Rotate an N × N matrix 90 degrees clockwise
Given the problem from "Cracking the coding interview"
Given an image represented by an NxN matrix, where each pixel in the
image is 4 bytes, write a method to rotate the image by 90 degrees clockwise.
For simplicity I assumed a matrix of strings for my example so
[["a", "a", "a", "a"],
["b", "b", "b", "b"],
["c", "c", "c", "c"],
["d", "d", "d", "d"]]
becomes
[["d", "c", "b", "a"],
["d", "c", "b", "a"],
["d", "c", "b", "a"],
["d", "c", "b", "a"]]
Here's my JavaScript solution
var rotate = function(matrix) {
// Copy the original matrix
var origMatrix = matrix.slice();
for(var i=0; i < matrix.length; i++) {
// Map each row entry to its rotated value
var row = matrix[i].map(function(x, j) {
var k = (matrix.length - 1) - j;
return origMatrix[k][i];
});
matrix[i] = row;
}
return matrix;
};
javascript matrix
add a comment |
Given the problem from "Cracking the coding interview"
Given an image represented by an NxN matrix, where each pixel in the
image is 4 bytes, write a method to rotate the image by 90 degrees clockwise.
For simplicity I assumed a matrix of strings for my example so
[["a", "a", "a", "a"],
["b", "b", "b", "b"],
["c", "c", "c", "c"],
["d", "d", "d", "d"]]
becomes
[["d", "c", "b", "a"],
["d", "c", "b", "a"],
["d", "c", "b", "a"],
["d", "c", "b", "a"]]
Here's my JavaScript solution
var rotate = function(matrix) {
// Copy the original matrix
var origMatrix = matrix.slice();
for(var i=0; i < matrix.length; i++) {
// Map each row entry to its rotated value
var row = matrix[i].map(function(x, j) {
var k = (matrix.length - 1) - j;
return origMatrix[k][i];
});
matrix[i] = row;
}
return matrix;
};
javascript matrix
add a comment |
Given the problem from "Cracking the coding interview"
Given an image represented by an NxN matrix, where each pixel in the
image is 4 bytes, write a method to rotate the image by 90 degrees clockwise.
For simplicity I assumed a matrix of strings for my example so
[["a", "a", "a", "a"],
["b", "b", "b", "b"],
["c", "c", "c", "c"],
["d", "d", "d", "d"]]
becomes
[["d", "c", "b", "a"],
["d", "c", "b", "a"],
["d", "c", "b", "a"],
["d", "c", "b", "a"]]
Here's my JavaScript solution
var rotate = function(matrix) {
// Copy the original matrix
var origMatrix = matrix.slice();
for(var i=0; i < matrix.length; i++) {
// Map each row entry to its rotated value
var row = matrix[i].map(function(x, j) {
var k = (matrix.length - 1) - j;
return origMatrix[k][i];
});
matrix[i] = row;
}
return matrix;
};
javascript matrix
Given the problem from "Cracking the coding interview"
Given an image represented by an NxN matrix, where each pixel in the
image is 4 bytes, write a method to rotate the image by 90 degrees clockwise.
For simplicity I assumed a matrix of strings for my example so
[["a", "a", "a", "a"],
["b", "b", "b", "b"],
["c", "c", "c", "c"],
["d", "d", "d", "d"]]
becomes
[["d", "c", "b", "a"],
["d", "c", "b", "a"],
["d", "c", "b", "a"],
["d", "c", "b", "a"]]
Here's my JavaScript solution
var rotate = function(matrix) {
// Copy the original matrix
var origMatrix = matrix.slice();
for(var i=0; i < matrix.length; i++) {
// Map each row entry to its rotated value
var row = matrix[i].map(function(x, j) {
var k = (matrix.length - 1) - j;
return origMatrix[k][i];
});
matrix[i] = row;
}
return matrix;
};
javascript matrix
javascript matrix
edited Feb 5 '18 at 12:39
200_success
128k15152414
128k15152414
asked Feb 5 '18 at 11:11
PDStatPDStat
210310
210310
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
Are you up to date?
There is really not much to the problem and there are a variety of solutions.
It is also hard to know what the interviews are looking for. For some it is knowledge of latest language features, for others it may be judged on efficiency.
These very simple tests are generally just to see if you can actually write code. You would be amazed at how many people apply for jobs they are not able to do.
Assessing the code.
Looking at your code as a conservative HR interested in your code writing style, and up to date language knowledge.
It is a pass, you can code, you have the job but will have you in supervision for a while..You also need to catch up on the language as you are not using ES6 to the full.
General points
- You need to use const for constants, and let for block scope variables.
- You have not used any arrow functions.
- I would question why you created the function using an expression, rather than a function statement, to make sure that you know the difference. (You should play it safe an use function statements)
- Also maybe a little to much code, but not really an issue.
- Code logic.. i would ask why you chose to duplicate the array rather than just create a new one as you went.
A quick rewrite
I am assuming the rotate is in place (original array) with new rows
function rotate(matrix) { // function statement
const N = matrix.length - 1; // use a constant
// use arrow functions and nested map;
const result = matrix.map((row, i) =>
row.map((val, j) => matrix[N - j][i])
);
matrix.length = 0; // hold original array reference
matrix.push(...result); // Spread operator
return matrix;
}
Some extras
The problem states "image" and "pixel" which give a hint that this may be judged on performance. They would tell you if it was. But if you get a similar challenge and performance is important then its best to use good old for
loops, avoiding any iterators that use callbacks.
Also a performance oriented function would swap in place rather than create new arrays. If you had to do that realtime with images for modern displays your code would be making big GC hits (Garbage Collection).
An alternative
Another way it can be done by rotating 4 pixels at a time can reduce memory overheads and processing time. Destructing lets you swap 4 corners in one go without the need to create temp variables.
I would never recommend you submit the following unless there was a clear directive for performance, and thinking outside the box.
