# CSharp solution with O(m)

• First, let's get the total column number. Assume the input character array is in length m and the input row number is n. The offset between two characters in two full columns will be n + 1. Also, the minimum value for column number will be 1. Then the column number should be m / (n + 1) + 1.
There will be a letter in the middle between two columns. The row index for the middle letter will be n / 2.

``````        private char[] ZigZag(string input, int row)
{
int offset = row + 1;
int colNum = input.Length / offset + 1;
int zigZagIndex = row / 2;
char[] result = new char[input.Length];
int resultIndex = 0;
for (int i = 0; i < row; i++)
{
for (int j = 0; j < colNum; j++)
{
int index = i + j * offset;
if (index >= input.Length)
{
break;
}

result[resultIndex] = input[index];
resultIndex++;

if (i == zigZagIndex)
{
int nextIndex = index + (row - zigZagIndex);
if (nextIndex < input.Length)
{
result[resultIndex] = input[nextIndex];
resultIndex++;
}
}
}
}

return result;
}
``````

The time complexcity will be O(n) = n * m / n = m. Space complexity is O(m).

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