In this problem, we are given an array of intervals, where each interval is an object with a start and end property. The intervals are not necessarily sorted. Our task is to merge any overlapping intervals and return the resulting array of intervals.

Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

Example 1:

Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlap, merge them into [1,6].

Example 2:

Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

Constraints:

• 1 <= intervals.length <= 104
• intervals[i].length == 2
• 0 <= starti <= endi <= 104 

To solve this problem, you can use a greedy approach, where you first sort the intervals based on the start time, and then you can merge the overlapping intervals one by one. Here is a JavaScript solution that demonstrates this approach:

function merge(intervals) {
  if (!intervals.length) return intervals;

  // sort intervals by start time
  intervals.sort((a, b) => a[0] - b[0]);

  // initialize the merged intervals with the first interval
  let merged = [intervals[0]];

  for (let i = 1; i < intervals.length; i++) {
    let currentInterval = intervals[i];
    let lastMergedInterval = merged[merged.length - 1];

    // if the current interval overlaps with the last merged interval, merge them
    if (currentInterval[0] <= lastMergedInterval[1]) {
      lastMergedInterval[1] = Math.max(lastMergedInterval[1], currentInterval[1]);
    } else {
      // otherwise, add the current interval to the merged intervals
      merged.push(currentInterval);
    }
  }

  return merged;
}