// Commented version
// N*N is square
// pixels are rotated inplace
// 4 by 4 rotated in 4 iterations
// 5 by 5 rotated in 7 iterations
function rotatePixels(image) {
var x, y, x1, y1, edge;
// Solve by swapping 4 at a time in rings from the outside in
const N = image.length; // size of array
const N1 = N - 1; // position of end item
const N2 = N / 2; // Half way position
// x,y hold the a cell coordinate
x = y = 0;
// x,y hold the diagonally opposite cell
x1 = y1 = N1;
// length of current edge
edge = N1;
// while not at the center
while (y < N2) {
// for each item on the current edge
while (x < edge) { // rotate points at outer edge
// swap 4 corner items
// using array destructed assignment
[
image[x ][y1],
image[y1][x1],
image[x1][N1 - y1],
image[y ][x ]
] = [
image[y ][x ],
image[x ][y1],
image[y1][x1],
image[x1][N1 - y1]
];
x += 1; // move top pos forward one
x1 -= 1; // move bottom pos back one
}
y += 1; // diagonally into array
x = y; // x same as y
y1 = x1 = N1-x; // and diagonal opposite
edge -= 1; // shorten the end
}
return image;
}
// How I would present it
function rotatePixels(image) {
var x, y, x1, y1, edge;
const N = image.length;
const N1 = N - 1;
const N2 = N / 2;
x = y = 0;
edge = x1 = y1 = N1;
while (y < N2) {
while (x < edge) {
[image[x][y1], image[y1][x1], image[x1][N1-y1], image[y][x]] =
[image[y][x] , image[x ][y1], image[y1][x1] , image[x1][N1-y1]];
x += 1;
x1 -= 1;
}
x = y += 1;
y1 = x1 = N1-x;
edge -= 1;
}
return image;
}
// At time of writing I was unsure as to the performance of the swap using destructuring
// Turns out it is very bad
// The next version is a more conventional swap and runs 3 time faster than the above version
// and 8 times faster than the conventional solution at top of answer
function rotatePixels(image) {
var x, y, x1, y1, edge;
const N = image.length;
const N1 = N - 1;
const N2 = N / 2;
x = y = 0;
edge = x1 = y1 = N1;
while (y < N2) {
while (x < edge) {
const a = image[y][x];
image[y][x] = image[x1][N1-y1];
image[x1][N1-y1] = image[y1][x1];
image[y1][x1] = image[x][y1];
image[x][y1] = a;
x += 1;
x1 -= 1;
}
x = y += 1;
y1 = x1 = N1-x;
edge -= 1;
}
return image;
}
Thanks that's interesting and I had been wondering how to do an in place version. As for ES6 I've been intentionally ignoring it for now as my work is based around IE11 use.
– PDStat
Feb 5 '18 at 15:42
3
@PaulStatham Don't let MS's IE11 pull you down. It will be until after 2020 when MS stops support for win7, that IE11 will go away and it will never support ECMAScript6+. You can use transpilers such as Babel to provide legacy support and have your code base kept up to date, and most important your skill set on the consumer's leading edge.
– Blindman67
Feb 6 '18 at 11:38
add a comment |
Cloning the matrix
.slice
is a good way to copy the values of an array, instead of getting a reference to the same array. Unfortunately all your values are themselves arrays, which you are still using a reference to and not a copy. In this code you are not changing individual values, but setting entire rows at a time, so it still works, but this could easily lead to unexpected behavior if you are not careful.
You don't mention if you have to modify the original matrix or return a new one (you are doing both). But if you leave the original unmodified, you don't even need to make a copy. Just start out with empty array and .push
each new row to it.
Mapping the values
Your .map
is a bit weird. You use it on one array, but then you don't even use the value and just use the index to access a different array. This can also be done simpler. The Nth row in the result is the Nth column in the input. So using .map
on the input and getting the right value from each row will give you the new row (you have to reverse it too).
var row = matrix.map(function(e) {
return e[i]
}).reverse();
ES6
A lot of new features was introduced in ES6, which was finalized back in 2015. You should be using them by now. This means using let
instead of var
, and arrow functions, which are excellent for short inline function like the one above.
The final code
Another small change I made was to change to the loop to loop over the inner array. This way it also works for matrices of different dimensions.
function rotate(matrix) {
let result = ;
for(let i = 0; i < matrix[0].length; i++) {
let row = matrix.map(e => e[i]).reverse();
result.push(row);
}
return result;
};
add a comment |
If performance is important, as mentioned above, it would be good to use for
.
Also it is nice if you show an example of how you would structure a basic class. E.g:
class Matrix {
constructor(n){
this.m = ;
var z = 0;
for (let i = 0; i < n; ++i) {
this.m[i] = ;
for (let j = 0; j < n; ++j) {
this.m[i][j] = z;
z++;
}
}
}
printMatrix() {
console.log(this.m);
}
rotateMatrix() {
const n = this.m[0].length;
let res =
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
if (!res[j])
res[j] =
res[j][i] = this.m[n-1-i][j];
}
}
return res;
}
}
let m = new Matrix(5);
m.printMatrix();
let t0 = performance.now();
console.log(m.rotateMatrix());
let t1 = performance.now();
console.log("Call to rotate took " + (t1 - t0) + " milliseconds.");
It seems to perform 20% faster than using the algorithms above. I am pretty sure this can be done even more efficient.
add a comment |
Matrix N × M
var m = [
[11, 12, 13],
[21, 22, 23],
[31, 32, 33],
[41, 42, 43],
[51, 52, 53],
];
var rn = m.length
,cn = m[0].length
,ARRC = Array.from(Array(cn))
,ARRR = Array.from(Array(rn));
function print_m(m) {
console.log('----------------');
m.forEach(r => console.log(r));
}
//90 clockwise
var m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[rn - 1 - r][c])
);
print_m(m1);
//90 counter clockwise
m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[r][cn - 1 - c])
);
print_m(m1);
//180
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[rn - 1 - r][cn - 1 - c])
);
print_m(m1);
// flip vertical
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[rn - 1 - r][c])
);
print_m(m1);
// flip horizontal
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[r][cn - 1 - c])
);
print_m(m1);
// flip horizontal & 90 counter clockwise
m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[r][c])
);
New contributor
We are looking for answers that provide insightful observations about the code in the question. Answers that consist of independent solutions with no justification do not constitute a code review, and may be removed.
2
You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process.
– Mast
2 days ago
My solution can be apply for nxm matrix, not only nxn, it also solve for other case such as roate 180, flip, counter clockwise. I'm not sure it better than the original about performance or not, but obviously it's concise.
– KevinBui
2 days ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Are you up to date?
There is really not much to the problem and there are a variety of solutions.
It is also hard to know what the interviews are looking for. For some it is knowledge of latest language features, for others it may be judged on efficiency.
These very simple tests are generally just to see if you can actually write code. You would be amazed at how many people apply for jobs they are not able to do.
Assessing the code.
Looking at your code as a conservative HR interested in your code writing style, and up to date language knowledge.
It is a pass, you can code, you have the job but will have you in supervision for a while..You also need to catch up on the language as you are not using ES6 to the full.
General points
- You need to use const for constants, and let for block scope variables.
- You have not used any arrow functions.
- I would question why you created the function using an expression, rather than a function statement, to make sure that you know the difference. (You should play it safe an use function statements)
- Also maybe a little to much code, but not really an issue.
- Code logic.. i would ask why you chose to duplicate the array rather than just create a new one as you went.
A quick rewrite
I am assuming the rotate is in place (original array) with new rows
function rotate(matrix) { // function statement
const N = matrix.length - 1; // use a constant
// use arrow functions and nested map;
const result = matrix.map((row, i) =>
row.map((val, j) => matrix[N - j][i])
);
matrix.length = 0; // hold original array reference
matrix.push(...result); // Spread operator
return matrix;
}
Some extras
The problem states "image" and "pixel" which give a hint that this may be judged on performance. They would tell you if it was. But if you get a similar challenge and performance is important then its best to use good old for
loops, avoiding any iterators that use callbacks.
