Minimum Number of Removals to Make Mountain Array LeetCode solution in JavaScript

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var minimumMountainRemovals = function (nums) {
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  let n = nums.length;
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  let previous = Array.from({ length: n }, (item) => 1);
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  let previous2 = Array.from({ length: n }, (item) => 1);
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  //calcultaing left and right side LIS in single iteration
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  for (let i = 0; i < n; i++) {
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    for (let j = 0; j < i; j++) {
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      //reverse indexes
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      let i2 = n - 1 - i;
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      let j2 = n - 1 - j;
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      //calculating first dp serially
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      if (nums[i] > nums[j] && previous[i] < previous[j] + 1) {
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        previous[i] = previous[j] + 1;
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      }
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      //calculating second dp reversely
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      if (nums[i2] > nums[j2] && previous2[i2] < previous2[j2] + 1) {
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        previous2[i2] = previous2[j2] + 1;
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      }
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    }
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  }
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  let max = 0;
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  for (let i = 0; i < n; i++) {
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    //golden condition// to avoid only increasing and decreasing
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    //because if a combination of 1 indicates its increasing fully or decreasing !
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    if (previous[i] > 1 && previous2[i] > 1) {
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      //at a specific index highest increasing will come from left LIS
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      // and highest decreasing will come from right side LIS
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      max = Math.max(max, previous[i] + previous2[i] - 1);
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    }
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  }
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  return n - max; //we need to remove the rest from total
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};