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fill out a spiral matrix in javascript
Genetic Drift simulatorJavaScript match scheduling (fixture generation) algorithmClimbing the Leaderboard: HackerranK, Terminated due to timeoutFrequency Queries hackerrankCount duplicates in a JavaScript arrayLeetcode Mountain ArrayTwo-sum solution in JavaScriptPrint out N by N Spiral Matrix in javascriptDeepest pit of an arrayJavaScript Spiral Matrix Coding Challenge
$begingroup$
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
$endgroup$
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
$endgroup$
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
$endgroup$
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
asked 7 mins ago
NinjaGNinjaG
817632
817632
add a comment |
add a comment |
0
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