Also a performance oriented function would swap in place rather than create new arrays. If you had to do that realtime with images for modern displays your code would be making big GC hits (Garbage Collection).
An alternative
Another way it can be done by rotating 4 pixels at a time can reduce memory overheads and processing time. Destructing lets you swap 4 corners in one go without the need to create temp variables.
I would never recommend you submit the following unless there was a clear directive for performance, and thinking outside the box.
// Commented version
// N*N is square
// pixels are rotated inplace
// 4 by 4 rotated in 4 iterations
// 5 by 5 rotated in 7 iterations
function rotatePixels(image) {
var x, y, x1, y1, edge;
// Solve by swapping 4 at a time in rings from the outside in
const N = image.length; // size of array
const N1 = N - 1; // position of end item
const N2 = N / 2; // Half way position
// x,y hold the a cell coordinate
x = y = 0;
// x,y hold the diagonally opposite cell
x1 = y1 = N1;
// length of current edge
edge = N1;
// while not at the center
while (y < N2) {
// for each item on the current edge
while (x < edge) { // rotate points at outer edge
// swap 4 corner items
// using array destructed assignment
[
image[x ][y1],
image[y1][x1],
image[x1][N1 - y1],
image[y ][x ]
] = [
image[y ][x ],
image[x ][y1],
image[y1][x1],
image[x1][N1 - y1]
];
x += 1; // move top pos forward one
x1 -= 1; // move bottom pos back one
}
y += 1; // diagonally into array
x = y; // x same as y
y1 = x1 = N1-x; // and diagonal opposite
edge -= 1; // shorten the end
}
return image;
}
// How I would present it
function rotatePixels(image) {
var x, y, x1, y1, edge;
const N = image.length;
const N1 = N - 1;
const N2 = N / 2;
x = y = 0;
edge = x1 = y1 = N1;
while (y < N2) {
while (x < edge) {
[image[x][y1], image[y1][x1], image[x1][N1-y1], image[y][x]] =
[image[y][x] , image[x ][y1], image[y1][x1] , image[x1][N1-y1]];
x += 1;
x1 -= 1;
}
x = y += 1;
y1 = x1 = N1-x;
edge -= 1;
}
return image;
}
// At time of writing I was unsure as to the performance of the swap using destructuring
// Turns out it is very bad
// The next version is a more conventional swap and runs 3 time faster than the above version
// and 8 times faster than the conventional solution at top of answer
function rotatePixels(image) {
var x, y, x1, y1, edge;
const N = image.length;
const N1 = N - 1;
const N2 = N / 2;
x = y = 0;
edge = x1 = y1 = N1;
while (y < N2) {
while (x < edge) {
const a = image[y][x];
image[y][x] = image[x1][N1-y1];
image[x1][N1-y1] = image[y1][x1];
image[y1][x1] = image[x][y1];
image[x][y1] = a;
x += 1;
x1 -= 1;
}
x = y += 1;
y1 = x1 = N1-x;
edge -= 1;
}
return image;
}
Thanks that's interesting and I had been wondering how to do an in place version. As for ES6 I've been intentionally ignoring it for now as my work is based around IE11 use.
– PDStat
Feb 5 '18 at 15:42
3
@PaulStatham Don't let MS's IE11 pull you down. It will be until after 2020 when MS stops support for win7, that IE11 will go away and it will never support ECMAScript6+. You can use transpilers such as Babel to provide legacy support and have your code base kept up to date, and most important your skill set on the consumer's leading edge.
– Blindman67
Feb 6 '18 at 11:38
add a comment |
Are you up to date?
There is really not much to the problem and there are a variety of solutions.
It is also hard to know what the interviews are looking for. For some it is knowledge of latest language features, for others it may be judged on efficiency.
These very simple tests are generally just to see if you can actually write code. You would be amazed at how many people apply for jobs they are not able to do.
Assessing the code.
Looking at your code as a conservative HR interested in your code writing style, and up to date language knowledge.
It is a pass, you can code, you have the job but will have you in supervision for a while..You also need to catch up on the language as you are not using ES6 to the full.
General points
- You need to use const for constants, and let for block scope variables.
- You have not used any arrow functions.
- I would question why you created the function using an expression, rather than a function statement, to make sure that you know the difference. (You should play it safe an use function statements)
- Also maybe a little to much code, but not really an issue.
- Code logic.. i would ask why you chose to duplicate the array rather than just create a new one as you went.
A quick rewrite
I am assuming the rotate is in place (original array) with new rows
function rotate(matrix) { // function statement
const N = matrix.length - 1; // use a constant
// use arrow functions and nested map;
const result = matrix.map((row, i) =>
row.map((val, j) => matrix[N - j][i])
);
matrix.length = 0; // hold original array reference
matrix.push(...result); // Spread operator
return matrix;
}
Some extras
The problem states "image" and "pixel" which give a hint that this may be judged on performance. They would tell you if it was. But if you get a similar challenge and performance is important then its best to use good old for
loops, avoiding any iterators that use callbacks.
Also a performance oriented function would swap in place rather than create new arrays. If you had to do that realtime with images for modern displays your code would be making big GC hits (Garbage Collection).
An alternative
Another way it can be done by rotating 4 pixels at a time can reduce memory overheads and processing time. Destructing lets you swap 4 corners in one go without the need to create temp variables.
I would never recommend you submit the following unless there was a clear directive for performance, and thinking outside the box.
// Commented version
// N*N is square
// pixels are rotated inplace
// 4 by 4 rotated in 4 iterations
// 5 by 5 rotated in 7 iterations
function rotatePixels(image) {
var x, y, x1, y1, edge;
// Solve by swapping 4 at a time in rings from the outside in
const N = image.length; // size of array
const N1 = N - 1; // position of end item
const N2 = N / 2; // Half way position
// x,y hold the a cell coordinate
x = y = 0;
// x,y hold the diagonally opposite cell
x1 = y1 = N1;
// length of current edge
edge = N1;
// while not at the center
while (y < N2) {
// for each item on the current edge
while (x < edge) { // rotate points at outer edge
// swap 4 corner items
// using array destructed assignment
[
image[x ][y1],
image[y1][x1],
image[x1][N1 - y1],
image[y ][x ]
] = [
image[y ][x ],
image[x ][y1],
image[y1][x1],
image[x1][N1 - y1]
];
x += 1; // move top pos forward one
x1 -= 1; // move bottom pos back one
}
y += 1; // diagonally into array
x = y; // x same as y
y1 = x1 = N1-x; // and diagonal opposite
edge -= 1; // shorten the end
}
return image;
}
// How I would present it
function rotatePixels(image) {
var x, y, x1, y1, edge;
const N = image.length;
const N1 = N - 1;
const N2 = N / 2;
x = y = 0;
edge = x1 = y1 = N1;
while (y < N2) {
while (x < edge) {
[image[x][y1], image[y1][x1], image[x1][N1-y1], image[y][x]] =
[image[y][x] , image[x ][y1], image[y1][x1] , image[x1][N1-y1]];
x += 1;
x1 -= 1;
}
x = y += 1;
y1 = x1 = N1-x;
edge -= 1;
}
return image;
}
// At time of writing I was unsure as to the performance of the swap using destructuring
// Turns out it is very bad
// The next version is a more conventional swap and runs 3 time faster than the above version
// and 8 times faster than the conventional solution at top of answer
function rotatePixels(image) {
var x, y, x1, y1, edge;
const N = image.length;
const N1 = N - 1;
const N2 = N / 2;
x = y = 0;
edge = x1 = y1 = N1;
while (y < N2) {
while (x < edge) {
const a = image[y][x];
image[y][x] = image[x1][N1-y1];
image[x1][N1-y1] = image[y1][x1];
image[y1][x1] = image[x][y1];
image[x][y1] = a;
x += 1;
x1 -= 1;
}
x = y += 1;
y1 = x1 = N1-x;
edge -= 1;
}
return image;
}
Thanks that's interesting and I had been wondering how to do an in place version. As for ES6 I've been intentionally ignoring it for now as my work is based around IE11 use.
