Last Day Where You Can Still Cross LeetCode solution in JavaScript

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/**
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 * @param {number} row
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 * @param {number} col
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 * @param {number[][]} cells
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 * @return {number}
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 */
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var latestDayToCross = function (row, col, cells) {
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  //Create empty ground
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  const emptyArr = () =>
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    Array(row)
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      .fill(0)
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      .map((el) => Array(col).fill(0));
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  let ground;
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  // Check if path exists
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  function isExistPath() {
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    //Depth First Search
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    function dfs(i, j) {
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      if (i === row || j === col || j < 0 || i < 0) return false;
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      if (ground[i][j] === 1) return false;
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      if (i === row - 1) return true;
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      ground[i][j] = 1;
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      return dfs(i + 1, j) || dfs(i, j - 1) || dfs(i, j + 1) || dfs(i - 1, j);
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    }
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    for (let j = 0; j < col; j++) if (dfs(0, j)) return true;
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    return false;
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  }
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  //fill the ground and check paths
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  function isGoodGround(end) {
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    ground = emptyArr();
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    for (let i = 0; i <= end; i++) ground[cells[i][0] - 1][cells[i][1] - 1] = 1;
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    return isExistPath();
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  }
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  //binary search by day
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  let l = -1,
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    r = cells.length;
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  while (r - l > 1) {
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    const mid = (l + r) >>> 1;
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    if (isGoodGround(mid)) l = mid;
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    else r = mid;
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  }
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  return l + 1;
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};