Code Review (Easy): Finding the Middle Ground

Story

Lena was working on a dashboard feature that compares two sets of sorted performance scores collected from different server clusters. Eva noticed that the current tool struggled when the datasets grew, so she asked Lena to find the exact middle value of both lists combined without needing to fully re-organize everything. For instance, if one cluster reported scores of 1 and 3 and the other reported 2, the tool should identify 2 as the central point. Once implemented, this utility will help the team pinpoint typical system behavior across multiple data streams much faster.

Your Task

Read the story and code, then decide if the code is ready for production or needs improvement.

Hints

What would block an approval? Give me a hint.

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function findMedianSortedArrays(nums1: number[], nums2: number[]): number {
  // Make nums1 the larger array to simplify partitioning
  if (nums1.length > nums2.length) return findMedianSortedArrays(nums2, nums1);

const m = nums1.length;
  const n = nums2.length;
  let left = 0;
  let right = m;

while (left <= right) {
    const partitionX = Math.floor((left + right) / 2);
    const partitionY = Math.floor((m + n + 1) / 2) - partitionX;

const maxLeftX = partitionX === 0 ? -Infinity : nums1[partitionX - 1];
    const minRightX = partitionX === m ? Infinity : nums1[partitionX];

const maxLeftY = partitionY === 0 ? -Infinity : nums2[partitionY - 1];
    const minRightY = partitionY === n ? Infinity : nums2[partitionY];

if (maxLeftX <= minRightY && maxLeftY <= minRightX) {
      // When total length is odd, average the two middle values
      if ((m + n) % 2 === 0) {
        return (Math.max(maxLeftX, maxLeftY) + Math.min(minRightX, minRightY)) / 2;
      } else {
        return Math.max(maxLeftX, maxLeftY);
      }
    } else if (maxLeftX > minRightY) {
      right = partitionX - 1;
    } else {
      left = partitionX + 1;
    }
  }

throw new Error("Input arrays are not sorted or invalid.");
}