– PDStat
Feb 5 '18 at 15:42
3
@PaulStatham Don't let MS's IE11 pull you down. It will be until after 2020 when MS stops support for win7, that IE11 will go away and it will never support ECMAScript6+. You can use transpilers such as Babel to provide legacy support and have your code base kept up to date, and most important your skill set on the consumer's leading edge.
– Blindman67
Feb 6 '18 at 11:38
add a comment |
Are you up to date?
There is really not much to the problem and there are a variety of solutions.
It is also hard to know what the interviews are looking for. For some it is knowledge of latest language features, for others it may be judged on efficiency.
These very simple tests are generally just to see if you can actually write code. You would be amazed at how many people apply for jobs they are not able to do.
Assessing the code.
Looking at your code as a conservative HR interested in your code writing style, and up to date language knowledge.
It is a pass, you can code, you have the job but will have you in supervision for a while..You also need to catch up on the language as you are not using ES6 to the full.
General points
- You need to use const for constants, and let for block scope variables.
- You have not used any arrow functions.
- I would question why you created the function using an expression, rather than a function statement, to make sure that you know the difference. (You should play it safe an use function statements)
- Also maybe a little to much code, but not really an issue.
- Code logic.. i would ask why you chose to duplicate the array rather than just create a new one as you went.
A quick rewrite
I am assuming the rotate is in place (original array) with new rows
function rotate(matrix) { // function statement
const N = matrix.length - 1; // use a constant
// use arrow functions and nested map;
const result = matrix.map((row, i) =>
row.map((val, j) => matrix[N - j][i])
);
matrix.length = 0; // hold original array reference
matrix.push(...result); // Spread operator
return matrix;
}
Some extras
The problem states "image" and "pixel" which give a hint that this may be judged on performance. They would tell you if it was. But if you get a similar challenge and performance is important then its best to use good old for
loops, avoiding any iterators that use callbacks.
Also a performance oriented function would swap in place rather than create new arrays. If you had to do that realtime with images for modern displays your code would be making big GC hits (Garbage Collection).
An alternative
Another way it can be done by rotating 4 pixels at a time can reduce memory overheads and processing time. Destructing lets you swap 4 corners in one go without the need to create temp variables.
I would never recommend you submit the following unless there was a clear directive for performance, and thinking outside the box.
// Commented version
// N*N is square
// pixels are rotated inplace
// 4 by 4 rotated in 4 iterations
// 5 by 5 rotated in 7 iterations
function rotatePixels(image) {
var x, y, x1, y1, edge;
// Solve by swapping 4 at a time in rings from the outside in
const N = image.length; // size of array
const N1 = N - 1; // position of end item
const N2 = N / 2; // Half way position
// x,y hold the a cell coordinate
x = y = 0;
// x,y hold the diagonally opposite cell
x1 = y1 = N1;
// length of current edge
edge = N1;
// while not at the center
while (y < N2) {
// for each item on the current edge
while (x < edge) { // rotate points at outer edge
// swap 4 corner items
// using array destructed assignment
[
image[x ][y1],
image[y1][x1],
image[x1][N1 - y1],
image[y ][x ]
] = [
image[y ][x ],
image[x ][y1],
image[y1][x1],
image[x1][N1 - y1]
];
x += 1; // move top pos forward one
x1 -= 1; // move bottom pos back one
}
y += 1; // diagonally into array
x = y; // x same as y
y1 = x1 = N1-x; // and diagonal opposite
edge -= 1; // shorten the end
}
return image;
}
// How I would present it
function rotatePixels(image) {
var x, y, x1, y1, edge;
const N = image.length;
const N1 = N - 1;
const N2 = N / 2;
x = y = 0;
edge = x1 = y1 = N1;
while (y < N2) {
while (x < edge) {
[image[x][y1], image[y1][x1], image[x1][N1-y1], image[y][x]] =
[image[y][x] , image[x ][y1], image[y1][x1] , image[x1][N1-y1]];
x += 1;
x1 -= 1;
}
x = y += 1;
y1 = x1 = N1-x;
edge -= 1;
}
return image;
}
// At time of writing I was unsure as to the performance of the swap using destructuring
// Turns out it is very bad
// The next version is a more conventional swap and runs 3 time faster than the above version
// and 8 times faster than the conventional solution at top of answer
function rotatePixels(image) {
var x, y, x1, y1, edge;
const N = image.length;
const N1 = N - 1;
const N2 = N / 2;
x = y = 0;
edge = x1 = y1 = N1;
while (y < N2) {
while (x < edge) {
const a = image[y][x];
image[y][x] = image[x1][N1-y1];
image[x1][N1-y1] = image[y1][x1];
image[y1][x1] = image[x][y1];
image[x][y1] = a;
x += 1;
x1 -= 1;
}
x = y += 1;
y1 = x1 = N1-x;
edge -= 1;
}
return image;
}
Are you up to date?
There is really not much to the problem and there are a variety of solutions.
It is also hard to know what the interviews are looking for. For some it is knowledge of latest language features, for others it may be judged on efficiency.
These very simple tests are generally just to see if you can actually write code. You would be amazed at how many people apply for jobs they are not able to do.
Assessing the code.
Looking at your code as a conservative HR interested in your code writing style, and up to date language knowledge.
It is a pass, you can code, you have the job but will have you in supervision for a while..You also need to catch up on the language as you are not using ES6 to the full.
General points
- You need to use const for constants, and let for block scope variables.
- You have not used any arrow functions.
- I would question why you created the function using an expression, rather than a function statement, to make sure that you know the difference. (You should play it safe an use function statements)
- Also maybe a little to much code, but not really an issue.
- Code logic.. i would ask why you chose to duplicate the array rather than just create a new one as you went.
