C++ Is a matrix triangular?

Question

So we have a nxn matrix (n <= 20). If a[i][j] == 0 when i > j and a[i][j] != 0 when i <= j ... or ... a[i][j] == 0 when i < j and a[i][j] != 0 when i >= j ... the matrix is called triangular. I suppose in a 3x3 triangular matrixes look like this

1 1 1     1 0 0 
0 1 1     1 1 0 
0 0 1     1 1 1

Here's my code:

    bool flag;
    int count = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
        {
            if ((i > j && a[i][j] != 0) || (i <= j && a[i][j] == 0))
            {
                count++;
                break;
            }
        }

    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
        {
            if ((i < j && a[i][j] != 0) || (i >= j && a[i][j] == 0))
            {
                count++;
                break;
            }
        }

    if (count == 2) flag = false;
    else flag = true;
    cout << flag;

Now, about the issue - the two examples of matrixes I gave before the code should always return 1 (true) as they are triangular by definition, while, say, this:

0 1 0
0 0 5
1 3 4

should always return 0 (false) as it is by no means triangular. My problem is that no matter what I input, fore some reason it always returns 1 (true) telling me that the matrix is triangular while it clearly is not. My question is how can I possibly fix this? I have absolutely no idea what my mistake is though I suspect it has something to do with my count approach, which is basically - if some of the conditions for a triangular matrix is not met - break the cycle and add +1 to the counter, then check if it's a reversed triangular matrix and if it still isn't add one more to the counter. In the end, the counter will be either 1 or 2, which will indicate whether the matrix is triangular or not. There's a leak somewhere and I can't seem to be able to find it. If it turns out it's all wrong, I'm open for suggestions to do it some other way.