Minimum Space Wasted From Packaging LeetCode solution in JavaScript

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/**
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 * @param {number[]} packages
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 * @param {number[][]} boxes
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 * @return {number}
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 */
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var minWastedSpace = function (packages, boxes) {
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  let count = 0,
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    b,
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    minWastage,
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    totalPackagesSize = 0n,
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    minBoxesAreaForAnySupplier = BigInt(Number.MAX_SAFE_INTEGER),
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    boxesArea = 0n,
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    p,
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    coverdPackagesIndex,
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    totalBoxesAreaForSupplier;
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  packages.sort(function (a, b) {
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    return a - b;
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  });
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  for (let i = 0; i < packages.length; i++) {
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    totalPackagesSize += BigInt(packages[i]);
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  }
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  for (let i = 0; i < boxes.length; i++) {
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    boxes[i].sort(function (a, b) {
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      return a - b;
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    });
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    p = 0;
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    totalBoxesAreaForSupplier = 0n;
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    if (boxes[i][boxes[i].length - 1] < packages[packages.length - 1]) {
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      //This supplier is not big enough box for the largest package, hence this supplier can't be used.
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      continue;
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    }
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    for (let b = 0; b < boxes[i].length; b++) {
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      coverdPackagesIndex = binaryLowerBound(
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        packages,
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        p,
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        packages.length - 1,
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        boxes[i][b]
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      );
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      if (coverdPackagesIndex !== -1) {
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        totalBoxesAreaForSupplier +=
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          BigInt(boxes[i][b]) * BigInt(coverdPackagesIndex - p + 1);
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        p = coverdPackagesIndex + 1;
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        if (p === packages.length) {
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          break;
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        }
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      }
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    }
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    if (p === packages.length) {
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      //Check if the current supplier was able to pack all the packages.
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      if (totalBoxesAreaForSupplier < minBoxesAreaForAnySupplier) {
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        minBoxesAreaForAnySupplier = totalBoxesAreaForSupplier;
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      }
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    }
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  }
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  if (minBoxesAreaForAnySupplier !== BigInt(Number.MAX_SAFE_INTEGER)) {
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    return (minBoxesAreaForAnySupplier - totalPackagesSize) % 1000000007n;
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  } else {
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    return -1;
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  }
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  function binaryLowerBound(arr, left, right, key) {
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    let mid,
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      ans = -1;
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    while (left <= right) {
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      mid = left + Math.floor((right - left) / 2);
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      if (arr[mid] <= key) {
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        ans = mid;
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        left = mid + 1;
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      } else {
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        right = mid - 1;
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      }
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    }
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    return ans;
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  }
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};