Insert Interval: Merging Intervals in Arrays

The "Insert Interval" problem is a common algorithmic challenge, involving the insertion and merging of intervals in a sorted list.

Problem Statement

Given a list of non-overlapping intervals sorted by their start times, insert a new interval into the list so that the list remains sorted and any overlapping intervals are merged.

Example

• Input: Intervals [[1,3],[6,9]], New Interval [2,5] Output: [[1,5],[6,9]] Explanation: The new interval [2,5] overlaps with [1,3] and should be merged into [1,5].

Solution Approach

Breaking Down the Solution

• Initial Non-Overlapping Intervals: Add all intervals that end before the new interval starts to the result.
• Merging Overlapping Intervals: Iterate through all intervals that overlap with the new interval and merge them into a single interval.
• Add Remaining Intervals: Finally, add the remaining intervals that start after the new interval ends.

Conclusion

The Insert Interval problem is an excellent example of interval manipulation and requires a careful approach to handle overlapping and merging. It is a common scenario in calendar and scheduling applications, making it a practical problem in software development.

Binary Search Solution

Breaking Down the Solution

• Binary Search for Insertion Point: Use binary search to find the correct position or overlapping interval for the new interval.
• Insert and Merge: Insert the new interval at the found position. Then, iterate through the list to merge any overlapping intervals resulting from the insertion.
• Optimized Overlapping Check: Post-insertion, check only nearby intervals for any potential overlap, reducing the number of comparisons.

Conclusion

Using binary search in the Insert Interval problem significantly enhances the efficiency of finding the correct position for insertion, especially in cases with a large number of intervals. This approach exemplifies the power of combining binary search with interval merging for optimized solutions in array manipulation tasks.

Divide and Conquer solution

In this solution:

1. The list of intervals is recursively split into two halves.
2. The mergeIntervals helper function is used to merge intervals if they overlap.
3. The mergeSortedIntervals function merges the two halves back together, ensuring that the result remains sorted and that all intervals are properly merged.

This approach is more complex than iterative solutions but showcases an alternative way of thinking about the problem.

Rust Solution

Rust's [i32; 2] fixed-size array encodes that every interval is exactly two ints at the type level, where the Go and JS versions lean on dynamically-sized arrays. Because [i32; 2] is a Copy type, result.push(intervals[i]) copies the pair outright with no .clone() or borrow needed. The parameter is declared mut new_interval so it can be widened in place via new_interval[0].min(intervals[i][0]) and new_interval[1].max(intervals[i][1]), and the bare final result expression returns the Vec<[i32; 2]> without a return keyword.

Go Solution

Go models each interval as a []int slice inside [][]int, so nothing pins the length to two the way a fixed-size array would. Lacking a method form of min/max, the merge widens newInterval through explicit if intervals[i][0] < newInterval[0] and if intervals[i][1] > newInterval[1] comparisons rather than .min()/.max() calls. The result grows via append(result, intervals[i]), and the output loop uses for _, iv := range result with a blank _ index since only the value is needed.