https://github.com/dnyaneshwargiri/matrixoperations

https://github.com/dnyaneshwargiri/matrixoperations

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• Host: GitHub
• URL: https://github.com/dnyaneshwargiri/matrixoperations
• Owner: dnyaneshwargiri
• Created: 2022-09-23T05:46:02.000Z (over 2 years ago)
• Default Branch: main
• Last Pushed: 2022-09-24T13:32:50.000Z (over 2 years ago)
• Last Synced: 2023-03-06T08:32:41.047Z (almost 2 years ago)
• Language: JavaScript
• Size: 1.86 MB
• Stars: 0
• Watchers: 1
• Forks: 0
• Open Issues: 0
• Metadata Files:
  • Readme: README.md

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README

        
# matrixOperations

**To run**

node cli.js inputFileName.csv > outputFile.csv

OR

node cli.js inputFileName.csv

**Test cases-**

Run test cases using

npm run test

1. Square matrix 

		Input=

		

		id,json

		1,"[1, 2, 3, 4, 5, 6, 7, 8, 9]"

		

		 Workflow=

		 i] Make matrix from list.

		  

			input matrix=	

						[[1, 2, 3],

						[ 4, 5, 6], 

						[7, 8, 9]]

		

		ii] Rotate clockwise by 1 step.

		

			rotated matrix=	

						[[4, 1, 2],

						[ 7, 5, 3], 

						[8, 9, 6]]

		

		iii] Matrix to list conversion= [4,1,2,7,5,3,8,9,6]

		  Expected Output=

		

		  1,"[4,1,2,7,5,3,8,9,6]",true

		

 

2. Rectangular matrix=

3. Invalid matrix=

		  Input=

		  8,"[1, 1, 1, 1, 1]"

		  Workflow=

		  1. Check for possible matrix formation- couldn’t form matrix of length 5 for which result is 1 column  & 5 rows

		  2. For single column matrix can’t perform edge rotation as edge has length=1

		  Output=

		  8,"[]", false

**Rotation Algorithm**

1. Start

2. let top=0, bottom=rows-1

3. Traverse the matrix in spiral order that is

   left 	=>		right         

   ||				 ||

   left 	<=		bottom

   and store it in list namely A.

4. Set k=1 as we have to rotate by 1 step

5. Do

   if length(A) <=k break

   let index= length(A)-k   

   

   Start assigning values to matrix from list A in below order

   i. left => right, for Each element do {update index++; index=index% length(A)}

   ii. right => bottom, for Each element do {update index++; index=index% length(A)}

   iii. bottom => top, for Each element do {update index++; index=index% length(A)}

   update below parameter as 

   top++

   bottom--

   left++

   right--

6. while top <= bottom repeat step 3 to 5 else break.

    

**Performance handling-**

Used readStream to process each row as below.



fs.createReadStream(__dirname+inputFile).pipe(csvParser).on('data', (element) => {

	//rotate each table

	stream.write(rotateTable(element.json,Number(element.id)))

})



**Time Complexity=**

The time complexity of the above algorithm is O(N * M), where N and M are the dimensions of the input matrix. It is because we are looping through the matrix once.

**Space Complexity=**

The Space Complexity of the above approach is O(N + M), where N and M are the dimensions of the input matrix. It is because we are creating a array to store the elements of the rings and the maximum possible size of a ring < 2 * (N + M).

	

**Data Cases-**

1. Square matrix- checking if square root of length of list is whole positive number

2. Rectangular matrix- Calculating various possibilities of matrix can be formed from given length 'l' 

   

**Error Cases-**

1. No file provided- handled at beginning of program by checking number of process arguments

2. No file present- 'ENOENT: no such file or directory' error throw

3. Matrix not possible- output= id,[], false