--description--

In day four of Space Week, you are given a matrix of numbers (an array of arrays), representing potential landing spots for your rover. Find the safest landing spot based on the following rules:

• Each spot in the matrix will contain a number from 0-9, inclusive.
• Any 0 represents a potential landing spot.
• Any number other than 0 is too dangerous to land. The higher the number, the more dangerous.
• The safest spot is defined as the 0 cell whose surrounding cells (up to 4 neighbors, ignore diagonals) have the lowest total danger.
• Ignore out-of-bounds neighbors (corners and edges just have fewer neighbors).
• Return the indices of the safest landing spot. There will always only be one safest spot.

For instance, given:

[
  [1, 0],
  [2, 0]
]

Return [0, 1], the indices for the 0 in the first array.

--hints--

findLandingSpot([[1, 0], [2, 0]]) should return [0, 1].

assert.deepEqual(findLandingSpot([[1, 0], [2, 0]]), [0, 1]);

findLandingSpot([[9, 0, 3], [7, 0, 4], [8, 0, 5]]) should return [1, 1].

assert.deepEqual(findLandingSpot([[9, 0, 3], [7, 0, 4], [8, 0, 5]]), [1, 1]);

findLandingSpot([[1, 2, 1], [0, 0, 2], [3, 0, 0]]) should return [2, 2].

assert.deepEqual(findLandingSpot([[1, 2, 1], [0, 0, 2], [3, 0, 0]]), [2, 2]);

findLandingSpot([[9, 6, 0, 8], [7, 1, 1, 0], [3, 0, 3, 9], [8, 6, 0, 9]]) should return [2, 1].

assert.deepEqual(findLandingSpot([[9, 6, 0, 8], [7, 1, 1, 0], [3, 0, 3, 9], [8, 6, 0, 9]]), [2, 1]);

--seed--

--seed-contents--

function findLandingSpot(matrix) {

  return matrix;
}

--solutions--

function findLandingSpot(matrix) {
  let bestSpot = null;
  let lowestNeighborSum = Infinity;

  for (let i = 0; i < matrix.length; i++) {
    for (let j = 0; j < matrix[i].length; j++) {

      if (matrix[i][j] === 0) {
        let currentNeighborSum = 0;

        if (i > 0) currentNeighborSum += matrix[i - 1][j];
        if (j < matrix[i].length - 1) currentNeighborSum += matrix[i][j + 1];
        if (i < matrix.length - 1) currentNeighborSum += matrix[i + 1][j];
        if (j > 0) currentNeighborSum += matrix[i][j - 1];

        if (currentNeighborSum < lowestNeighborSum) {
          lowestNeighborSum = currentNeighborSum;
          bestSpot = [i, j];
        }
      }
    }
  }

  return bestSpot;
}