#coding question.

Matrix has 1's followed by 0's. Find the row with the maximum 1

int GetRowWithMost1(int[,] matrix, int row, int column)
{
int maxCount = 0;
int ret  = 0;
for (int I =0; I < row;i++)
{
int count = 0;
for (int j = 0; j < col; j++)
{
   if (matrix[I,j] == 1) count++;
}
if (count > maxCount){
    maxCount = count;
    ret = I;
}
}
return ret;
}

Q: For any integer n >= 2, let N(n) be the maximal number of triples (ai, bi, ci), i = 1, ...., N(n), consisting of nonnegative integers ai; bi and ci such that the following two conditions are satisfied:

(1) ai + bi + ci = n for all i = 1, ...., N(n);

(2) If i != j; then ai != aj ; bi != bj and ci != cj .

Determine N(n) for all n >= 2.

A: The second condition above suggests that a-coordinates in all the triples are pairwise distinct.
If we sum the a-coordinates we get a value that has to be greater than or equal to the sum of the suffixes i-1 from 1 to N. This is because ai has to be less than n and and i can only range from 1 to N(n)
Therefore the sum of a-coordinates has to be greater than or equal to the sum of consecutive numbers which is equal to N(N-1)/2
This same inequality holds for all the coordinates. Consequently, we have 3N(N-1)/2 <= Sum of all coordinates from 1 to N = nN
Therefore 3(N-1)/2 <= n
and N has to be less than or equal to the floor of (2n/3) + 1
If we take different examples we can see that N can be equal to the floor of (2n/3) + 1 and this is the value we are required to determine.

There are three different cases for n that we can take. These are
n=3k-1
n=3k
n=3k+1
and for each we can draw the examples up to floor(2n/3) + 1 and we will see that the limiting case of floor(2n/3) + 1 also holds the above two conditions as given. Hence the determination of N(n) is correct.