# How to Create a circular matrix table in C# .NET Here will discuss how to create a circular matrix in the C#. matrix is a two-dimensional array of numeric values, which contain any type of data. below is the 3×3 matrix: 3 23 44 65 32 43 87 6 7 Matrices contain columns and rows and start with a 0 index. below are the coordinate of reading a matrix 0,0 0,1 0,2 1,0 1,1 1,2 2,0 2,1 2,2  The matrix starts with (0, 0) coordinates, generally 0 or 1,. below is an example of a 3×3  matrix: 1 2 3 11 12 13 21 22 23 So we start with 1 and then till row 3. Then we continue to 23 (2, 2 coordinates), followed by the bottom row, and then again we do the same.the top row is 1, 2, and 3 but the second row is 11, 12, and 13 which doesn’t reveal anything to me. Looking at the columns is even less promising: 1, 12, 11, 10, and then 2, 13, 16, and 9. Frankly, I don’t see anything that catches the eye, no symmetries, no patterns that can be modeled in a loop, etc.Therefore a more viable solution is to follow the index as it’s incremented and in the outer circle. We then do the same in the inner circle and continue inwards for any remaining circles.So in the first iteration, we want to fill the matrix as below:            In the second iteration, we follow the same route with the remaining, smaller inner matrix:  stop This iteration will be an outer one in the code. We’ll also need to iterate through the various sections:§  top row rightwards§  right column downwards§  bottom row leftwards§  left column upwardsbelow is the code for the sameusing System;using System.Collections.Generic;using System.Linq;using System.Text;using System.Threading.Tasks;namespace circularMatrixInCSharp{    public class Class1    {        public class circularMatrix        {            public List<List<int>> circularMatrixFun(int n)            {                var output = new List<List<int>>();                var tempArray = new int[n][];                for (int i = 0; i < n; i++)                {                    tempArray[i] = new int[n];                    output.Add(new List<int>());                }                 for (int i = 0; i < tempArray.Length; i++)                {                    output[i].AddRange(tempArray[i]);                }                return output;            }        }    }}here we are returning a temporary jagged array and then filling them with some empty values. We need to see that the array contents are copied to the list and then returned from the given function.We’ll need many counters. We need to know where we start and end each row and the column in the main outer loop. need to Remember, the main outer loop serves to identify the dimensions of the array. 1 1 1 1 2 2 1 2 2 In a 3×3 matrix, we have 2 circles. They are denoted by purple 1’s and reds 2’s in the above example. A 3×3 matrix will also have 2 circles but the inner circle has only one element, the one in the very middle of the matrix. A 2×2 matric has only a single circle.The counter starts at 1 in row 0 and column 0, those are 3 variables that we can identify right there. Those are valid for the first circle. When we’re done with the first circle then the counter will have whatever value it has got up to in the iteration, whereas the start row and start column will be 1. (1, 1) is the starting point for the inner circle of a 4×4 matrix – and for any other matrix of any size for that matter.We also need to see how far out we’re allowed to go in each circular iteration. In the first iteration, we’ll start at (0, 0) and the farthest we reach is at (3, 3), i.e. the bottom right-hand cell of the matrix. In the second circle, we start at (1, 1) and the max reach will be at (2, 2) otherwise we overwrite some of the values we have already calculated. It sounds like it’s not enough to know where we start, i.e. start column and start row, but where we end as well, i.e. the end column and end row.If you refer back to the table with the arrows then we’ll need to set these row and column pointers as follows:§  start row: 0§  end row: n – 1, i.e. 3 in a 4×4 matrix to comply with the 0-based array indexing§  start column: 0§  end column: n – 1, same as for end rowWe start filling up the top row, i.e. from the start row to the end column, (0, 0) to (0, 3) in a 4×4 matrix. When we’re done then we increment the start row to 1. At this point, we’re filling up the right-hand column from (1, 3) to (3, 3). Then we decrement the end column and continue filling up the bottom row from (3, 2) to (3, 0). Finally, we move up, i.e. decrement the end row and fill in the cells from (2, 0) to (1, 0). Recall that we incremented the start row to 1 so we don’t return to (0, 0) when filling up the left-hand column.At this point start row and start column are 1, end row and end column are 2, we’re ready for the inner circle where the same steps are taken.So we’ll need the following:§  the main loop for each circle in the matrix§  inner loops for each section of the circle§  top row§  right column§  bottom row§  left columnAlso, the bottom row and left column are iterated backward, i.e. we need a for-loop where the iterator integer decreases to make sure we’re moving in the right direction.We break the main outer loop when we have reached the end column and the end row, i.e. the start row and the start columns have converged to the end equivalents.This is a lot of talking that’s difficult to visualize so let’s see a partial implementation with only the counter values and the various loops:using System;using System.Collections.Generic;using System.Linq;using System.Runtime.Remoting.Metadata.W3cXsd2001;using System.Text;using System.Threading.Tasks;namespace circularMatrixInCSharp{    public class circularMatrix    {        public List<List<int>> circularMatrixFun(int n)        {            var output = new List<List<int>>();            var tempArray = new int[n][];            for (int i = 0; i < n; i++)            {                tempArray[i] = new int[n];                