Count Artifacts That Can Be Extracted LeetCode solution in JavaScript

1
/**
2
 * @param {number} n
3
 * @param {number[][]} artifacts
4
 * @param {number[][]} dig
5
 * @return {number}
6
 */
7
var digArtifacts = function (n, artifacts, dig) {
8
  // Step 1
9
  const virtualMap = []; // 0 = null, n = artifact_n
10
  for (let x = 0; x < n; x++) {
11
    const row = [];
12
    for (let y = 0; y < n; y++) {
13
      row.push(0);
14
    }
15
    virtualMap.push(row);
16
  }
17

18
  // Step 2
19
  const artifactSize = [];
20
  for (let artifactIdx in artifacts) {
21
    let [artY1, artX1, artY2, artX2] = artifacts[artifactIdx];
22
    artifactSize.push((artX2 + 1 - artX1) * (artY2 + 1 - artY1));
23
    while (artY1 <= artY2) {
24
      while (artX1 <= artX2) {
25
        virtualMap[artY1][artX1] = Number(artifactIdx) + 1;
26
        artX1++;
27
      }
28
      artY1++;
29
      artX1 = artifacts[artifactIdx][1];
30
    }
31
  }
32

33
  // Step 3
34
  for (let digComp of dig) {
35
    const [digY, digX] = digComp;
36
    const content = virtualMap[digY][digX];
37
    if (content !== 0) {
38
      artifactSize[content - 1] -= 1;
39
    }
40
  }
41

42
  // Step 4
43
  return artifactSize.reduce((acc, remainder) => {
44
    if (remainder === 0) {
45
      acc++;
46
    }
47
    return acc;
48
  }, 0);
49
};