A quick rewrite
I am assuming the rotate is in place (original array) with new rows
function rotate(matrix) { // function statement
const N = matrix.length - 1; // use a constant
// use arrow functions and nested map;
const result = matrix.map((row, i) =>
row.map((val, j) => matrix[N - j][i])
);
matrix.length = 0; // hold original array reference
matrix.push(...result); // Spread operator
return matrix;
}
Some extras
The problem states "image" and "pixel" which give a hint that this may be judged on performance. They would tell you if it was. But if you get a similar challenge and performance is important then its best to use good old for
loops, avoiding any iterators that use callbacks.
Also a performance oriented function would swap in place rather than create new arrays. If you had to do that realtime with images for modern displays your code would be making big GC hits (Garbage Collection).
An alternative
Another way it can be done by rotating 4 pixels at a time can reduce memory overheads and processing time. Destructing lets you swap 4 corners in one go without the need to create temp variables.
I would never recommend you submit the following unless there was a clear directive for performance, and thinking outside the box.
// Commented version
// N*N is square
// pixels are rotated inplace
// 4 by 4 rotated in 4 iterations
// 5 by 5 rotated in 7 iterations
function rotatePixels(image) {
var x, y, x1, y1, edge;
// Solve by swapping 4 at a time in rings from the outside in
const N = image.length; // size of array
const N1 = N - 1; // position of end item
const N2 = N / 2; // Half way position
// x,y hold the a cell coordinate
x = y = 0;
// x,y hold the diagonally opposite cell
x1 = y1 = N1;
// length of current edge
edge = N1;
// while not at the center
while (y < N2) {
// for each item on the current edge
while (x < edge) { // rotate points at outer edge
// swap 4 corner items
// using array destructed assignment
[
image[x ][y1],
image[y1][x1],
image[x1][N1 - y1],
image[y ][x ]
] = [
image[y ][x ],
image[x ][y1],
image[y1][x1],
image[x1][N1 - y1]
];
x += 1; // move top pos forward one
x1 -= 1; // move bottom pos back one
}
y += 1; // diagonally into array
x = y; // x same as y
y1 = x1 = N1-x; // and diagonal opposite
edge -= 1; // shorten the end
}
return image;
}
// How I would present it
function rotatePixels(image) {
var x, y, x1, y1, edge;
const N = image.length;
const N1 = N - 1;
const N2 = N / 2;
x = y = 0;
edge = x1 = y1 = N1;
while (y < N2) {
while (x < edge) {
[image[x][y1], image[y1][x1], image[x1][N1-y1], image[y][x]] =
[image[y][x] , image[x ][y1], image[y1][x1] , image[x1][N1-y1]];
x += 1;
x1 -= 1;
}
x = y += 1;
y1 = x1 = N1-x;
edge -= 1;
}
return image;
}
// At time of writing I was unsure as to the performance of the swap using destructuring
// Turns out it is very bad
// The next version is a more conventional swap and runs 3 time faster than the above version
// and 8 times faster than the conventional solution at top of answer
function rotatePixels(image) {
var x, y, x1, y1, edge;
const N = image.length;
const N1 = N - 1;
const N2 = N / 2;
x = y = 0;
edge = x1 = y1 = N1;
while (y < N2) {
while (x < edge) {
const a = image[y][x];
image[y][x] = image[x1][N1-y1];
image[x1][N1-y1] = image[y1][x1];
image[y1][x1] = image[x][y1];
image[x][y1] = a;
x += 1;
x1 -= 1;
}
x = y += 1;
y1 = x1 = N1-x;
edge -= 1;
}
return image;
}
// Commented version
// N*N is square
// pixels are rotated inplace
// 4 by 4 rotated in 4 iterations
// 5 by 5 rotated in 7 iterations
function rotatePixels(image) {
var x, y, x1, y1, edge;
// Solve by swapping 4 at a time in rings from the outside in
const N = image.length; // size of array
const N1 = N - 1; // position of end item
const N2 = N / 2; // Half way position
// x,y hold the a cell coordinate
x = y = 0;
// x,y hold the diagonally opposite cell
x1 = y1 = N1;
// length of current edge
edge = N1;
// while not at the center
while (y < N2) {
// for each item on the current edge
while (x < edge) { // rotate points at outer edge
// swap 4 corner items
// using array destructed assignment
[
image[x ][y1],
image[y1][x1],
image[x1][N1 - y1],
image[y ][x ]
] = [
image[y ][x ],
image[x ][y1],
image[y1][x1],
image[x1][N1 - y1]
];
x += 1; // move top pos forward one
x1 -= 1; // move bottom pos back one
}
y += 1; // diagonally into array
x = y; // x same as y
y1 = x1 = N1-x; // and diagonal opposite
edge -= 1; // shorten the end
}
return image;
}
// How I would present it
function rotatePixels(image) {
var x, y, x1, y1, edge;
const N = image.length;
const N1 = N - 1;
const N2 = N / 2;
x = y = 0;
edge = x1 = y1 = N1;
while (y < N2) {
while (x < edge) {
[image[x][y1], image[y1][x1], image[x1][N1-y1], image[y][x]] =
[image[y][x] , image[x ][y1], image[y1][x1] , image[x1][N1-y1]];
x += 1;
x1 -= 1;
}
x = y += 1;
y1 = x1 = N1-x;
edge -= 1;
}
return image;
}
// At time of writing I was unsure as to the performance of the swap using destructuring
// Turns out it is very bad
// The next version is a more conventional swap and runs 3 time faster than the above version
// and 8 times faster than the conventional solution at top of answer
function rotatePixels(image) {
var x, y, x1, y1, edge;
const N = image.length;
const N1 = N - 1;
const N2 = N / 2;
x = y = 0;
edge = x1 = y1 = N1;
while (y < N2) {
while (x < edge) {
const a = image[y][x];
image[y][x] = image[x1][N1-y1];
image[x1][N1-y1] = image[y1][x1];
image[y1][x1] = image[x][y1];
image[x][y1] = a;
x += 1;
x1 -= 1;
}
x = y += 1;
y1 = x1 = N1-x;
edge -= 1;
}
return image;
}
// Commented version
// N*N is square
// pixels are rotated inplace
// 4 by 4 rotated in 4 iterations
// 5 by 5 rotated in 7 iterations
function rotatePixels(image) {
var x, y, x1, y1, edge;
// Solve by swapping 4 at a time in rings from the outside in
const N = image.length; // size of array
const N1 = N - 1; // position of end item
const N2 = N / 2; // Half way position
// x,y hold the a cell coordinate
x = y = 0;
// x,y hold the diagonally opposite cell
x1 = y1 = N1;
// length of current edge
edge = N1;
// while not at the center
while (y < N2) {
// for each item on the current edge
while (x < edge) { // rotate points at outer edge
// swap 4 corner items
// using array destructed assignment
[
image[x ][y1],
image[y1][x1],
image[x1][N1 - y1],
image[y ][x ]
] = [
image[y ][x ],
image[x ][y1],
image[y1][x1],
image[x1][N1 - y1]
];
x += 1; // move top pos forward one
x1 -= 1; // move bottom pos back one
}
y += 1; // diagonally into array
x = y; // x same as y
y1 = x1 = N1-x; // and diagonal opposite
edge -= 1; // shorten the end
}
return image;
}
// How I would present it
function rotatePixels(image) {
var x, y, x1, y1, edge;
const N = image.length;
const N1 = N - 1;
const N2 = N / 2;
x = y = 0;
edge = x1 = y1 = N1;
while (y < N2) {
while (x < edge) {
[image[x][y1], image[y1][x1], image[x1][N1-y1], image[y][x]] =
[image[y][x] , image[x ][y1], image[y1][x1] , image[x1][N1-y1]];
x += 1;
x1 -= 1;
}
x = y += 1;
y1 = x1 = N1-x;
edge -= 1;
}
return image;
}
// At time of writing I was unsure as to the performance of the swap using destructuring
// Turns out it is very bad
// The next version is a more conventional swap and runs 3 time faster than the above version
// and 8 times faster than the conventional solution at top of answer
function rotatePixels(image) {
var x, y, x1, y1, edge;
const N = image.length;
const N1 = N - 1;
const N2 = N / 2;
x = y = 0;
edge = x1 = y1 = N1;
while (y < N2) {
while (x < edge) {
const a = image[y][x];
image[y][x] = image[x1][N1-y1];
image[x1][N1-y1] = image[y1][x1];
image[y1][x1] = image[x][y1];
image[x][y1] = a;
x += 1;
x1 -= 1;
}
x = y += 1;
y1 = x1 = N1-x;
edge -= 1;
}
return image;
}
edited Feb 6 '18 at 11:28
answered Feb 5 '18 at 15:14
Blindman67Blindman67
7,2011521
7,2011521
Thanks that's interesting and I had been wondering how to do an in place version. As for ES6 I've been intentionally ignoring it for now as my work is based around IE11 use.
