Find Minimum Time to Finish All Jobs LeetCode solution in JavaScript

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var minimumTimeRequired = function (jobs, k) {
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  let n = jobs.length,
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    maskSum = [...Array(1 << n)].map((d) => 0);
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  for (
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    let mask = 0;
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    mask < 1 << n;
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    mask++ //pre-store the sums of every mask
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  )
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    for (let i = 0; i < n; i++)
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      maskSum[mask] += Number(((1 << i) & mask) != 0) * jobs[i];
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  let dp = [...Array(k + 1)].map((d) =>
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    [...Array(1 << n)].map((d) => Infinity)
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  );
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  dp[0][0] = 0;
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  for (
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    let i = 1;
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    i <= k;
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    i++ //for each new person
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  )
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    for (
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      let curmask = 0;
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      curmask < 1 << n;
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      curmask++ //guess what his mask can be, what items can he select
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    )
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      for (
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        let prevmask = 0;
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        prevmask < 1 << n;
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        prevmask++ // but also, guess what the previous i-1 persons took already
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      )
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        if ((curmask & prevmask) === 0)
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          //obviously, 2 people can't take the same job
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          dp[i][curmask | prevmask] = Math.min(
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            dp[i][curmask | prevmask],
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            Math.max(dp[i - 1][prevmask], maskSum[curmask])
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          );
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  return dp[k][(1 << n) - 1];
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};