Largest Plus Sign LeetCode solution in JavaScript

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/**
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 * @param {number} n
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 * @param {number[][]} mines
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 * @return {number}
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 */
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var orderOfLargestPlusSign = function (n, mines) {
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  // create n * n as a table, and each grid fill the maximum number
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  const t = new Array(n).fill(0).map(() => new Array(n).fill(n));
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  // insert `0` to the grid in table according the position in mines
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  mines.forEach((a) => (t[a[0]][a[1]] = 0));
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  // loop each rows and cols
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  // - for rows: calculate non-zero grid from left to right
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  // - for cols: calculate non-zero grid from top to bottom
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  // - when the loop is completed, all of non-zero values will be calculated by four directions
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  // - and these grids value will be updated by comparing. then we can obtain the minimum value of the four directions caculation, which is the maximum of the grid
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  for (let i = 0; i < n; i++) {
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    // save the maximum value of non-zeros with for directions
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    let [l, r, u, d] = [0, 0, 0, 0];
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    // l: left,  loop row i from left to right
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    // r: right, loop row i from right to left
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    // u: up,    loop col i from top to bottom
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    // d: down,  loop col i from bottom to top
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    for (let j = 0, k = n - 1; j < n; j++, k--) {
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      // if the value is `0`, then set variable to `0`, and indicates it's broken
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      l = t[i][j] && l + 1;
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      r = t[i][k] && r + 1;
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      u = t[j][i] && u + 1;
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      d = t[k][i] && d + 1;
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      // if current value is less than origin value
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      // one possibility: the origin value is default
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      // another possibility: the length of non-zero in a centain direction is langer than in the current direction, which the minimum value we need
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      if (l < t[i][j]) t[i][j] = l;
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      if (r < t[i][k]) t[i][k] = r;
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      if (u < t[j][i]) t[j][i] = u;
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      if (d < t[k][i]) t[k][i] = d;
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    }
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  }
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  // return maximum value be saved by all cells
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  return Math.max(...t.map((v) => Math.max(...v)));
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};