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The Question
Coding

Merge Overlapping Intervals

Given an array of intervals where intervals[i] = [start_i, end_i], merge all overlapping intervals and return an array of the non-overlapping intervals that cover all the intervals in the input. Two intervals [a, b] and [c, d] overlap if there is any point shared between them (e.g., [1, 2] and [2, 3] overlap). Constraints: - 1 <= intervals.length <= 10^4 - intervals[i].length == 2 - 0 <= start_i <= end_i <= 10^4
Java
Sorting
Greedy
Questions & Insights

Clarifying Questions

Are the input intervals sorted by their start times? (Assuming no, as it's a common requirement to sort them first).
How should touching intervals be handled? For example, should [1, 2] and [2, 3] be merged into [1, 3]? (Assuming yes, intervals are inclusive).
What are the constraints on the number of intervals and the range of the coordinates? (Assuming N up to 10^5 and coordinates within standard integer bounds).
Is it possible to receive an empty array or an array with null elements? (Assuming a non-null array of valid pairs).
Assumptions:
The input is an int[][] where each inner array has exactly two elements.
The output should be a new int[][] containing the merged intervals.
The intervals are not necessarily sorted.
Touching intervals are considered overlapping.

Thinking Process

Sort by Start Time: To merge intervals efficiently, we must process them in chronological order based on their start points. This ensures that when we consider a new interval, we only need to compare it against the "current" merged interval we are building.
Greedy Merging: Iterate through the sorted intervals. If the start of the current interval is less than or equal to the end of the last merged interval, they overlap. Extend the end of the last merged interval to be the maximum of its current end and the new interval's end.
Result Construction: Since we don't know the final number of intervals beforehand, use a dynamic list (like ArrayList) to collect the results and convert it back to a 2D array at the end.
Edge Cases: Handle the case where the input array is empty by returning an empty array immediately.
Implementation Breakdown

Problem Set

Functional Requirements: Merge all overlapping intervals and return a new array of disjoint intervals.
Constraints:
1 \le intervals.length \le 10^4
intervals[i].length == 2
0 \le start_i \le end_i \le 10^4
Time Complexity: O(N \log N) due to sorting.
Space Complexity: O(\log N) or O(N) depending on the sorting implementation and the space required for the output.

Approach

Algorithm: Sorting + Greedy Linear Scan
Data Structure: ArrayList<int[]> for dynamic result collection, Arrays.sort.
Complexity: Time O(N \log N) | Space O(N) (to store the result).

Implementation

Wrap Up

Advanced Topics

Readability vs Performance: Using Arrays.sort with a lambda is concise. In performance-critical systems, a custom Comparator or a primitive-optimized sort could reduce overhead, though O(N \log N) remains the bottleneck.
Streaming/Iterative Version: If intervals were coming from a stream, we could maintain a sorted set or an interval tree to merge in O(\log N) per insertion.
In-place Merging: In Java, modifying the input array size is impossible, so a new list is standard. However, we could move merged intervals to the front of the input array to save space and return a slice, achieving O(1) extra space (excluding sort overhead).
Parallelization: Sorting can be parallelized using Arrays.parallelSort() if the dataset is extremely large (N > 10^6), which can significantly improve performance on multi-core machines.