Count of Smaller Numbers After Self LeetCode solution in JavaScript

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var countSmaller = function (nums) {
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  if (nums === null || nums.length == 0) return [];
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  let N = nums.length;
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  const result = new Array(N).fill(0);
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  const numsWithIdx = new Array(N).fill(0);
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  for (let i = 0; i < nums.length; i++) {
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    const curr = nums[i];
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    numsWithIdx[i] = { val: curr, originalIdx: i };
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  }
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  divideAndConqueuer(0, nums.length - 1);
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  const resultList = [];
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  result.forEach((el) => resultList.push(el));
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  return resultList;
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  function divideAndConqueuer(left, right) {
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    if (left < right) {
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      const mid = Math.floor(left + (right - left) / 2);
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      divideAndConqueuer(left, mid);
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      divideAndConqueuer(mid + 1, right);
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      merge(left, mid, right);
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    }
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  }
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  function merge(left, mid, right) {
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    const leftLen = mid - left + 1;
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    const rightLen = right - mid;
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    let rightSmallers = 0;
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    const leftAux = new Array(leftLen).fill(0);
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    const rightAux = new Array(rightLen).fill(0);
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    for (let i = 0; i < leftLen; i++) {
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      leftAux[i] = numsWithIdx[left + i];
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    }
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    for (let i = 0; i < rightLen; i++) {
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      rightAux[i] = numsWithIdx[mid + 1 + i];
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    }
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    let leftPointer = 0;
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    let rightPointer = 0;
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    let insertAtPointer = left;
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    while (leftPointer < leftLen && rightPointer < rightLen) {
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      if (leftAux[leftPointer].val <= rightAux[rightPointer].val) {
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        result[leftAux[leftPointer].originalIdx] += rightSmallers;
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        numsWithIdx[insertAtPointer++] = leftAux[leftPointer++];
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      } else {
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        numsWithIdx[insertAtPointer++] = rightAux[rightPointer++];
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        rightSmallers++;
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      }
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    }
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    while (rightPointer < rightLen) {
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      numsWithIdx[insertAtPointer++] = rightAux[rightPointer++];
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    }
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    while (leftPointer < leftLen) {
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      result[leftAux[leftPointer].originalIdx] += rightSmallers;
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      numsWithIdx[insertAtPointer++] = leftAux[leftPointer++];
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    }
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  }
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};