Fill out a spiral matrix












1












$begingroup$


I got this question during my interview:




Given an integer N, output an N x N spiral matrix with integers 1 through N.



Examples: Input: 3



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



Input: 1



output: matrix filled out as a spiral [[1]]




/*PSEUDO:

rowMin = 0

rowMax = n - 1

colMin = 0

colMax = n - 1

counter = 1

matrix = 

create the matrix:

loop from 0 to n - 1

  array 

  loop from 0 to n-1

    push 0 into array

while rowMin <= rowMax and colMin <= colMax

  loop on rowMin from colMin to colMax.  col

    matrix[rowMin][col] becomes counter++

  rowMin++

  loop on colMax and from rowMin to rowMax.  row

    matrix[row][colMax] becomes counter++

  colMax--

  loop on rowMax from colMax to colMin.  col

    matrix[rowMax][col] becomes counter++

  rowMax--

  loop on colMin from rowMax to rowMin.  row

    matrix[row][colMin] becomes counter++

  colMin++

return matrix

*/





const spiralMatrix = (n) => {

  const matrix = ;

  let rowMin = 0,

    rowMax = n - 1,

    colMin = 0,

    colMax = n - 1,

    counter = 1;



  for (let i = 0; i < n; i++) {

    matrix.push(new Array(n).fill(0));

  }



  while (rowMin <= rowMax && colMin <= colMax) {

    for (let col = colMin; col <= colMax; col++) {

      matrix[rowMin][col] = counter++;

    }

    rowMin++;

    for (let row = rowMin; row <= rowMax; row++) {

      matrix[row][colMax] = counter++;

    }

    colMax--;

    for (let col = colMax; col >= colMin; col--) {

      matrix[rowMax][col] = counter++;

    }

    rowMax--;

    for (let row = rowMax; row >= rowMin; row--) {

      matrix[row][colMin] = counter++;

    }

    colMin++;

  }



  return matrix;

}



console.log(spiralMatrix(10));





javascript







share|improve this question











$endgroup$












  • $begingroup$
    it would be interesting to see this problem solved without random access. Left to right, top to bottom, in the same order you'd output to a terminal.
    $endgroup$
    – Oh My Goodness
    18 hours ago
















1












$begingroup$


I got this question during my interview:




Given an integer N, output an N x N spiral matrix with integers 1 through N.



Examples: Input: 3



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



Input: 1



output: matrix filled out as a spiral [[1]]




/*PSEUDO:

rowMin = 0

rowMax = n - 1

colMin = 0

colMax = n - 1

counter = 1

matrix = 

create the matrix:

loop from 0 to n - 1

  array 

  loop from 0 to n-1

    push 0 into array

while rowMin <= rowMax and colMin <= colMax

  loop on rowMin from colMin to colMax.  col

    matrix[rowMin][col] becomes counter++

  rowMin++

  loop on colMax and from rowMin to rowMax.  row

    matrix[row][colMax] becomes counter++

  colMax--

  loop on rowMax from colMax to colMin.  col

    matrix[rowMax][col] becomes counter++

  rowMax--

  loop on colMin from rowMax to rowMin.  row

    matrix[row][colMin] becomes counter++

  colMin++

return matrix

*/





const spiralMatrix = (n) => {

  const matrix = ;

  let rowMin = 0,

    rowMax = n - 1,

    colMin = 0,

    colMax = n - 1,

    counter = 1;



  for (let i = 0; i < n; i++) {

    matrix.push(new Array(n).fill(0));

  }



  while (rowMin <= rowMax && colMin <= colMax) {

    for (let col = colMin; col <= colMax; col++) {

      matrix[rowMin][col] = counter++;

    }

    rowMin++;

    for (let row = rowMin; row <= rowMax; row++) {

      matrix[row][colMax] = counter++;

    }

    colMax--;

    for (let col = colMax; col >= colMin; col--) {

      matrix[rowMax][col] = counter++;

    }

    rowMax--;

    for (let row = rowMax; row >= rowMin; row--) {

      matrix[row][colMin] = counter++;

    }

    colMin++;

  }



  return matrix;

}



console.log(spiralMatrix(10));





javascript







share|improve this question











$endgroup$












  • $begingroup$
    it would be interesting to see this problem solved without random access. Left to right, top to bottom, in the same order you'd output to a terminal.
    $endgroup$
    – Oh My Goodness
    18 hours ago














1












1








1





$begingroup$


I got this question during my interview:




Given an integer N, output an N x N spiral matrix with integers 1 through N.