– PDStat
Feb 5 '18 at 15:42
3
@PaulStatham Don't let MS's IE11 pull you down. It will be until after 2020 when MS stops support for win7, that IE11 will go away and it will never support ECMAScript6+. You can use transpilers such as Babel to provide legacy support and have your code base kept up to date, and most important your skill set on the consumer's leading edge.
– Blindman67
Feb 6 '18 at 11:38
add a comment |
Thanks that's interesting and I had been wondering how to do an in place version. As for ES6 I've been intentionally ignoring it for now as my work is based around IE11 use.
– PDStat
Feb 5 '18 at 15:42
3
@PaulStatham Don't let MS's IE11 pull you down. It will be until after 2020 when MS stops support for win7, that IE11 will go away and it will never support ECMAScript6+. You can use transpilers such as Babel to provide legacy support and have your code base kept up to date, and most important your skill set on the consumer's leading edge.
– Blindman67
Feb 6 '18 at 11:38
Thanks that's interesting and I had been wondering how to do an in place version. As for ES6 I've been intentionally ignoring it for now as my work is based around IE11 use.
– PDStat
Feb 5 '18 at 15:42
Thanks that's interesting and I had been wondering how to do an in place version. As for ES6 I've been intentionally ignoring it for now as my work is based around IE11 use.
– PDStat
Feb 5 '18 at 15:42
3
3
@PaulStatham Don't let MS's IE11 pull you down. It will be until after 2020 when MS stops support for win7, that IE11 will go away and it will never support ECMAScript6+. You can use transpilers such as Babel to provide legacy support and have your code base kept up to date, and most important your skill set on the consumer's leading edge.
– Blindman67
Feb 6 '18 at 11:38
@PaulStatham Don't let MS's IE11 pull you down. It will be until after 2020 when MS stops support for win7, that IE11 will go away and it will never support ECMAScript6+. You can use transpilers such as Babel to provide legacy support and have your code base kept up to date, and most important your skill set on the consumer's leading edge.
– Blindman67
Feb 6 '18 at 11:38
add a comment |
Cloning the matrix
.slice
is a good way to copy the values of an array, instead of getting a reference to the same array. Unfortunately all your values are themselves arrays, which you are still using a reference to and not a copy. In this code you are not changing individual values, but setting entire rows at a time, so it still works, but this could easily lead to unexpected behavior if you are not careful.
You don't mention if you have to modify the original matrix or return a new one (you are doing both). But if you leave the original unmodified, you don't even need to make a copy. Just start out with empty array and .push
each new row to it.
Mapping the values
Your .map
is a bit weird. You use it on one array, but then you don't even use the value and just use the index to access a different array. This can also be done simpler. The Nth row in the result is the Nth column in the input. So using .map
on the input and getting the right value from each row will give you the new row (you have to reverse it too).
var row = matrix.map(function(e) {
return e[i]
}).reverse();
ES6
A lot of new features was introduced in ES6, which was finalized back in 2015. You should be using them by now. This means using let
instead of var
, and arrow functions, which are excellent for short inline function like the one above.
The final code
Another small change I made was to change to the loop to loop over the inner array. This way it also works for matrices of different dimensions.
function rotate(matrix) {
let result = ;
for(let i = 0; i < matrix[0].length; i++) {
let row = matrix.map(e => e[i]).reverse();
result.push(row);
}
return result;
};
add a comment |
Cloning the matrix
.slice
is a good way to copy the values of an array, instead of getting a reference to the same array. Unfortunately all your values are themselves arrays, which you are still using a reference to and not a copy. In this code you are not changing individual values, but setting entire rows at a time, so it still works, but this could easily lead to unexpected behavior if you are not careful.
You don't mention if you have to modify the original matrix or return a new one (you are doing both). But if you leave the original unmodified, you don't even need to make a copy. Just start out with empty array and .push
each new row to it.
Mapping the values
Your .map
is a bit weird. You use it on one array, but then you don't even use the value and just use the index to access a different array. This can also be done simpler. The Nth row in the result is the Nth column in the input. So using .map
on the input and getting the right value from each row will give you the new row (you have to reverse it too).
var row = matrix.map(function(e) {
return e[i]
}).reverse();
ES6
A lot of new features was introduced in ES6, which was finalized back in 2015. You should be using them by now. This means using let
instead of var
, and arrow functions, which are excellent for short inline function like the one above.
The final code
Another small change I made was to change to the loop to loop over the inner array. This way it also works for matrices of different dimensions.
function rotate(matrix) {
let result = ;
for(let i = 0; i < matrix[0].length; i++) {
let row = matrix.map(e => e[i]).reverse();
result.push(row);
}
return result;
};
add a comment |
Cloning the matrix
.slice
is a good way to copy the values of an array, instead of getting a reference to the same array. Unfortunately all your values are themselves arrays, which you are still using a reference to and not a copy. In this code you are not changing individual values, but setting entire rows at a time, so it still works, but this could easily lead to unexpected behavior if you are not careful.
