Demystifying the Merge: A Deep Dive into Merging Time Ranges

The Initial Hurdle: Perceived Complexity

The process of merging time ranges can appear daunting at first, mostly due to misconceptions about how the merging unfolds. Yet, in reality, there's a clear workflow to follow. To build intuition, it's helpful to start with the most simplified example and dissect it step by step.

Setting the Stage: A Simple Example

Imagine we are given just two time ranges to merge: [1, 4] and [2, 5]. It's evident that these ranges overlap, but how exactly should we merge them?

Finding the Start Time

We observe that 1 precedes 2. So, the merged range must start at the earliest time, which in this case is 1. It's like having two runners on a track; naturally, the one who starts first sets the pace. This choice is relatively straightforward if you sort the intervals by their start times.

Deciding the End Time

Deciding the end time is a bit more nuanced. While 4 comes before 5, we want our merged range to encapsulate all overlapping time. So, the end time must be 5. Think of this as ensuring the last runner crosses the finish line.

The Backbone: Sorting as a First Step

Sorting the ranges helps us establish a chronological order, making it easier to merge them. The lowest starting time will always be the beginning of our merged range, and from there, we can focus on the end times. Here's how to sort these ranges in both JavaScript and Python:

1const ranges = [[1, 4], [2, 5]];
2
3ranges.sort(function(a, b) {
4  return a[0] - b[0] || a[1] - b[1];
5});

A Roadmap for Merging Time Ranges

Merging time ranges is a task that becomes much more manageable once you understand the underlying structure:

1. Sort the Ranges: Establish a chronological order.
2. Determine the Start Time: Choose the earliest start time among overlapping intervals.
3. Decide the End Time: Make sure to cover all time within the overlapping intervals.

By following this methodical approach, you'll find that what seemed like a challenging problem is, in fact, quite solvable.

Complexity Breakdown

Given that the input intervals are already sorted, our solution exhibits a linear time complexity of (O(n)) and linear space complexity. This is because we iterate through each of the (n) intervals exactly once.

The Final Logic: A Code Snapshot

The final logic could be understood as below (actual code not shown here). While iterating through the sorted array:

• Check for overlap with the last interval.
• Extend last if there's an overlap.
• Append to result in case of a clear separation.

By following this systematic approach, merging time intervals becomes a straightforward task, completely doable within reasonable time and space constraints.

Merging Intervals: A Detailed Walkthrough

Let's get our hands dirty by stepping through the actual code to merge the given list of intervals: [[1, 4], [2, 5], [7, 10], [12, 16]].

Preparing the Ground: Initial Sort and Setup

Before jumping into merging, we sort our intervals by their start times.

1intervals = [[1, 4], [2, 5], [7, 10], [12, 16]]
2intervals.sort(key=lambda x: x[0])

Also, initialize an empty array called result to store the merged intervals.

1result = []

Step 1: Seed the result Array

We start by taking the first interval [1, 4] from our sorted array and appending it to result.

1result.append(intervals[0])

At this point, result = [[1, 4]].

Step 2: Enter the Loop: First Iteration

Now, we loop through the remaining intervals in the sorted array, starting with [2, 5].

1current = [2, 5]
2last = result[-1]  # [1, 4]

Here, last is [1, 4], and current is [2, 5]. We notice that current[0] <= last[1], which indicates an overlap. So, we extend last to cover current.

1if current[0] <= last[1]:
2    last[1] = max(current[1], last[1])  # max(5, 4) = 5

After this step, result = [[1, 5]].

Step 3: Second Iteration

Next, we consider the interval [7, 10].

1current = [7, 10]
2last = result[-1]  # [1, 5]

Here, current[0] > last[1], meaning there's a gap. So, we simply append current to result.

1result.append(current)

Now, result = [[1, 5], [7, 10]].

Step 4: Final Iterations

We repeat the above steps for [12, 16]. Since it doesn't overlap with the last interval [7, 10], we add it to the result as is.

1result.append([12, 16])

Our final result array becomes [[1, 5], [7, 10], [12, 16]].

And there you have it! We've successfully merged the intervals using Python code. This step-by-step walkthrough should give you a detailed understanding of how interval merging works in practice.

One Pager Cheat Sheet

• We can write a method that merges any overlapping meetings in a given array of time intervals with a time complexity of O(n) and a space complexity of O(n).
• Merging time ranges is relatively straightforward, as long as the intervals are properly sorted; we just need to take the smallest beginning/start time, and follow the chronological order to find the largest end/finishing time.
• We iterate through the sorted array of intervals, comparing the last interval and referencing a result list to store modified intervals from clear separations or overlaps.

This is our final solution.

To visualize the solution and step through the below code, click Visualize the Solution on the right-side menu or the VISUALIZE button in Interactive Mode.

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}

var assert = require('assert');

​

function mergeIntervals(ranges) {

  ranges.sort(function (a, b) {

    return a[0] - b[0] || a[1] - b[1];

  });

​

  let result = [],

    last;

​

  ranges.forEach(function (r) {

    if (!last || r[0] > last[1]) {

      last = r;

      result.push(last);

    } else if (r[1] > last[1]) last[1] = r[1];

  });

​

  return result;

}

​

// console.log(mergeIntervals([[1, 3], [2, 9], [3, 10], [1, 16]]))

​

try {

  assertIsFunction(mergeIntervals, '`mergeIntervals` is a function');

​

  console.log('PASSED: ' + '`mergeIntervals` is a function');

} catch (err) {

  console.log(err);

}

​

try {

  assert.deepEqual(

Great job getting through this. Let's move on.

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