Merge Intervals
Given an array of intervals [start, end], merge all overlapping intervals and return the result.
Input: intervals as start:end comma-separated. Output: merged intervals, one per line.
Example:
1:3,2:6,8:10,15:18
1 6 8 10 15 18
- The input intervals are first split into individual intervals: [1:3], [2:6], [8:10], [15:18]
- Overlapping intervals are merged: [1:3] and [2:6] overlap, resulting in [1:6], while [8:10] and [15:18] do not overlap with any other intervals
- The merged intervals are then checked for any further overlaps, but since [1:6], [8:10], and [15:18] do not overlap, they remain as separate intervals
- The final merged intervals are output in the required format: 1 6 8 10 15 18
Constraints:
- 1 <= len(intervals) <= 10^4
- 0 <= start <= end <= 10^4
Background Knowledge
The Merge Intervals problem is a classic example of an interval scheduling problem, which involves arranging and optimizing a set of intervals to achieve a specific goal. In this case, the goal is to merge all overlapping intervals. To understand this problem, it's essential to have a solid grasp of algorithmic thinking, data structures, and sorting. The problem requires analyzing the given intervals, identifying overlaps, and combining them into a new set of non-overlapping intervals.
The key concept here is the idea of interval overlap, where two intervals are considered overlapping if they share a common point. For example, the intervals [1, 3] and [2, 4] overlap because they both contain the point 2. To determine if two intervals overlap, we can use a simple comparison: if the start value of one interval is less than or equal to the end value of another interval, and the start value of the second interval is less than or equal to the end value of the first interval, then the intervals overlap.
To solve this problem, we'll need to use a combination of sorting and iteration. The idea is to sort the intervals based on their start values and then iterate through the sorted list, merging any overlapping intervals we find. This approach requires a good understanding of array manipulation and conditional statements.
Algorithm/Approach
The general approach to solving this problem involves using a greedy algorithm, which makes the locally optimal choice at each step with the hope of finding a global optimum solution. In this case, the locally optimal choice is to merge any overlapping intervals we find. The algorithm pattern we'll use is a combination of sorting and scanning, where we sort the intervals and then scan through the sorted list, merging any overlapping intervals.
Step-by-Step Strategy
To implement the solution, we can follow these steps:
- Sort the intervals based on their start values
- Initialize an empty list to store the merged intervals
- Iterate through the sorted list of intervals
- For each interval, check if it overlaps with the last interval in the merged list
- If it overlaps, merge the two intervals by updating the end value of the last interval in the merged list
- If it doesn't overlap, add the current interval to the merged list
- Return the merged list of intervals
Common Pitfalls
Some common pitfalls to watch out for when implementing this solution include:
- Failing to handle edge cases, such as an empty input list or a list with a single interval
- Incorrectly implementing the overlap check, which can lead to incorrect merging or failure to merge overlapping intervals
- Not properly updating the end value of the last interval in the merged list when merging overlapping intervals
Time & Space Complexity
The expected time complexity for this solution is O(nlogn), where n is the number of intervals, due to the sorting step. The space complexity is O(n), as we need to store the merged list of intervals. Note that the time complexity can be improved to O(n) if we use a more efficient sorting algorithm or a different approach, such as using a heap or priority queue. However, the basic greedy algorithm approach with sorting is a simple and effective solution to this problem.