You don't mention if you have to modify the original matrix or return a new one (you are doing both). But if you leave the original unmodified, you don't even need to make a copy. Just start out with empty array and .push
each new row to it.
Mapping the values
Your .map
is a bit weird. You use it on one array, but then you don't even use the value and just use the index to access a different array. This can also be done simpler. The Nth row in the result is the Nth column in the input. So using .map
on the input and getting the right value from each row will give you the new row (you have to reverse it too).
var row = matrix.map(function(e) {
return e[i]
}).reverse();
ES6
A lot of new features was introduced in ES6, which was finalized back in 2015. You should be using them by now. This means using let
instead of var
, and arrow functions, which are excellent for short inline function like the one above.
The final code
Another small change I made was to change to the loop to loop over the inner array. This way it also works for matrices of different dimensions.
function rotate(matrix) {
let result = ;
for(let i = 0; i < matrix[0].length; i++) {
let row = matrix.map(e => e[i]).reverse();
result.push(row);
}
return result;
};
Cloning the matrix
.slice
is a good way to copy the values of an array, instead of getting a reference to the same array. Unfortunately all your values are themselves arrays, which you are still using a reference to and not a copy. In this code you are not changing individual values, but setting entire rows at a time, so it still works, but this could easily lead to unexpected behavior if you are not careful.
You don't mention if you have to modify the original matrix or return a new one (you are doing both). But if you leave the original unmodified, you don't even need to make a copy. Just start out with empty array and .push
each new row to it.
Mapping the values
Your .map
is a bit weird. You use it on one array, but then you don't even use the value and just use the index to access a different array. This can also be done simpler. The Nth row in the result is the Nth column in the input. So using .map
on the input and getting the right value from each row will give you the new row (you have to reverse it too).
var row = matrix.map(function(e) {
return e[i]
}).reverse();
ES6
A lot of new features was introduced in ES6, which was finalized back in 2015. You should be using them by now. This means using let
instead of var
, and arrow functions, which are excellent for short inline function like the one above.
The final code
Another small change I made was to change to the loop to loop over the inner array. This way it also works for matrices of different dimensions.
function rotate(matrix) {
let result = ;
for(let i = 0; i < matrix[0].length; i++) {
let row = matrix.map(e => e[i]).reverse();
result.push(row);
}
return result;
};
answered Feb 5 '18 at 15:12
KrugaKruga
77829
77829
add a comment |
add a comment |
If performance is important, as mentioned above, it would be good to use for
.
Also it is nice if you show an example of how you would structure a basic class. E.g:
class Matrix {
constructor(n){
this.m = ;
var z = 0;
for (let i = 0; i < n; ++i) {
this.m[i] = ;
for (let j = 0; j < n; ++j) {
this.m[i][j] = z;
z++;
}
}
}
printMatrix() {
console.log(this.m);
}
rotateMatrix() {
const n = this.m[0].length;
let res =
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
if (!res[j])
res[j] =
res[j][i] = this.m[n-1-i][j];
}
}
return res;
}
}
let m = new Matrix(5);
m.printMatrix();
let t0 = performance.now();
console.log(m.rotateMatrix());
let t1 = performance.now();
console.log("Call to rotate took " + (t1 - t0) + " milliseconds.");
It seems to perform 20% faster than using the algorithms above. I am pretty sure this can be done even more efficient.
add a comment |
If performance is important, as mentioned above, it would be good to use for
.
Also it is nice if you show an example of how you would structure a basic class. E.g:
class Matrix {
constructor(n){
this.m = ;
var z = 0;
for (let i = 0; i < n; ++i) {
this.m[i] = ;
for (let j = 0; j < n; ++j) {
this.m[i][j] = z;
z++;
}
}
}
printMatrix() {
console.log(this.m);
}
rotateMatrix() {
const n = this.m[0].length;
let res =
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
if (!res[j])
res[j] =
res[j][i] = this.m[n-1-i][j];
}
}
return res;
}
}
let m = new Matrix(5);
m.printMatrix();
let t0 = performance.now();
console.log(m.rotateMatrix());
let t1 = performance.now();
console.log("Call to rotate took " + (t1 - t0) + " milliseconds.");
It seems to perform 20% faster than using the algorithms above. I am pretty sure this can be done even more efficient.
add a comment |
If performance is important, as mentioned above, it would be good to use for
.
Also it is nice if you show an example of how you would structure a basic class. E.g:
class Matrix {
constructor(n){
this.m = ;
var z = 0;
for (let i = 0; i < n; ++i) {
this.m[i] = ;
for (let j = 0; j < n; ++j) {
this.m[i][j] = z;
z++;
}
}
}
printMatrix() {
console.log(this.m);
}
rotateMatrix() {
const n = this.m[0].length;
let res =
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
if (!res[j])
res[j] =
res[j][i] = this.m[n-1-i][j];
}
}
return res;
}
}
let m = new Matrix(5);
m.printMatrix();
let t0 = performance.now();
console.log(m.rotateMatrix());
let t1 = performance.now();
console.log("Call to rotate took " + (t1 - t0) + " milliseconds.");
It seems to perform 20% faster than using the algorithms above. I am pretty sure this can be done even more efficient.
If performance is important, as mentioned above, it would be good to use for
.
Also it is nice if you show an example of how you would structure a basic class. E.g:
class Matrix {
constructor(n){
this.m = ;
var z = 0;
for (let i = 0; i < n; ++i) {
this.m[i] = ;
for (let j = 0; j < n; ++j) {
this.m[i][j] = z;
z++;
}
}
}
printMatrix() {
console.log(this.m);
}
rotateMatrix() {
const n = this.m[0].length;
let res =
for (let i = 0; i < n; ++i) {
for (let j = 0; j < n; ++j) {
if (!res[j])
res[j] =
res[j][i] = this.m[n-1-i][j];
}
}
return res;
}
}
let m = new Matrix(5);
m.printMatrix();
let t0 = performance.now();
console.log(m.rotateMatrix());
let t1 = performance.now();
console.log("Call to rotate took " + (t1 - t0) + " milliseconds.");
It seems to perform 20% faster than using the algorithms above. I am pretty sure this can be done even more efficient.
edited Apr 27 '18 at 6:38
Billal Begueradj
1
1
answered Apr 27 '18 at 6:08
agonza1agonza1
211
211
add a comment |
add a comment |
Matrix N × M
var m = [
[11, 12, 13],
[21, 22, 23],
[31, 32, 33],
[41, 42, 43],
[51, 52, 53],
];
var rn = m.length
,cn = m[0].length
,ARRC = Array.from(Array(cn))
,ARRR = Array.from(Array(rn));
function print_m(m) {
console.log('----------------');
m.forEach(r => console.log(r));
}
//90 clockwise
var m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[rn - 1 - r][c])
);
print_m(m1);
//90 counter clockwise
m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[r][cn - 1 - c])
);
print_m(m1);
//180
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[rn - 1 - r][cn - 1 - c])
);
print_m(m1);
// flip vertical
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[rn - 1 - r][c])
);
print_m(m1);
// flip horizontal
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[r][cn - 1 - c])
);
print_m(m1);
// flip horizontal & 90 counter clockwise
m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[r][c])
);
New contributor
We are looking for answers that provide insightful observations about the code in the question. Answers that consist of independent solutions with no justification do not constitute a code review, and may be removed.