Examples: Input: 3



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



Input: 1



output: matrix filled out as a spiral [[1]]




/*PSEUDO:

rowMin = 0

rowMax = n - 1

colMin = 0

colMax = n - 1

counter = 1

matrix = 

create the matrix:

loop from 0 to n - 1

  array 

  loop from 0 to n-1

    push 0 into array

while rowMin <= rowMax and colMin <= colMax

  loop on rowMin from colMin to colMax.  col

    matrix[rowMin][col] becomes counter++

  rowMin++

  loop on colMax and from rowMin to rowMax.  row

    matrix[row][colMax] becomes counter++

  colMax--

  loop on rowMax from colMax to colMin.  col

    matrix[rowMax][col] becomes counter++

  rowMax--

  loop on colMin from rowMax to rowMin.  row

    matrix[row][colMin] becomes counter++

  colMin++

return matrix

*/





const spiralMatrix = (n) => {

  const matrix = ;

  let rowMin = 0,

    rowMax = n - 1,

    colMin = 0,

    colMax = n - 1,

    counter = 1;



  for (let i = 0; i < n; i++) {

    matrix.push(new Array(n).fill(0));

  }



  while (rowMin <= rowMax && colMin <= colMax) {

    for (let col = colMin; col <= colMax; col++) {

      matrix[rowMin][col] = counter++;

    }

    rowMin++;

    for (let row = rowMin; row <= rowMax; row++) {

      matrix[row][colMax] = counter++;

    }

    colMax--;

    for (let col = colMax; col >= colMin; col--) {

      matrix[rowMax][col] = counter++;

    }

    rowMax--;

    for (let row = rowMax; row >= rowMin; row--) {

      matrix[row][colMin] = counter++;

    }

    colMin++;

  }



  return matrix;

}



console.log(spiralMatrix(10));





javascript







share|improve this question











$endgroup$




I got this question during my interview:




Given an integer N, output an N x N spiral matrix with integers 1 through N.



Examples: Input: 3



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



Input: 1



output: matrix filled out as a spiral [[1]]




/*PSEUDO:

rowMin = 0

rowMax = n - 1

colMin = 0

colMax = n - 1

counter = 1

matrix = 

create the matrix:

loop from 0 to n - 1

  array 

  loop from 0 to n-1

    push 0 into array

while rowMin <= rowMax and colMin <= colMax

  loop on rowMin from colMin to colMax.  col

    matrix[rowMin][col] becomes counter++

  rowMin++

  loop on colMax and from rowMin to rowMax.  row

    matrix[row][colMax] becomes counter++

  colMax--

  loop on rowMax from colMax to colMin.  col

    matrix[rowMax][col] becomes counter++

  rowMax--

  loop on colMin from rowMax to rowMin.  row

    matrix[row][colMin] becomes counter++

  colMin++

return matrix

*/





const spiralMatrix = (n) => {

  const matrix = ;

  let rowMin = 0,

    rowMax = n - 1,

    colMin = 0,

    colMax = n - 1,

    counter = 1;



  for (let i = 0; i < n; i++) {

    matrix.push(new Array(n).fill(0));

  }



  while (rowMin <= rowMax && colMin <= colMax) {

    for (let col = colMin; col <= colMax; col++) {

      matrix[rowMin][col] = counter++;

    }

    rowMin++;

    for (let row = rowMin; row <= rowMax; row++) {

      matrix[row][colMax] = counter++;

