Fill out a spiral matrix
$begingroup$
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
edited 11 mins ago
Jamal♦
30.4k11121227
asked 22 hours ago
NinjaGNinjaG
825632
$endgroup$
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
edited 11 mins ago
Jamal♦
30.4k11121227
asked 22 hours ago
NinjaGNinjaG
825632
$endgroup$
$begingroup$
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
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
edited 11 mins ago
Jamal♦
30.4k11121227
asked 22 hours ago
NinjaGNinjaG
825632
$endgroup$
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
edited 11 mins ago
Jamal♦
30.4k11121227
asked 22 hours ago
NinjaGNinjaG
825632
edited 11 mins ago
Jamal♦
30.4k11121227
asked 22 hours ago
NinjaGNinjaG
825632
edited 11 mins ago
Jamal♦
30.4k11121227
edited 11 mins ago
Jamal♦
30.4k11121227
edited 11 mins ago
Jamal♦
30.4k11121227
30.4k11121227
asked 22 hours ago
NinjaGNinjaG
825632
asked 22 hours ago
NinjaGNinjaG
825632
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered 15 hours ago
RoToRaRoToRa
6,3981336
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 15 hours ago
RoToRaRoToRa
6,3981336
$endgroup$
add a comment |
$begingroup$
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.
answered 15 hours ago
RoToRaRoToRa
6,3981336
$endgroup$
add a comment |
$begingroup$
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.
answered 15 hours ago
RoToRaRoToRa
6,3981336
$endgroup$
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.
answered 15 hours ago
RoToRaRoToRa
6,3981336
answered 15 hours ago
RoToRaRoToRa
6,3981336
answered 15 hours ago
RoToRaRoToRa
6,3981336
answered 15 hours ago
RoToRaRoToRa
6,3981336
6,3981336
add a comment |
add a comment |
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$begingroup$
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