2
You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process.
– Mast
2 days ago
My solution can be apply for nxm matrix, not only nxn, it also solve for other case such as roate 180, flip, counter clockwise. I'm not sure it better than the original about performance or not, but obviously it's concise.
– KevinBui
2 days ago
add a comment |
Matrix N × M
var m = [
[11, 12, 13],
[21, 22, 23],
[31, 32, 33],
[41, 42, 43],
[51, 52, 53],
];
var rn = m.length
,cn = m[0].length
,ARRC = Array.from(Array(cn))
,ARRR = Array.from(Array(rn));
function print_m(m) {
console.log('----------------');
m.forEach(r => console.log(r));
}
//90 clockwise
var m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[rn - 1 - r][c])
);
print_m(m1);
//90 counter clockwise
m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[r][cn - 1 - c])
);
print_m(m1);
//180
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[rn - 1 - r][cn - 1 - c])
);
print_m(m1);
// flip vertical
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[rn - 1 - r][c])
);
print_m(m1);
// flip horizontal
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[r][cn - 1 - c])
);
print_m(m1);
// flip horizontal & 90 counter clockwise
m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[r][c])
);
New contributor
We are looking for answers that provide insightful observations about the code in the question. Answers that consist of independent solutions with no justification do not constitute a code review, and may be removed.
2
You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process.
– Mast
2 days ago
My solution can be apply for nxm matrix, not only nxn, it also solve for other case such as roate 180, flip, counter clockwise. I'm not sure it better than the original about performance or not, but obviously it's concise.
– KevinBui
2 days ago
add a comment |
Matrix N × M
var m = [
[11, 12, 13],
[21, 22, 23],
[31, 32, 33],
[41, 42, 43],
[51, 52, 53],
];
var rn = m.length
,cn = m[0].length
,ARRC = Array.from(Array(cn))
,ARRR = Array.from(Array(rn));
function print_m(m) {
console.log('----------------');
m.forEach(r => console.log(r));
}
//90 clockwise
var m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[rn - 1 - r][c])
);
print_m(m1);
//90 counter clockwise
m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[r][cn - 1 - c])
);
print_m(m1);
//180
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[rn - 1 - r][cn - 1 - c])
);
print_m(m1);
// flip vertical
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[rn - 1 - r][c])
);
print_m(m1);
// flip horizontal
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[r][cn - 1 - c])
);
print_m(m1);
// flip horizontal & 90 counter clockwise
m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[r][c])
);
New contributor
Matrix N × M
var m = [
[11, 12, 13],
[21, 22, 23],
[31, 32, 33],
[41, 42, 43],
[51, 52, 53],
];
var rn = m.length
,cn = m[0].length
,ARRC = Array.from(Array(cn))
,ARRR = Array.from(Array(rn));
function print_m(m) {
console.log('----------------');
m.forEach(r => console.log(r));
}
//90 clockwise
var m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[rn - 1 - r][c])
);
print_m(m1);
//90 counter clockwise
m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[r][cn - 1 - c])
);
print_m(m1);
//180
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[rn - 1 - r][cn - 1 - c])
);
print_m(m1);
// flip vertical
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[rn - 1 - r][c])
);
print_m(m1);
// flip horizontal
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[r][cn - 1 - c])
);
print_m(m1);
// flip horizontal & 90 counter clockwise
m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[r][c])
);
var m = [
[11, 12, 13],
[21, 22, 23],
[31, 32, 33],
[41, 42, 43],
[51, 52, 53],
];
var rn = m.length
,cn = m[0].length
,ARRC = Array.from(Array(cn))
,ARRR = Array.from(Array(rn));
function print_m(m) {
console.log('----------------');
m.forEach(r => console.log(r));
}
//90 clockwise
var m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[rn - 1 - r][c])
);
print_m(m1);
//90 counter clockwise
m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[r][cn - 1 - c])
);
print_m(m1);
//180
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[rn - 1 - r][cn - 1 - c])
);
print_m(m1);
// flip vertical
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[rn - 1 - r][c])
);
print_m(m1);
// flip horizontal
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[r][cn - 1 - c])
);
print_m(m1);
// flip horizontal & 90 counter clockwise
m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[r][c])
);
var m = [
[11, 12, 13],
[21, 22, 23],
[31, 32, 33],
[41, 42, 43],
[51, 52, 53],
];
var rn = m.length
,cn = m[0].length
,ARRC = Array.from(Array(cn))
,ARRR = Array.from(Array(rn));
function print_m(m) {
console.log('----------------');
m.forEach(r => console.log(r));
}
//90 clockwise
var m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[rn - 1 - r][c])
);
print_m(m1);
//90 counter clockwise
m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[r][cn - 1 - c])
);
print_m(m1);
//180
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[rn - 1 - r][cn - 1 - c])
);
print_m(m1);
// flip vertical
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[rn - 1 - r][c])
);
print_m(m1);
// flip horizontal
m1 = ARRR.map((_,r) =>
ARRC.map((_,c) => m[r][cn - 1 - c])
);
print_m(m1);
// flip horizontal & 90 counter clockwise
m1 = ARRC.map((_,c) =>
ARRR.map((_,r) => m[r][c])
);
New contributor
New contributor
answered 2 days ago
KevinBuiKevinBui
97
97
New contributor
New contributor
We are looking for answers that provide insightful observations about the code in the question. Answers that consist of independent solutions with no justification do not constitute a code review, and may be removed.
We are looking for answers that provide insightful observations about the code in the question. Answers that consist of independent solutions with no justification do not constitute a code review, and may be removed.
2
You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process.
– Mast
2 days ago
My solution can be apply for nxm matrix, not only nxn, it also solve for other case such as roate 180, flip, counter clockwise. I'm not sure it better than the original about performance or not, but obviously it's concise.
– KevinBui
2 days ago
add a comment |
2
You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process.
– Mast
2 days ago
My solution can be apply for nxm matrix, not only nxn, it also solve for other case such as roate 180, flip, counter clockwise. I'm not sure it better than the original about performance or not, but obviously it's concise.
– KevinBui
2 days ago
2
2
You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process.
– Mast
2 days ago
You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process.
– Mast
2 days ago
My solution can be apply for nxm matrix, not only nxn, it also solve for other case such as roate 180, flip, counter clockwise. I'm not sure it better than the original about performance or not, but obviously it's concise.
– KevinBui
2 days ago
My solution can be apply for nxm matrix, not only nxn, it also solve for other case such as roate 180, flip, counter clockwise. I'm not sure it better than the original about performance or not, but obviously it's concise.
– KevinBui
2 days ago
add a comment |
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