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fill out a spiral matrix in javascript

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0

$begingroup$

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

asked 16 mins ago

NinjaGNinjaG

817632

$endgroup$

add a comment | 

0

$begingroup$

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

asked 16 mins ago

NinjaGNinjaG

817632

$endgroup$

add a comment | 

$begingroup$

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

asked 16 mins ago

NinjaGNinjaG

817632

$endgroup$

/*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));

share|improve this question

asked 16 mins ago

NinjaGNinjaG

817632

share|improve this question

asked 16 mins ago

NinjaGNinjaG

817632

asked 16 mins ago

NinjaGNinjaG

817632

asked 16 mins ago

NinjaGNinjaG

817632