Insert Interval
Given a sorted list of non-overlapping intervals and a new interval, insert and merge if necessary. Return the result.
Input: first line = intervals as start:end comma-separated (or 'none'), second = new interval start:end.
Example:
1:3,6:9 2:5
1 5 6 9
- The given intervals are [1,3] and [6,9], and the new interval is [2,5].
- We compare the new interval with the existing ones and find that it overlaps with [1,3], so we merge them to get [1,5].
- The merged interval [1,5] does not overlap with [6,9], so we keep [6,9] as is.
- The resulting merged intervals are [1,5] and [6,9], which are then formatted as the output: 1 5 and 6 9.
Constraints:
- 0 <= len(intervals) <= 10^4
- 0 <= start <= end <= 10^5
Background Knowledge
The problem of inserting and merging intervals is a classic example of a scheduling problem, which involves managing and optimizing the allocation of resources over time. In this case, the resources are intervals of time, and the goal is to insert a new interval into a list of existing, non-overlapping intervals while minimizing the number of intervals. This problem requires an understanding of interval arithmetic, which involves performing operations on intervals, such as merging and intersecting. The key concept here is to recognize that two intervals can be merged if they overlap, i.e., if the end of one interval is greater than or equal to the start of the other.
To approach this problem, it's essential to understand the properties of sorted lists and how to iterate through them efficiently. The input list of intervals is sorted, which means that the start time of each interval is less than or equal to the start time of the next interval. This property can be exploited to simplify the insertion and merging process. Additionally, the problem requires an understanding of conditional statements and looping constructs, which will be used to iterate through the list of intervals and perform the necessary operations.
The problem also involves comparing intervals, which can be done by comparing their start and end times. This comparison can be used to determine whether two intervals overlap or not. The merge operation involves combining two overlapping intervals into a single interval, which can be done by updating the start and end times of the resulting interval. These concepts will be crucial in developing an efficient solution to the problem.
Algorithm/Approach
The general approach to solving this problem involves using a linear scan algorithm, which iterates through the list of intervals and performs the necessary operations to insert and merge the new interval. The algorithm will use a combination of conditional statements and looping constructs to compare intervals and perform the merge operation. The key idea is to find the correct position to insert the new interval and then merge any overlapping intervals.
Step-by-Step Strategy
To implement the solution, follow these steps:
- Parse the input list of intervals and the new interval from the input string
- Initialize an empty list to store the result
- Iterate through the input list of intervals and compare each interval with the new interval
- If the current interval ends before the new interval starts, add it to the result list
- If the current interval overlaps with the new interval, merge them and update the new interval
- If the current interval starts after the new interval ends, add the new interval to the result list and reset the new interval to the current interval
- Add any remaining intervals to the result list
Common Pitfalls
When implementing the solution, watch out for the following pitfalls:
- Failing to handle edge cases, such as an empty input list or a new interval that overlaps with multiple intervals
- Incorrectly comparing intervals, which can lead to incorrect merging or insertion
- Failing to update the new interval correctly after merging with an overlapping interval
Time & Space Complexity
The expected time complexity of the solution is O(n), where n is the number of intervals in the input list, since we are iterating through the list once. The expected space complexity is also O(n), since we are storing the result in a new list. Note that the space complexity can be reduced to O(1) if we modify the input list in-place, but this may not be possible depending on the specific requirements of the problem.