```ts
// mergeIntervals.ts

export function mergeIntervals(intervals: number[][]): number[][] {
  if (intervals.length === 0) {
    return [];
  }

  intervals.sort((a, b) => a[0] - b[0]);

  const mergedIntervals: number[][] = [intervals[0]];

  for (let i = 1; i < intervals.length; i++) {
    const currentInterval = intervals[i];
    const lastMerged = mergedIntervals[mergedIntervals.length - 1];

    if (currentInterval[0] <= lastMerged[1]) {
      // Merge the two intervals
      mergedIntervals[mergedIntervals.length - 1][1] = Math.max(lastMerged[1], currentInterval[1]);
    } else {
      // Add the unmerged interval to the result array
      mergedIntervals.push(currentInterval);
    }
  }

  return mergedIntervals;
}
```

**Description of Solution:**

The `mergeIntervals` function takes an array of number arrays (`intervals`) as input. It first checks if the input array is empty and returns it immediately in that case.

Here’s a step-by-step breakdown of how the function works:

1. **Sort Input Intervals:** The function sorts the intervals based on their start values using `Array.prototype.sort()`. This ensures that we process intervals in ascending order, which simplifies merging tasks.

2. **Initialize Result Array:** A variable `mergedIntervals` is initialized with the first interval from the input sorted array. This array will store the merged intervals.

3. **Iterate Through Remaining Intervals:** The function iterates through the sorted array starting from the second interval. For each interval, it checks if it overlaps with the last interval in the `mergedIntervals` array.

   - If an overlap is detected:
     - It updates the end of the last merged interval to be the maximum of its current end and the end of the current interval.
   - If no overlap is detected:
     - It adds the current interval as a new merged interval to the `mergedIntervals` array.

4. **Return Result:** After processing all intervals, the function returns the `mergedIntervals` array containing none overwriting or modifying the input array.

This approach ensures that each interval in `intervals` is processed in constant time (`O(n)`) where `n` is the number of intervals, and the space complexity is also O(n) due to the additional space needed for the `mergedIntervals` array.