][2] = -1;
	mat [1][0] = 6; mat [1][1] = 6; mat [1][2] = 5;
	mat [2][0] = -2; mat [2][1] = 7; mat [2][2] = 10;
	
	//Procedure:
	for (int i = 0; i<n; i++) {
		for (int j = 0; j<n; j++) {
			if (mat[i][j] > vecmax[i]) {
				vecmax[i] = mat[i][j];
			}
		}
	}
	
	for (int i = 0; i<n; i++) {
		vecmin[i] = mat[i][i];
		for (int j = 0; j<n; j++) {
			if (mat[j][i] < vecmin[i]) {
				vecmin[i] = mat[j][i];
			}
		}
	}
	
	//Output:
	for (int i = 0; i<n; i++) {
		cout<<vecmax[i]<<" ";
	}
	cout<<"\n";
	for (int i = 0; i<n; i++) {
		cout<<vecmin[i]<<" ";
	}
	
	//Termination:
	return 0;
}

In the third point I should find the position of each max and min number. How do I do that?

[LB]

Don't just store the min/max itself, also store the position you found it.

[Kernul]

How? Should I create two arrays of position?

[shadowCODE]

1 2 3
4 5 6
7 0 1

Let X = maximum of each line
Let Y = minimum of each column

Then

X = {3,6,7}

Y = {1,0,1}

Calculate the difference between the relative maximum and minimum of the same location;

For each index in the matrix M(i,j), Difference = X(i) - Y(j).
Then print the table of X(i) | Y(j) | Difference

[Kernul]

What do you mean by

For each index in the matrix M(i,j), Difference = X(i) - Y(j).

? I don't understand it. The way you wrote it would be like:

for (int i = 0; i<n; i++) {
		for (int j = 0; j<n; j++) {
			difference = X(i) - Y(j);
		}
	}

And it simply do the difference without taking into account "the same location".