Hey there, fellow coders! Today, we're going to tackle a classic problem that often pops up in coding interviews and real-world applications: the Merge Intervals problem. Don't worry if it sounds intimidating – by the end of this blog, you'll be merging intervals like a pro!
Imagine you have a list of time intervals, each represented by a start and end time. Your task is to merge all overlapping intervals and return a new list of non-overlapping intervals. Sounds simple, right? Well, it can get tricky when you have to handle various edge cases and optimize for performance.
Here's a quick example to get you started:
Input: [[1,3], [2,6], [8,10], [15,18]] Output: [[1,6], [8,10], [15,18]]
In this case, [1,3] and [2,6] overlap, so we merge them into [1,6]. The other intervals don't overlap, so they remain unchanged.
Before we dive into solving the problem, let's talk about why this matters. Merge Intervals isn't just a theoretical exercise – it has plenty of practical applications:
Now that we understand the problem and its importance, let's break down how to solve it. We'll start with a straightforward approach and then optimize it step by step.
The first thing we need to do is sort the intervals based on their start times. This step is crucial because it allows us to compare adjacent intervals easily.
def merge_intervals(intervals): # Sort intervals based on start time intervals.sort(key=lambda x: x[0]) # Rest of the code...
Next, we'll iterate through the sorted intervals and merge overlapping ones. We'll use a result list to store our merged intervals.
def merge_intervals(intervals): intervals.sort(key=lambda x: x[0]) merged = [] for interval in intervals: # If the result list is empty or if the current interval # doesn't overlap with the previous, append it if not merged or merged[-1][1] < interval[0]: merged.append(interval) else: # Otherwise, there is overlap, so we merge the current and previous intervals merged[-1][1] = max(merged[-1][1], interval[1]) return merged
Let's break down what's happening here:
merged
list.merged
list.merged
.merged
to be the maximum of its current end time and the end time of the current interval.The solution we've come up with works well, but let's talk about its time and space complexity:
Can we do better? In terms of time complexity, not really – we need to sort the intervals, which gives us a lower bound of O(n log n). However, we can optimize our space usage in some cases.
If we're allowed to modify the input list, we can perform the merging in-place:
def merge_intervals_in_place(intervals): intervals.sort(key=lambda x: x[0]) i = 0 for j in range(1, len(intervals)): if intervals[i][1] >= intervals[j][0]: intervals[i][1] = max(intervals[i][1], intervals[j][1]) else: i += 1 intervals[i] = intervals[j] return intervals[:i+1]
This approach modifies the input list directly and uses two pointers to keep track of the merged intervals. It can save some space, especially when there are many overlapping intervals.
When implementing this solution, don't forget to handle these edge cases:
Make sure to test your code with these scenarios to ensure robustness.
Once you've mastered the basic Merge Intervals problem, you can explore some variations:
The Merge Intervals problem is a fantastic exercise in algorithmic thinking and optimization. It teaches you to:
Remember, the key to mastering this problem (and coding challenges in general) is practice. Try implementing the solution in different programming languages, experiment with various inputs, and time your code to see how it performs.
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