Contours Shifting

Description

Mark got a rectangular array matrix for his birthday, and now he’s thinking about all the fun things he can do with it. He likes shifting a lot, so he decides to shift all of its i-contours in a clockwise direction if i is even, and counterclockwise if i is odd.

Here is how Mark defines i-contours:

the 0-contour of a rectangular array as the union of left and right columns as well as top and bottom rows;
consider the initial matrix without the 0-contour: its 0-contour is the 1-contour of the initial matrix;
define 2-contour, 3-contour, etc. in the same manner by removing 0-contours from the obtained arrays.

Implement a function that does exactly what Mark wants to do to his matrix.

Example

For

matrix = [[ 1, 2, 3, 4], [ 5, 6, 7, 8], [ 9, 10, 11, 12], [13, 14, 15, 16], [17, 18, 19, 20]]

the output should be

contoursShifting(matrix) = [[ 5, 1, 2, 3], [ 9, 7, 11, 4], [13, 6, 15, 8], [17, 10, 14, 12], [18, 19, 20, 16]]
For

matrix = [[238, 239, 240, 241, 242, 243, 244, 245]]

the output should be

contoursShifting(matrix) = [[245, 238, 239, 240, 241, 242, 243, 244]]

Note, that if a contour is represented by a 1 × n array, its center is considered to be below it.
For

matrix = [[238], [239], [240], [241], [242], [243], [244], [245]]

the output should be

contoursShifting(matrix) = [[245], [238], [239], [240], [241], [242], [243], [244]]

If a contour is represented by an n × 1 array, its center is considered to be to the left of it.

Input/Output

[execution time limit] 4 seconds (js)
[input] array.array.integer matrix

Guaranteed constraints:
1 ≤ matrix.length ≤ 100,
1 ≤ matrix[i].length ≤ 100,
1 ≤ matrix[i][j] ≤ 1000.
[output] array.array.integer
- The given matrix with its contours shifted.

[JavaScript (ES6)] Syntax Tips

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// Prints help message to the console
// Returns a string
function helloWorld(name) {
console.log("This prints to the console when you Run Tests");
return "Hello, " + name;
}

Solution

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function contoursShifting(matrix) {
  var maxC = Math.floor((Math.min(matrix.length, matrix[0].length) + 1) / 2);
  var q = [];
  var next;
  for (var i = 0; i < maxC; i++) {
    var it = contourIterator(i);
    next = it.next();
    while (!next.done) {
      q.push(matrix[next.value[0]][next.value[1]]);
      next = it.next();
    }
    if (i % 2) {
      q = q.slice(1).concat(q.slice(0, 1));
    } else {
      q = q.slice(-1).concat(q.slice(0, -1));
    }
    it = contourIterator(i);
    next = it.next();
    while (q.length) {
      matrix[next.value[0]][next.value[1]] = q.shift();
      next = it.next();
    }
  }

  return matrix;

  function* contourIterator(contour) {
    var directions = [
      ([i, j]) => [i, j + 1],
      ([i, j]) => [i + 1, j],
      ([i, j]) => [i, j - 1],
      ([i, j]) => [i - 1, j],
    ];
    var startPos = [contour, contour];
    var curPos = [contour, contour];
    var curDirIndex = 0;
    var curDir = directions[curDirIndex];
    var nextPos = [];
    while (nextPos[0] != startPos[0] || nextPos[1] !== startPos[1]) {
      yield curPos;
      nextPos = curDir(curPos);
      if (
        nextPos[0] < contour ||
        nextPos[1] < contour ||
        nextPos[0] >= matrix.length - contour ||
        nextPos[1] >= matrix[0].length - contour
      ) {
        curDirIndex = (curDirIndex + 1) % 4;
        curDir = directions[curDirIndex];
        nextPos = curDir(curPos);
      }
      if (
        nextPos[0] < contour ||
        nextPos[1] < contour ||
        nextPos[0] >= matrix.length - contour ||
        nextPos[1] >= matrix[0].length - contour
      ) {
        break;
      }
      curPos = nextPos;
    }
  }
}

Codefights

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Contours Shifting

Description

Solution