    }

    colMax--;

    for (let col = colMax; col >= colMin; col--) {

      matrix[rowMax][col] = counter++;

    }

    rowMax--;

    for (let row = rowMax; row >= rowMin; row--) {

      matrix[row][colMin] = counter++;

    }

    colMin++;

  }



  return matrix;

}



console.log(spiralMatrix(10));





javascript




javascript






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 11 mins ago









Jamal♦

30.4k11121227




30.4k11121227










asked 22 hours ago









NinjaGNinjaG

825632




825632












  • $begingroup$
    it would be interesting to see this problem solved without random access. Left to right, top to bottom, in the same order you'd output to a terminal.
    $endgroup$
    – Oh My Goodness
    18 hours ago


















  • $begingroup$
    it would be interesting to see this problem solved without random access. Left to right, top to bottom, in the same order you'd output to a terminal.
    $endgroup$
    – Oh My Goodness
    18 hours ago
















$begingroup$
it would be interesting to see this problem solved without random access. Left to right, top to bottom, in the same order you'd output to a terminal.
$endgroup$
– Oh My Goodness
18 hours ago




$begingroup$
it would be interesting to see this problem solved without random access. Left to right, top to bottom, in the same order you'd output to a terminal.
$endgroup$
– Oh My Goodness
18 hours ago










1 Answer
1






active

oldest

votes


















2












$begingroup$

Seems mostly solid to me. I just have two small remarks:





const spiralMatrix = (n) => {



I'm not a big fan of using the arrow syntax here. The arrow syntax is (mostly) used for inline functions. For a "top level" function I'd prefer a normal, more readable (and hoisted) function declaration:



function spiralMatrix(n) {


The only disadvantage I see is that const prevents accidental overwriting, which I don't see as an serious problem in this case.




And .fill(0) during initialization of the arrays is unnecessary.






share|improve this answer









$endgroup$













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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    2












    $begingroup$

    Seems mostly solid to me. I just have two small remarks:





    const spiralMatrix = (n) => {



    I'm not a big fan of using the arrow syntax here. The arrow syntax is (mostly) used for inline functions. For a "top level" function I'd prefer a normal, more readable (and hoisted) function declaration:



    function spiralMatrix(n) {


    The only disadvantage I see is that const prevents accidental overwriting, which I don't see as an serious problem in this case.




    And .fill(0) during initialization of the arrays is unnecessary.






    share|improve this answer









    $endgroup$


















      2












      $begingroup$

      Seems mostly solid to me. I just have two small remarks:





      const spiralMatrix = (n) => {



      I'm not a big fan of using the arrow syntax here. The arrow syntax is (mostly) used for inline functions. For a "top level" function I'd prefer a normal, more readable (and hoisted) function declaration:



      function spiralMatrix(n) {


      The only disadvantage I see is that const prevents accidental overwriting, which I don't see as an serious problem in this case.




      And .fill(0) during initialization of the arrays is unnecessary.






      share|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Seems mostly solid to me. I just have two small remarks:





        const spiralMatrix = (n) => {



        I'm not a big fan of using the arrow syntax here. The arrow syntax is (mostly) used for inline functions. For a "top level" function I'd prefer a normal, more readable (and hoisted) function declaration:



        function spiralMatrix(n) {


        The only disadvantage I see is that const prevents accidental overwriting, which I don't see as an serious problem in this case.




        And .fill(0) during initialization of the arrays is unnecessary.






        share|improve this answer









        $endgroup$



        Seems mostly solid to me. I just have two small remarks:





        const spiralMatrix = (n) => {



        I'm not a big fan of using the arrow syntax here. The arrow syntax is (mostly) used for inline functions. For a "top level" function I'd prefer a normal, more readable (and hoisted) function declaration:



        function spiralMatrix(n) {


        The only disadvantage I see is that const prevents accidental overwriting, which I don't see as an serious problem in this case.




        And .fill(0) during initialization of the arrays is unnecessary.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 15 hours ago









        RoToRaRoToRa

        6,3981336




        6,3981336






























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