output.Add(new List<int>());            }            int counter = 1;            int ColStart = 0;            int Colend = n - 1;            int startRow = 0;            int endRow = n - 1;            while (ColStart <= Colend && startRow <= endRow)            {                for (int i = ColStart; i <= Colend; i++)                {                }                //this row completed, now increase the row pointer, ex move down in the matrix                startRow++;                //right column                               for (int i = startRow; i <= endRow; i++)                {                }                //done with the column, move to the left                Colend--;                for (int i = Colend; i >= ColStart; i--)                {                }                //move it up again to the right hand side of the matrix                endRow--;                for (int i = endRow; i >= startRow; i--)                {                }                //move to the right for the next loop                ColStart++;            }            for (int i = 0; i < tempArray.Length; i++)            {                output[i].AddRange(tempArray[i]);            }            return output;        }    }}Those loops pave the way for us to keep the index right, i.e. that we’re in the correct cell all the time. The while loop represents the circles in the matrix and is broken when the start and end indexes of the rows and columns have converged, i.e. we have reached the final cell in the matrix.The 4 for-loops represent the 4 directions we’re taking, as outlined above. Also, notice the decreasing direction of the second two loops that we also hinted at above.Now that we have the correct pointers we can increase the counter variable within each for-loop, that’s simple:for (int i = ColStart; i <= Colend; i++){      counter++;}    startRow++; for (int i = startRow; i <= endRow; i++){       counter++;}Colend--;for (int i = endColumn; i >= ColStart; i--){    counter++;}endRow--;for (int i = endRow; i >= startRow; i--){    counter++;}ColStart++;    We have the pointers to the cell coordinates and the integer values we want to put in them. Now we need to use those to fill in the jagged array. This is where it’s absolutely necessary to use arrays and not lists because we fill the arrays the same way as we’re traversing the matrix. Each array within the temporary array represents a row and when we’re done with the first for-loop we’ll have the following jagged array: 1 2 3 x x x x x x here ‘x’ means “not yet properly filled in”.Then we move on and fill in the rightmost column: 1 2 3 x x x x x x So we need to access the last spots of arrays 2, 3, and 4 of the 2-dimensional array. We could not do the same with a list, there’s no equivalent accessor to a list where we can set the value at position n, we can only add new values to the end of the list. So this is the main reason why it’s better to go for an array.Next up we’ll fill up the bottom row: 1 2 3 x x x x x x …and finally the left-hand column: 1 2 3 12 x x 11 x x We then do the same in the “inner” matrix: 1 2 3 12 13 14 11 16 15 Now we need to select the correct indexes for the jagged array in the 4 for-loops.In the first loop, we move from left to right, i.e. from (0, 0) to (0, 3). So the row is constant, it will be equal to startRow. The column increases from 0 to endColumn. Hence we’ll use the following accessors: 1 temporaryArray[startRow][i] In the second for loop, we move from (1, 3) to (3, 3). The column is equal to the end column and the row moves up from startRow to and including endRow. Recall that we incremented startRow by 1 after the first loop. Therefore we won’t accidentally overwrite cells (0, 3). Here’s the accessor in the second loop: 1 temporaryArray[i][endColumn] In the third loop, we move to the left from (3, 2) to (3, 0) and in the fourth loop upwards from (2, 0) to (1, 0).Here’s the full solution:using System;using System.Collections.Generic;using System.Linq;using System.Runtime.Remoting.Metadata.W3cXsd2001;using System.Text;using System.Threading.Tasks;namespace circularMatrixInCSharp{    public class circularMatrix    {        public List<List<int>> circularMatrixFun(int n)        {            var output = new List<List<int>>();            var tempArray = new int[n][];            for (int i = 0; i < n; i++)            {                tempArray[i] = new int[n];                output.Add(new List<int>());            }            int counter = 1;            int Colstart = 0;            int colEnd = n - 1;            int startRow = 0;            int endRow = n - 1;            while (Colstart <= colEnd && startRow <= endRow)            {                for (int i = Colstart; i <= colEnd; i++)                {                    tempArray[startRow][i] = counter;                    counter++;                }                startRow++;                for (int i = startRow; i <= endRow; i++)                {                    tempArray[i][colEnd] = counter;                    counter++;                }                colEnd--;                for (int i = colEnd; i >= Colstart; i--)                {                    tempArray[endRow][i] = counter;                    counter++;                }                endRow--;                for (int i = endRow; i >= startRow; i--)                {                    tempArray[i][Colstart] = counter;                    counter++;                }                Colstart++;            }            for (int i = 0; i < tempArray.Length; i++)            {                output[i].AddRange(tempArray[i]);            }            return output;        }    }}I think that was quite a challenging problem. Make sure you study it before your next job interview so that you can quickly present your solution and move on to the next problem.Reference:- https://dotnetcodr.com/2020/10/21/how-to-build-a-circular-matrix-with-c- net/

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