Question

C#

1. Given two lengths between 0 and 9, create an rowLength by colLength matrix with each element representing its column and row value, starting from 1. So the element at the first column and the first row will be 11. If either length is out of the range, simply return a null. For exmaple, if colLength = 5 and rowLength = 4, you will see:

11 12 13 14 15
21 22 23 24 25
31 32 33 34 35
41 42 43 44 45

public static int[][] CreateJaggedMatrix(int rowLength, int colLength)
{
  
}

2.

Given two lengths between 0 and 9, create an rowLength by colLength matrix with each element representing its column and row value, starting from 1. So the element at the first column and the first row will be 11. If either length is out of the range, simply return a null. For exmaple, if colLength = 5 and rowLength = 4, you will see:

11 12 13 14 15
21 22 23 24 25
31 32 33 34 35
41 42 43 44 45

public static int[,] CreateMatrix(int rowLength, int colLength)
{

}

3. Produce a new one-dimensional array that maps all the diagonals from upper-left to lower-right elements. The first element will start from the bottom-left corner of the array. For example, if the value of a is
11 12 13 14
21 22 23 24
31 32 33 34

The new array will be: { 31, 21, 32, 11, 22, 33, 12, 23, 34, 13, 24, 14 }.

public static int[] Diagonals_NWtoSE(int[,] a)
{
  
}

4. generate a rectangular array by specifying the number of rows and the number of columns. A sequence of numbers, starting from a number, of the array will then be populated into the rectangular array in clockwise loops, starting from the top-left corner of the outer loop to the last inner loop. If the row's length or the col's length is invalid, simply return a null.
/// For 8 by 11 array and start = 1, you will have
/// 1 2 3 4 5 6 7 8 9 10 11
/// 34 35 36 37 38 39 40 41 42 43 12
/// 33 60 61 62 63 64 65 66 67 44 13
/// 32 59 78 79 80 81 82 83 68 45 14
/// 31 58 77 88 87 86 85 84 69 46 15
/// 30 57 76 75 74 73 72 71 70 47 16
/// 29 56 55 54 53 52 51 50 49 48 17
/// 28 27 26 25 24 23 22 21 20 19 18

public static int[,] CreateSpiral(int rowLength, int colLength, int start = 1)
{
  
}

Answer 1

Answer 1) Jagged Matrix is a matrix in which each row has different number of columns.

Since in this question each row has same number of columns, so we can use a simple 2-Dimensional array.

public static int[][] CreateJaggedMatrix(int rowLength, int colLength)
{
if ( rowLength <0 || rowLength >9) // If rowLength doesn't fall in range 0-9
return null;
  
if ( colLength <0 || colLength >9) // If colLength doesn't fall in range 0-9
return null;
  
int[, ] intarray = new int[rowLength, colLength]; // 2-Dimensional Array created with given number of //rows and columns
for (int i = 0; i < rowLength; i++)
{
for (int j = 0 ; j< colLength ; j++)
intarray[i, j] = (i+1)*10 + (j+1); //(i+1) & (j+1) bcoz indexing of array starts from 0
// This formula represents when we have to create a 2-digit no.
// For Example, to generate 11 we will write 1*10 + 1

}
return intarray;

}

Answer 2)

public static int[,] CreateMatrix(int rowLength, int colLength)
{

if ( rowLength <0 || rowLength >9)
return null;
  
if ( colLength <0 || colLength >9)
return null;
  
int[, ] intarray = new int[rowLength, colLength]; // 2-Dimensional Array created with given number of //rows and columns
for (int i = 0; i < rowLength; i++)
{
for (int j = 0 ; j< colLength ; j++)
intarray[i, j] = (i+1)*10 + (j+1); //(i+1) & (j+1) bcoz indexing of array starts from 0
// This formula represents when we have to create a 2-digit no.
// For Example, to generate 11 we will write 1*10 + 1

}
return intarray;

}

Answer 4)

public static int[,] CreateSpiral(int rowLength, int colLength, int start = 1)
{ // We will maintain four variables to toggle between rows and columns to achieve the required spiral form.
int i, k = 0, l = 0, m, n;
m = rowLength;
n = colLength;
int [, ] a = new int[m, n];
/* k - starting row index
m - ending row index
l - starting column index
n - ending column index
  i - variable used for iterating through rows in matrix  
*/

while (k < m && l < n)
{
// Storing the first row from the remaining rows
for (i = l; i < n; ++i)
{
a[k, i] = start++;
}
k++;

// Storing the last column from the remaining columns  
for (i = k; i < m; ++i)
{
a[i,n - 1] = start++;
}
n--;

// Storing the last row from the remaining rows */
if ( k < m)
{
for (i = n-1; i >= l; --i)
{
a[m - 1, i] = start++;
}
m--;
}

// Storing the first column from the remaining columns */
if (l < n)
{
for (i = m-1; i >= k; --i)
{
a[i, l] = start++;
}
l++;  
}   
}
return a;
}