Hello!

I'm trying to find the longest sequence of equal elements in a matrix M x N. I use two nested for-loops to search into the rows and compare the cells - same thing for the columns. When two adjacent elements are equal, I increment a counter. If the counter is greater than the previous one, I save the new counter value and the current element.

Now I should search in the diagonal directions. How can I do? If i and j are the indexes of a cell, the main diagonal (from left to right) has i == j. But for the secondary diagonals (from left to right and from right to left)? I also have a problem to reach the border of the matrix to avoid the IndexOutOfRangeException... Any help is appreciated.

Hi Rik30, welcome at DaniWeb. :)
Could you show us the method you're having trouble with?

You say you have IndexOutOfRangeExceptions could you explain what your Length method is doing?
And to traverse diagonal, do you mean
1 2 3
4 5 6
7 8 9
[3] [2,6] [1,5,9] [4,8] [7] ?

OK, Took me awhile but this code will print out the diagonal values of a MxN array.

``````for (int i = 0; i < M; i++)
{
int j = 0;
int k = i;

Console.WriteLine("Diagonal nr {0}", diagCnt);
diagCnt++;
while (j < N && k >= …``````

## All 10 Replies

Hi Rik30, welcome at DaniWeb. :)
Could you show us the method you're having trouble with?

Sure! :) I have written the following code:

``````for (int i=0; i<arr.Length(0)-1; i++) // check rows
{
for (int j=0; j<arr.Length(1)-1; j++)
{
if(arr[i,j]==arr[i,j+1])
k++;

if(k > temp)
{
temp = k;
value = arr[i,j];
}
}

k = 0;
}

for (int i=(arr.GetLength(0)-2); i>=0; i--) // check diagonals (left to right) ?
{
while (!border) // border ??
{
if (arr[i,j] == arr[i+1,j+1])
{
k++;
}

if (k > temp)
{
temp = k;
value = arr[i,j];
}

j++; // ?
}

counter = 0;
}
``````

But I cannot imagine a method for the diagonal directions... I also did some sketches on a sheet of paper to find a solution...

You say you have IndexOutOfRangeExceptions could you explain what your Length method is doing?
And to traverse diagonal, do you mean
1 2 3
4 5 6
7 8 9
[3] [2,6] [1,5,9] [4,8] [7] ?

Yes, I mean it. The matrix can be of any size N x M.

I'd start from the penultimate row, first column (the corners cannot be a sequence). Then, I'd compare the cell with another that has indexes i+1 and j+1 (as temporary values?), and so on... until the border is reached. But how can I be sure that I have reached the border? When the row is 0 (after decrements), I should increment the column... (?)

OK, Took me awhile but this code will print out the diagonal values of a MxN array.

``````for (int i = 0; i < M; i++)
{
int j = 0;
int k = i;

Console.WriteLine("Diagonal nr {0}", diagCnt);
diagCnt++;
while (j < N && k >= 0)
{
Console.WriteLine(">> {0}", R[j, k]);
k--;
j++;
}
}
for (int i = 1; i < M; i++)
{
int j = M - 1;
int k = i;

Console.WriteLine("Diagonal nr {0}", diagCnt);
diagCnt++;
while (j >= 0 && k < N)
{
Console.WriteLine(">> {0}", R[k, j]);
k++;
j--;
}
}
``````

Thank you! :)

It was quite tricky for me

Hmm... there is an issue. I solved it by noting that the sum of the indexes is the same (from right to left).

Example:
1 2 3
4 5 6
7 8 9

[2,4] has i + j = 1
[3,5,7] has i + j = 2
[6,8] has i + j = 3

But for the other diagonals (left-to-right)?

Still working on a solution in all directions. Been in hospital lately, I'll keep in touch.

No problem, thank you once again for your time :)

PS: I'm not English... I have written "issue" instead of "problem" thinking it is a synonym. I'm sorry for the misunderstanding

Resolved.

To find the right-to-left diagonals I have split the problem in two sub-problems: the lower left and the upper right of the matrix.

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