Minimum Operations to Make a Subsequence LeetCode solution in JavaScript

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var lengthOfLIS = function (A) {
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  if (!A.length) return 0;
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  let tails = [-Infinity]; //here -Infinity acts as a Sentinel, for cleaner code (or else I would ahve to place A[0] here)
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  for (let i = 0; i < A.length; i++) {
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    // If my curr element is bigger than all possible tails, i just need to create a new subarray, which of course will be of length tails.length-1 +1
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    if (A[i] > tails[tails.length - 1]) tails.push(A[i]);
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    else {
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      let lo = 0,
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        hi = tails.length - 1;
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      //binary search to find where to place my current element so i have more chances of creating a bigger subarray
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      while (lo < hi) {
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        let mid = (lo + hi) >> 1;
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        if (tails[mid] < A[i]) lo = mid + 1;
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        else hi = mid;
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      }
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      tails[lo] = A[i];
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    }
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  }
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  return tails.length - 1; // is the length of the longest possible subarray (-1 because of the -Infinity I added)
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};
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var minOperations = function (T, A) {
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  let seen = new Set(T),
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    res = 0,
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    originalIndex = {},
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    n = T.length;
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  //filter out all the unnecessary values
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  A = A.filter((d) => seen.has(d));
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  for (let i = 0; i < T.length; i++) originalIndex[T[i]] = i; //maintain the original indices of the values in T
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  A = A.map((d) => originalIndex[d]);
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  return T.length - lengthOfLIS(A);
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};